2. Example A.
a. We order pizzas from Pizza Grande. Each pizza is $3.
There is $10 delivery charge. How much would it cost
if we want x pizzas delivered?
Recall example A from the section on expressions.(–Link this)
Linear Equations I
3. Example A.
a. We order pizzas from Pizza Grande. Each pizza is $3.
There is $10 delivery charge. How much would it cost
if we want x pizzas delivered?
For x pizzas it would cost 3 * x = $3x.
Recall example A from the section on expressions.(–Link this)
Linear Equations I
4. Example A.
a. We order pizzas from Pizza Grande. Each pizza is $3.
There is $10 delivery charge. How much would it cost
if we want x pizzas delivered?
For x pizzas it would cost 3 * x = $3x.
To have them delivered, it would cost 3x + 10 ($) in total.
Recall example A from the section on expressions.(–Link this)
Linear Equations I
5. Example A.
a. We order pizzas from Pizza Grande. Each pizza is $3.
There is $10 delivery charge. How much would it cost
if we want x pizzas delivered?
For x pizzas it would cost 3 * x = $3x.
To have them delivered, it would cost 3x + 10 ($) in total.
Recall example A from the section on expressions.(–Link this)
Linear Equations I
b. Suppose the total is $34, how many pizzas did we order?
6. Example A.
a. We order pizzas from Pizza Grande. Each pizza is $3.
There is $10 delivery charge. How much would it cost
if we want x pizzas delivered?
For x pizzas it would cost 3 * x = $3x.
To have them delivered, it would cost 3x + 10 ($) in total.
Recall example A from the section on expressions.(–Link this)
Linear Equations I
b. Suppose the total is $34, how many pizzas did we order?
We backtrack the calculation by subtracting the $10 for delivery
so the cost for the pizzas is $24,
7. Example A.
a. We order pizzas from Pizza Grande. Each pizza is $3.
There is $10 delivery charge. How much would it cost
if we want x pizzas delivered?
For x pizzas it would cost 3 * x = $3x.
To have them delivered, it would cost 3x + 10 ($) in total.
Recall example A from the section on expressions.(–Link this)
Linear Equations I
b. Suppose the total is $34, how many pizzas did we order?
We backtrack the calculation by subtracting the $10 for delivery
so the cost for the pizzas is $24, each pizza is $3 so we must
have ordered 8 pizzas.
8. Example A.
a. We order pizzas from Pizza Grande. Each pizza is $3.
There is $10 delivery charge. How much would it cost
if we want x pizzas delivered?
For x pizzas it would cost 3 * x = $3x.
To have them delivered, it would cost 3x + 10 ($) in total.
Recall example A from the section on expressions.(–Link this)
Linear Equations I
b. Suppose the total is $34, how many pizzas did we order?
We backtrack the calculation by subtracting the $10 for delivery
so the cost for the pizzas is $24, each pizza is $3 so we must
have ordered 8 pizzas.
In symbols, we've the equation 3x + 10 = 34,
9. Example A.
a. We order pizzas from Pizza Grande. Each pizza is $3.
There is $10 delivery charge. How much would it cost
if we want x pizzas delivered?
For x pizzas it would cost 3 * x = $3x.
To have them delivered, it would cost 3x + 10 ($) in total.
Recall example A from the section on expressions.(–Link this)
Linear Equations I
b. Suppose the total is $34, how many pizzas did we order?
We backtrack the calculation by subtracting the $10 for delivery
so the cost for the pizzas is $24, each pizza is $3 so we must
have ordered 8 pizzas.
In symbols, we've the equation 3x + 10 = 34,
backtrack-calculation: 3x + 10 = 34 subtract 10
–10 –10
10. Example A.
a. We order pizzas from Pizza Grande. Each pizza is $3.
There is $10 delivery charge. How much would it cost
if we want x pizzas delivered?
For x pizzas it would cost 3 * x = $3x.
To have them delivered, it would cost 3x + 10 ($) in total.
Recall example A from the section on expressions.(–Link this)
Linear Equations I
b. Suppose the total is $34, how many pizzas did we order?
We backtrack the calculation by subtracting the $10 for delivery
so the cost for the pizzas is $24, each pizza is $3 so we must
have ordered 8 pizzas.
In symbols, we've the equation 3x + 10 = 34,
backtrack-calculation: 3x + 10 = 34 subtract 10
–10 –10
so 3x = 24
11. Example A.
a. We order pizzas from Pizza Grande. Each pizza is $3.
There is $10 delivery charge. How much would it cost
if we want x pizzas delivered?
For x pizzas it would cost 3 * x = $3x.
To have them delivered, it would cost 3x + 10 ($) in total.
Recall example A from the section on expressions.(–Link this)
Linear Equations I
b. Suppose the total is $34, how many pizzas did we order?
We backtrack the calculation by subtracting the $10 for delivery
so the cost for the pizzas is $24, each pizza is $3 so we must
have ordered 8 pizzas.
In symbols, we've the equation 3x + 10 = 34,
backtrack-calculation: 3x + 10 = 34 subtract 10
–10 –10
so 3x = 24 divide by 3
so x = 8 (pizzas)
12. In the above examples, the symbolic method to find solution
may seem unnecessarily cumbersome but for complicated
problems, the symbolic versions are indispensable.
Linear Equations I
13. In the above examples, the symbolic method to find solution
may seem unnecessarily cumbersome but for complicated
problems, the symbolic versions are indispensable.
An equation is two expressions set equal to each other.
Equations look like:
left expression = right expression
or
LHS = RHS
Linear Equations I
14. In the above examples, the symbolic method to find solution
may seem unnecessarily cumbersome but for complicated
problems, the symbolic versions are indispensable.
An equation is two expressions set equal to each other.
Equations look like:
left expression = right expression
or
LHS = RHS
Linear Equations I
We want to solve equations, i.e. we want to find the value
(or values) for the variable x such that it makes both sides
equal.
15. In the above examples, the symbolic method to find solution
may seem unnecessarily cumbersome but for complicated
problems, the symbolic versions are indispensable.
An equation is two expressions set equal to each other.
Equations look like:
left expression = right expression
or
LHS = RHS
Linear Equations I
We want to solve equations, i.e. we want to find the value
(or values) for the variable x such that it makes both sides
equal. Such a value is called a solution of the equation.
16. In the above examples, the symbolic method to find solution
may seem unnecessarily cumbersome but for complicated
problems, the symbolic versions are indispensable.
An equation is two expressions set equal to each other.
Equations look like:
left expression = right expression
or
LHS = RHS
In the example above 3x + 10 = 34 is an equations and
x = 8 is the solution for this equations because 3(8) + 10 is 34.
Linear Equations I
We want to solve equations, i.e. we want to find the value
(or values) for the variable x such that it makes both sides
equal. Such a value is called a solution of the equation.
17. In the above examples, the symbolic method to find solution
may seem unnecessarily cumbersome but for complicated
problems, the symbolic versions are indispensable.
An equation is two expressions set equal to each other.
Equations look like:
left expression = right expression
or
LHS = RHS
Linear Equations I
We want to solve equations, i.e. we want to find the value
(or values) for the variable x such that it makes both sides
equal. Such a value is called a solution of the equation.
Where as we use an expression to calculate future outcomes,
we use an equation to help us to backtrack from known
outcomes to the original input x, the solution for the equation.
In the example above 3x + 10 = 34 is an equations and
x = 8 is the solution for this equations because 3(8) + 10 is 34.
18. Linear Equations I
A linear equation is an equation where both the
expressions on both sides are linear expressions such as
3x + 10 = 34, or
8 = 4x – 6.
19. Linear Equations I
A linear equation does not contain any higher powers of x
such as x2, x3; x2 – 3x = 2x – 3 is not a linear equation because
of the x2.
A linear equation is an equation where both the
expressions on both sides are linear expressions such as
3x + 10 = 34, or
8 = 4x – 6.
20. Linear equations are the easy to solve, i.e. it’s easy to
manipulate a linear equation, to backtrack the calculations,
to reveal what x is.
Linear Equations I
A linear equation does not contain any higher powers of x
such as x2, x3; x2 – 3x = 2x – 3 is not a linear equation because
of the x2.
A linear equation is an equation where both the
expressions on both sides are linear expressions such as
3x + 10 = 34, or
8 = 4x – 6.
21. Linear equations are the easy to solve, i.e. it’s easy to
manipulate a linear equation, to backtrack the calculations,
to reveal what x is. The easiest linear equations to solve are
the single–step equations such as the following ones,
Linear Equations I
A linear equation does not contain any higher powers of x
such as x2, x3; x2 – 3x = 2x – 3 is not a linear equation because
of the x2.
A linear equation is an equation where both the
expressions on both sides are linear expressions such as
3x + 10 = 34, or
8 = 4x – 6.
22. Linear equations are the easy to solve, i.e. it’s easy to
manipulate a linear equation, to backtrack the calculations,
to reveal what x is. The easiest linear equations to solve are
the single–step equations such as the following ones,
x – 3 = 12,
Linear Equations I
A linear equation does not contain any higher powers of x
such as x2, x3; x2 – 3x = 2x – 3 is not a linear equation because
of the x2.
A linear equation is an equation where both the
expressions on both sides are linear expressions such as
3x + 10 = 34, or
8 = 4x – 6.
23. Linear equations are the easy to solve, i.e. it’s easy to
manipulate a linear equation, to backtrack the calculations,
to reveal what x is. The easiest linear equations to solve are
the single–step equations such as the following ones,
x – 3 = 12,
12 = x + 3,
Linear Equations I
A linear equation does not contain any higher powers of x
such as x2, x3; x2 – 3x = 2x – 3 is not a linear equation because
of the x2.
A linear equation is an equation where both the
expressions on both sides are linear expressions such as
3x + 10 = 34, or
8 = 4x – 6.
24. Linear equations are the easy to solve, i.e. it’s easy to
manipulate a linear equation, to backtrack the calculations,
to reveal what x is. The easiest linear equations to solve are
the single–step equations such as the following ones,
x – 3 = 12,
12 = x + 3,
3*x = 12,
12 =
all four equation are one-step equations.
x
3
Linear Equations I
A linear equation does not contain any higher powers of x
such as x2, x3; x2 – 3x = 2x – 3 is not a linear equation because
of the x2.
A linear equation is an equation where both the
expressions on both sides are linear expressions such as
3x + 10 = 34, or
8 = 4x – 6.
25. Linear equations are the easy to solve, i.e. it’s easy to
manipulate a linear equation, to backtrack the calculations,
to reveal what x is. The easiest linear equations to solve are
the single–step equations such as the following ones,
x – 3 = 12,
12 = x + 3,
3*x = 12,
12 =
all four equation are one-step equations.
x
3
Linear Equations I
A linear equation does not contain any higher powers of x
such as x2, x3; x2 – 3x = 2x – 3 is not a linear equation because
of the x2.
A linear equation is an equation where both the
expressions on both sides are linear expressions such as
3x + 10 = 34, or
8 = 4x – 6.
12 = x – 3,
x + 3 = 12,
12 = 3*x,
x/3 = 12
These equations are the same,
i.e. it doesn’t matter it’s
A = B or B = A. Both versions
will lead to the answer for x.
27. Basic principle for solving one- step-equations:
To solve one-step-equations, isolate the x on one side by
applying the opposite operation to both sides of the equation.
Linear Equations I
28. Basic principle for solving one- step-equations:
To solve one-step-equations, isolate the x on one side by
applying the opposite operation to both sides of the equation.
Example B. Solve for x
a. x – 3 = 12
b. x + 3 = –12
c. 3x = 15
Linear Equations I
29. Basic principle for solving one- step-equations:
To solve one-step-equations, isolate the x on one side by
applying the opposite operation to both sides of the equation.
Example B. Solve for x
a. x – 3 = 12
b. x + 3 = –12
c. 3x = 15
Linear Equations I
This says
“x take away 3 gives 12”,
hence add 3 to get back to x.
30. Basic principle for solving one- step-equations:
To solve one-step-equations, isolate the x on one side by
applying the opposite operation to both sides of the equation.
Example B. Solve for x
a. x – 3 = 12 Add 3 to both sides
+ 3 + 3
b. x + 3 = –12
c. 3x = 15
Linear Equations I
This says
“x take away 3 gives 12”,
hence add 3 to get back to x.
31. Basic principle for solving one- step-equations:
To solve one-step-equations, isolate the x on one side by
applying the opposite operation to both sides of the equation.
Example B. Solve for x
a. x – 3 = 12 Add 3 to both sides
+ 3 + 3
x = 15
b. x + 3 = –12
c. 3x = 15
Linear Equations I
This says
“x take away 3 gives 12”,
hence add 3 to get back to x.
32. Basic principle for solving one- step-equations:
To solve one-step-equations, isolate the x on one side by
applying the opposite operation to both sides of the equation.
Example B. Solve for x
a. x – 3 = 12 Add 3 to both sides
+ 3 + 3
x = 15 check: 15 – 3 = 12
b. x + 3 = –12
c. 3x = 15
Linear Equations I
?
This says
“x take away 3 gives 12”,
hence add 3 to get back to x.
33. Basic principle for solving one- step-equations:
To solve one-step-equations, isolate the x on one side by
applying the opposite operation to both sides of the equation.
Example B. Solve for x
a. x – 3 = 12 Add 3 to both sides
+ 3 + 3
x = 15 check: 15 – 3 = 12
b. x + 3 = –12
c. 3x = 15
Linear Equations I
12 = 12 (yes)
?
This says
“x take away 3 gives 12”,
hence add 3 to get back to x.
34. Basic principle for solving one- step-equations:
To solve one-step-equations, isolate the x on one side by
applying the opposite operation to both sides of the equation.
Example B. Solve for x
a. x – 3 = 12 Add 3 to both sides
+ 3 + 3
x = 15 check: 15 – 3 = 12
b. x + 3 = –12
c. 3x = 15
Linear Equations I
12 = 12 (yes)
?
This says
“x take away 3 gives 12”,
hence add 3 to get back to x.
This says
“3 added to x gives –12”,
hence subtract 3 to get
back to x.
35. Basic principle for solving one- step-equations:
To solve one-step-equations, isolate the x on one side by
applying the opposite operation to both sides of the equation.
Example B. Solve for x
a. x – 3 = 12 Add 3 to both sides
+ 3 + 3
x = 15 check: 15 – 3 = 12
b. x + 3 = –12 Subtract 3 from both sides
–3 –3
c. 3x = 15
Linear Equations I
12 = 12 (yes)
?
This says
“x take away 3 gives 12”,
hence add 3 to get back to x.
This says
“3 added to x gives –12”,
hence subtract 3 to get
back to x.
36. Basic principle for solving one- step-equations:
To solve one-step-equations, isolate the x on one side by
applying the opposite operation to both sides of the equation.
Example B. Solve for x
a. x – 3 = 12 Add 3 to both sides
+ 3 + 3
x = 15 check: 15 – 3 = 12
b. x + 3 = –12 Subtract 3 from both sides
–3 –3
x = –15
c. 3x = 15
Linear Equations I
12 = 12 (yes)
?
This says
“x take away 3 gives 12”,
hence add 3 to get back to x.
This says
“3 added to x gives –12”,
hence subtract 3 to get
back to x.
37. Basic principle for solving one- step-equations:
To solve one-step-equations, isolate the x on one side by
applying the opposite operation to both sides of the equation.
Example B. Solve for x
a. x – 3 = 12 Add 3 to both sides
+ 3 + 3
x = 15 check: 15 – 3 = 12
b. x + 3 = –12 Subtract 3 from both sides
–3 –3
x = –15 check: –15 + 3 = –12
c. 3x = 15
Linear Equations I
12 = 12 (yes)
?
?
This says
“x take away 3 gives 12”,
hence add 3 to get back to x.
This says
“3 added to x gives –12”,
hence subtract 3 to get
back to x.
38. Basic principle for solving one- step-equations:
To solve one-step-equations, isolate the x on one side by
applying the opposite operation to both sides of the equation.
Example B. Solve for x
a. x – 3 = 12 Add 3 to both sides
+ 3 + 3
x = 15 check: 15 – 3 = 12
b. x + 3 = –12 Subtract 3 from both sides
–3 –3
x = –15 check: –15 + 3 = –12
c. 3x = 15
Linear Equations I
12 = 12 (yes)
?
–12 = –12 (yes)
?
This says
“x take away 3 gives 12”,
hence add 3 to get back to x.
This says
“3 added to x gives –12”,
hence subtract 3 to get
back to x.
39. Basic principle for solving one- step-equations:
To solve one-step-equations, isolate the x on one side by
applying the opposite operation to both sides of the equation.
Example B. Solve for x
a. x – 3 = 12 Add 3 to both sides
+ 3 + 3
x = 15 check: 15 – 3 = 12
b. x + 3 = –12 Subtract 3 from both sides
–3 –3
x = –15 check: –15 + 3 = –12
c. 3x = 15
Linear Equations I
12 = 12 (yes)
?
–12 = –12 (yes)
?
This says
“x take away 3 gives 12”,
hence add 3 to get back to x.
This says
“3 added to x gives –12”,
hence subtract 3 to get
back to x.
This says
“triple the x gives 15”,
hence divide by 3 to get
back to x.
40. Basic principle for solving one- step-equations:
To solve one-step-equations, isolate the x on one side by
applying the opposite operation to both sides of the equation.
Example B. Solve for x
a. x – 3 = 12 Add 3 to both sides
+ 3 + 3
x = 15 check: 15 – 3 = 12
b. x + 3 = –12 Subtract 3 from both sides
–3 –3
x = –15 check: –15 + 3 = –12
3x
3
15
3
=
c. 3x = 15 Both sides divided by 3
Linear Equations I
12 = 12 (yes)
?
–12 = –12 (yes)
?
This says
“x take away 3 gives 12”,
hence add 3 to get back to x.
This says
“3 added to x gives –12”,
hence subtract 3 to get
back to x.
This says
“triple the x gives 15”,
hence divide by 3 to get
back to x.
41. Basic principle for solving one- step-equations:
To solve one-step-equations, isolate the x on one side by
applying the opposite operation to both sides of the equation.
Example B. Solve for x
a. x – 3 = 12 Add 3 to both sides
+ 3 + 3
x = 15 check: 15 – 3 = 12
b. x + 3 = –12 Subtract 3 from both sides
–3 –3
x = –15 check: –15 + 3 = –12
3x
3
15
3
=
x = 5
c. 3x = 15 Both sides divided by 3
Linear Equations I
12 = 12 (yes)
?
–12 = –12 (yes)
?
This says
“x take away 3 gives 12”,
hence add 3 to get back to x.
This says
“3 added to x gives –12”,
hence subtract 3 to get
back to x.
This says
“triple the x gives 15”,
hence divide by 3 to get
back to x.
42. Basic principle for solving one- step-equations:
To solve one-step-equations, isolate the x on one side by
applying the opposite operation to both sides of the equation.
Example B. Solve for x
a. x – 3 = 12 Add 3 to both sides
+ 3 + 3
x = 15 check: 15 – 3 = 12
b. x + 3 = –12 Subtract 3 from both sides
–3 –3
x = –15 check: –15 + 3 = –12
3x
3
15
3
=
x = 5 check: 3(5) = 15
c. 3x = 15 Both sides divided by 3
Linear Equations I
12 = 12 (yes)
?
–12 = –12 (yes)
?
?
This says
“x take away 3 gives 12”,
hence add 3 to get back to x.
This says
“3 added to x gives –12”,
hence subtract 3 to get
back to x.
This says
“triple the x gives 15”,
hence divide by 3 to get
back to x.
43. Basic principle for solving one- step-equations:
To solve one-step-equations, isolate the x on one side by
applying the opposite operation to both sides of the equation.
Example B. Solve for x
a. x – 3 = 12 Add 3 to both sides
+ 3 + 3
x = 15 check: 15 – 3 = 12
b. x + 3 = –12 Subtract 3 from both sides
–3 –3
x = –15 check: –15 + 3 = –12
3x
3
15
3
=
x = 5 check: 3(5) = 15
c. 3x = 15 Both sides divided by 3
Linear Equations I
12 = 12 (yes)
?
–12 = –12 (yes)
?
15 = 15 (yes)
?
This says
“x take away 3 gives 12”,
hence add 3 to get back to x.
This says
“3 added to x gives –12”,
hence subtract 3 to get
back to x.
This says
“triple the x gives 15”,
hence divide by 3 to get
back to x.
45. x
3
–12=d. Multiply both sides by 3
x
3
–12=( (3))
Linear Equations I
This says
“x divided by 3 gives –12”,
hence multiply by 3 to get
back to x.
46. x
3
–12=d. Multiply both sides by 3
x
3
–12=( (3))
x = –36
Linear Equations I
This says
“x divided by 3 gives –12”,
hence multiply by 3 to get
back to x.
47. x
3
–12=d. Multiply both sides by 3
x
3
–12=( (3))
x = –36 Check:
3
–12=– 36
Linear Equations I
This says
“x divided by 3 gives –12”,
hence multiply by 3 to get
back to x.
48. x
3
–12=d. Multiply both sides by 3
x
3
–12=( (3))
x = –36 Check:
3
–12=– 36
Linear Equations I
This says
“x divided by 3 gives –12”,
hence multiply by 3 to get
back to x.
Fact: Given a linear equation if we +, –, * , /, to both sides by
the same quantity, the new equation will have the same
solution.
49. x
3
–12=d. Multiply both sides by 3
x
3
–12=( (3))
x = –36 Check:
3
–12=– 36
Linear Equations I
This says
“x divided by 3 gives –12”,
hence multiply by 3 to get
back to x.
Next we solve equations that require two steps. These are the
ones that we have to collect the x-terms (or the number–terms)
first with addition or subtraction, then multiply or divide to get x.
Fact: Given a linear equation if we +, –, * , /, to both sides by
the same quantity, the new equation will have the same
solution.
50. x
3
–12=d. Multiply both sides by 3
x
3
–12=( (3))
x = –36 Check:
3
–12=– 36
Linear Equations I
This says
“x divided by 3 gives –12”,
hence multiply by 3 to get
back to x.
Next we solve equations that require two steps. These are the
ones that we have to collect the x-terms (or the number–terms)
first with addition or subtraction, then multiply or divide to get x.
Example C. Solve for x
a. 4x – 6 = 30
Fact: Given a linear equation if we +, –, * , /, to both sides by
the same quantity, the new equation will have the same
solution.
51. x
3
–12=d. Multiply both sides by 3
x
3
–12=( (3))
x = –36 Check:
3
–12=– 36
Linear Equations I
This says
“x divided by 3 gives –12”,
hence multiply by 3 to get
back to x.
Next we solve equations that require two steps. These are the
ones that we have to collect the x-terms (or the number–terms)
first with addition or subtraction, then multiply or divide to get x.
Example C. Solve for x
a. 4x – 6 = 30 Collect the numbers by adding 6 to both sides
Fact: Given a linear equation if we +, –, * , /, to both sides by
the same quantity, the new equation will have the same
solution.
+6+6
52. x
3
–12=d. Multiply both sides by 3
x
3
–12=( (3))
x = –36 Check:
3
–12=– 36
Linear Equations I
This says
“x divided by 3 gives –12”,
hence multiply by 3 to get
back to x.
Next we solve equations that require two steps. These are the
ones that we have to collect the x-terms (or the number–terms)
first with addition or subtraction, then multiply or divide to get x.
Example C. Solve for x
a. 4x – 6 = 30 Collect the numbers by adding 6 to both sides
Fact: Given a linear equation if we +, –, * , /, to both sides by
the same quantity, the new equation will have the same
solution.
+6+6
4x = 36
53. x
3
–12=d. Multiply both sides by 3
x
3
–12=( (3))
x = –36 Check:
3
–12=– 36
Linear Equations I
This says
“x divided by 3 gives –12”,
hence multiply by 3 to get
back to x.
Next we solve equations that require two steps. These are the
ones that we have to collect the x-terms (or the number–terms)
first with addition or subtraction, then multiply or divide to get x.
Example C. Solve for x
a. 4x – 6 = 30 Collect the numbers by adding 6 to both sides
Fact: Given a linear equation if we +, –, * , /, to both sides by
the same quantity, the new equation will have the same
solution.
+6+6
4x = 36
4 4
Divide both sides by 4
54. x
3
–12=d. Multiply both sides by 3
x
3
–12=( (3))
x = –36 Check:
3
–12=– 36
Linear Equations I
This says
“x divided by 3 gives –12”,
hence multiply by 3 to get
back to x.
Next we solve equations that require two steps. These are the
ones that we have to collect the x-terms (or the number–terms)
first with addition or subtraction, then multiply or divide to get x.
Example C. Solve for x
a. 4x – 6 = 30 Collect the numbers by adding 6 to both sides
Fact: Given a linear equation if we +, –, * , /, to both sides by
the same quantity, the new equation will have the same
solution.
+6+6
x = 9
4x = 36
(Check this is the right answer.)
4 4
Divide both sides by 4
56. Linear Equations I
Example C. Solve for x
b. x – 6 = 3x Collect the x's by subtracting x from both sides
–x –x
57. Linear Equations I
Example C. Solve for x
b. x – 6 = 3x Collect the x's by subtracting x from both sides
–x –x
–6 2x=
58. Linear Equations I
Example C. Solve for x
b. x – 6 = 3x Collect the x's by subtracting x from both sides
–x –x
–3 = x
–6 2x=
2 2 Divide by 2
59. Linear Equations I
Example C. Solve for x
b. x – 6 = 3x Collect the x's by subtracting x from both sides
–x –x
–3 = x
–6 2x=
2 2
In real–life, we encounter linear equations often.
Divide by 2
60. Linear Equations I
Example C. Solve for x
b. x – 6 = 3x Collect the x's by subtracting x from both sides
–x –x
–3 = x
–6 2x=
2 2
In real–life, we encounter linear equations often.
Example D. To make a cheese sandwich, we use two slices of
bread each having 70 calories and slices of cheeses where
each slice of cheese is 90 calories
a. How many calories are there in the sandwich with 2 slices of
cheese?
Divide by 2
61. Linear Equations I
Example C. Solve for x
b. x – 6 = 3x Collect the x's by subtracting x from both sides
–x –x
–3 = x
–6 2x=
2 2
In real–life, we encounter linear equations often.
Example D. To make a cheese sandwich, we use two slices of
bread each having 70 calories and slices of cheeses where
each slice of cheese is 90 calories
a. How many calories are there in the sandwich with 2 slices of
cheese?
There are 140 cal in the bread and 2 * 90 = 180 cal to make a
total of 140 + 180 = 320 calories in the cheese.
Divide by 2
62. Linear Equations I
Example C. Solve for x
b. x – 6 = 3x Collect the x's by subtracting x from both sides
–x –x
–3 = x
–6 2x=
2 2
In real–life, we encounter linear equations often.
Example D. To make a cheese sandwich, we use two slices of
bread each having 70 calories and slices of cheeses where
each slice of cheese is 90 calories
a. How many calories are there in the sandwich with 2 slices of
cheese?
There are 140 cal in the bread and 2 * 90 = 180 cal to make a
total of 140 + 180 = 320 calories in the cheese.
b. What is the expression that calculate the number of calories
of a sandwich with x slices of cheese?
Divide by 2
63. Linear Equations I
Example C. Solve for x
b. x – 6 = 3x Collect the x's by subtracting x from both sides
–x –x
–3 = x
–6 2x=
2 2
In real–life, we encounter linear equations often.
Example D. To make a cheese sandwich, we use two slices of
bread each having 70 calories and slices of cheeses where
each slice of cheese is 90 calories
a. How many calories are there in the sandwich with 2 slices of
cheese?
There are 140 cal in the bread and 2 * 90 = 180 cal to make a
total of 140 + 180 = 320 calories in the cheese.
b. What is the expression that calculate the number of calories
of a sandwich with x slices of cheese?
There are 140 + 90x calories in the sandwich.
Divide by 2
65. Linear Equations I
c. How many slices of cheese are there in a 500–cal sandwich?
The total calories 14 + 90x is 500,
66. Linear Equations I
c. How many slices of cheese are there in a 500–cal sandwich?
The total calories 14 + 90x is 500, i.e.
140 + 90x = 500
67. Linear Equations I
c. How many slices of cheese are there in a 500–cal sandwich?
Subtract 140 from
both sides–140 –140
The total calories 14 + 90x is 500, i.e.
140 + 90x = 500
68. Linear Equations I
c. How many slices of cheese are there in a 500–cal sandwich?
Subtract 140 from
both sides–140 –140
90x = 360
The total calories 14 + 90x is 500, i.e.
140 + 90x = 500
69. Linear Equations I
c. How many slices of cheese are there in a 500–cal sandwich?
Subtract 140 from
both sides–140 –140
90x = 360 Divide both sides by 90
90 90
x = 4
The total calories 14 + 90x is 500, i.e.
140 + 90x = 500
70. Linear Equations I
c. How many slices of cheese are there in a 500–cal sandwich?
Subtract 140 from
both sides–140 –140
90x = 360 Divide both sides by 90
90 90
x = 4
So there are 4 slices of cheese in a 500–cal sandwich.
The total calories 14 + 90x is 500, i.e.
140 + 90x = 500
71. Linear Equations I
c. How many slices of cheese are there in a 500–cal sandwich?
Subtract 140 from
both sides–140 –140
90x = 360 Divide both sides by 90
90 90
x = 4
The more general linear equations have the form
#x ± # = #x ± #,
where # can be any number.
So there are 4 slices of cheese in a 500–cal sandwich.
The total calories 14 + 90x is 500, i.e.
140 + 90x = 500
72. Linear Equations I
c. How many slices of cheese are there in a 500–cal sandwich?
Subtract 140 from
both sides–140 –140
90x = 360 Divide both sides by 90
90 90
x = 4
The more general linear equations have the form
#x ± # = #x ± #,
where # can be any number. We solve it by following steps:
So there are 4 slices of cheese in a 500–cal sandwich.
The total calories 14 + 90x is 500, i.e.
140 + 90x = 500
73. Linear Equations I
c. How many slices of cheese are there in a 500–cal sandwich?
Subtract 140 from
both sides–140 –140
90x = 360 Divide both sides by 90
90 90
x = 4
The more general linear equations have the form
#x ± # = #x ± #,
where # can be any number. We solve it by following steps:
1. Add or subtract to move the x-term to one side of the
equation and get: #x ± # = # or # = #x ± #
So there are 4 slices of cheese in a 500–cal sandwich.
The total calories 14 + 90x is 500, i.e.
140 + 90x = 500
74. Linear Equations I
c. How many slices of cheese are there in a 500–cal sandwich?
Subtract 140 from
both sides–140 –140
90x = 360 Divide both sides by 90
90 90
x = 4
The more general linear equations have the form
#x ± # = #x ± #,
where # can be any number. We solve it by following steps:
1. Add or subtract to move the x-term to one side of the
equation and get: #x ± # = # or # = #x ± #
2. Add or subtract the # to separate the number-term from the
x-term to get: #x = # or # = #x.
So there are 4 slices of cheese in a 500–cal sandwich.
The total calories 14 + 90x is 500, i.e.
140 + 90x = 500
75. Linear Equations I
c. How many slices of cheese are there in a 500–cal sandwich?
Subtract 140 from
both sides–140 –140
90x = 360 Divide both sides by 90
90 90
x = 4
The more general linear equations have the form
#x ± # = #x ± #,
where # can be any number. We solve it by following steps:
1. Add or subtract to move the x-term to one side of the
equation and get: #x ± # = # or # = #x ± #
2. Add or subtract the # to separate the number-term from the
x-term to get: #x = # or # = #x.
3. Divide or multiply to get x:
x = solution or solution = x
So there are 4 slices of cheese in a 500–cal sandwich.
The total calories 14 + 90x is 500, i.e.
140 + 90x = 500
77. Example E.
Solve 3x – 4 = 5x + 2
–3x –3x
subtract 3x to remove
the x from one side.
Linear Equations I
78. Example E.
Solve 3x – 4 = 5x + 2
–3x –3x
– 4 = 2x + 2
subtract 3x to remove
the x from one side.
Linear Equations I
79. Example E.
Solve 3x – 4 = 5x + 2
–3x –3x
– 4 = 2x + 2
–2 –2
subtract 3x to remove
the x from one side.
subtract 2 to move the 2 to
the other side.
Linear Equations I
80. Example E.
Solve 3x – 4 = 5x + 2
–3x –3x
– 4 = 2x + 2
–2 –2
– 6 = 2x
subtract 3x to remove
the x from one side.
subtract 2 to move the 2 to
the other side.
Linear Equations I
81. Example E.
Solve 3x – 4 = 5x + 2
–3x –3x
– 4 = 2x + 2
–2 –2
– 6 = 2x
2
–6
2
2x
=
subtract 3x to remove
the x from one side.
subtract 2 to move the 2 to
the other side.
divide by 2 get x.
Linear Equations I
82. Example E.
Solve 3x – 4 = 5x + 2
–3x –3x
– 4 = 2x + 2
–2 –2
– 6 = 2x
2
–6
2
2x
=
–3 = x
subtract 3x to remove
the x from one side.
subtract 2 to move the 2 to
the other side.
divide by 2 get x.
Linear Equations I
83. Example E.
Solve 3x – 4 = 5x + 2
–3x –3x
– 4 = 2x + 2
–2 –2
– 6 = 2x
2
–6
2
2x
=
–3 = x
subtract 3x to remove
the x from one side.
subtract 2 to move the 2 to
the other side.
divide by 2 get x.
Linear Equations I
84. Example E.
Solve 3x – 4 = 5x + 2
–3x –3x
– 4 = 2x + 2
–2 –2
– 6 = 2x
2
–6
2
2x
=
–3 = x
subtract 3x to remove
the x from one side.
subtract 2 to move the 2 to
the other side.
divide by 2 get x.
Finally, all linear equations can be reduced to the format
#x ± # = #x ± #
by simplifying each side by itself first.
Linear Equations I
85. Example E.
Solve 3x – 4 = 5x + 2
–3x –3x
– 4 = 2x + 2
–2 –2
– 6 = 2x
2
–6
2
2x
=
–3 = x
subtract 3x to remove
the x from one side.
subtract 2 to move the 2 to
the other side.
divide by 2 get x.
Finally, all linear equations can be reduced to the format
#x ± # = #x ± #
by simplifying each side by itself first.
Linear Equations I
Example F. Solve the equation.
2x – 3(1 – 3x) = 3(x – 6) – 1
Simplify each side by combining like-term first.
86. 2x – 3(1 – 3x) = 3(x – 6) – 1
Linear Equations I
100. Below we make some observations and give some short cuts
about solving equations.
Linear Equations II
101. Below we make some observations and give some short cuts
about solving equations. Recall that to solve linear equations
we simplify each side of equation first then we separate the
x's from the numbers.
Linear Equations II
102. Below we make some observations and give some short cuts
about solving equations. Recall that to solve linear equations
we simplify each side of equation first then we separate the
x's from the numbers.
Linear Equations II
I. (Change-side-change-sign Rule)
103. Below we make some observations and give some short cuts
about solving equations. Recall that to solve linear equations
we simplify each side of equation first then we separate the
x's from the numbers.
Linear Equations II
I. (Change-side-change-sign Rule)
104. Below we make some observations and give some short cuts
about solving equations. Recall that to solve linear equations
we simplify each side of equation first then we separate the
x's from the numbers.
Linear Equations II
I. (Change-side-change-sign Rule)
When separating the x's from the number–terms to the two
sides of the equal sign, just move the items across the equal
sign to other side and switch their signs (instead of adding or
subtracting to remove terms).
x + a = b x – a = b
105. Below we make some observations and give some short cuts
about solving equations. Recall that to solve linear equations
we simplify each side of equation first then we separate the
x's from the numbers.
x = b – a
Linear Equations II
change-side
change-sign
I. (Change-side-change-sign Rule)
When separating the x's from the number–terms to the two
sides of the equal sign, just move the items across the equal
sign to other side and switch their signs (instead of adding or
subtracting to remove terms).
x + a = b x – a = b
106. Below we make some observations and give some short cuts
about solving equations. Recall that to solve linear equations
we simplify each side of equation first then we separate the
x's from the numbers.
x = b – a x = b + a
Linear Equations II
change-side
change-sign
change-side
change-signor
I. (Change-side-change-sign Rule)
When separating the x's from the number–terms to the two
sides of the equal sign, just move the items across the equal
sign to other side and switch their signs (instead of adding or
subtracting to remove terms).
x + a = b x – a = b
107. Below we make some observations and give some short cuts
about solving equations. Recall that to solve linear equations
we simplify each side of equation first then we separate the
x's from the numbers.
x = b – a x = b + a
Linear Equations II
change-side
change-sign
change-side
change-signor
There are two sides to group the x's, the right hand side or the
left hand side.
I. (Change-side-change-sign Rule)
When separating the x's from the number–terms to the two
sides of the equal sign, just move the items across the equal
sign to other side and switch their signs (instead of adding or
subtracting to remove terms).
x + a = b x – a = b
108. Below we make some observations and give some short cuts
about solving equations. Recall that to solve linear equations
we simplify each side of equation first then we separate the
x's from the numbers.
x = b – a x = b + a
Linear Equations II
change-side
change-sign
change-side
change-signor
There are two sides to group the x's, the right hand side or the
left hand side. It’s better to collect the x's to the side so that it
ends up with positive x-term.
I. (Change-side-change-sign Rule)
When separating the x's from the number–terms to the two
sides of the equal sign, just move the items across the equal
sign to other side and switch their signs (instead of adding or
subtracting to remove terms).
x + a = b x – a = b
111. Linear Equations II
Example G. Solve.
a. –5 = x + 6
–6 –5 = x
–11 = x
move +6 to the left,
it turns into –6
112. b. 3(x + 2) – x = 2(–2 + x) + x
Linear Equations II
Example G. Solve.
a. –5 = x + 6
–6 –5 = x
–11 = x
move +6 to the left,
it turns into –6
113. b. 3(x + 2) – x = 2(–2 + x) + x simplify each side
Linear Equations II
Example G. Solve.
a. –5 = x + 6
–6 –5 = x
–11 = x
move +6 to the left,
it turns into –6
114. b. 3(x + 2) – x = 2(–2 + x) + x
3x + 6 – x = –4 + 2x + x
simplify each side
Linear Equations II
Example G. Solve.
a. –5 = x + 6
–6 –5 = x
–11 = x
move +6 to the left,
it turns into –6
115. b. 3(x + 2) – x = 2(–2 + x) + x
3x + 6 – x = –4 + 2x + x
2x + 6 = –4 + 3x
simplify each side
Linear Equations II
Example G. Solve.
a. –5 = x + 6
–6 –5 = x
–11 = x
move +6 to the left,
it turns into –6
116. b. 3(x + 2) – x = 2(–2 + x) + x
3x + 6 – x = –4 + 2x + x
2x + 6 = –4 + 3x
4 + 6 = 3x – 2x
simplify each side
move 2x and –4 to the other
sides, change their signs
Linear Equations II
Example G. Solve.
a. –5 = x + 6
–6 –5 = x
–11 = x
move +6 to the left,
it turns into –6
117. b. 3(x + 2) – x = 2(–2 + x) + x
3x + 6 – x = –4 + 2x + x
2x + 6 = –4 + 3x
4 + 6 = 3x – 2x
10 = x
simplify each side
move 2x and –4 to the other
sides, change their signs
Linear Equations II
Example G. Solve.
a. –5 = x + 6
–6 –5 = x
–11 = x
move +6 to the left,
it turns into –6
118. b. 3(x + 2) – x = 2(–2 + x) + x
3x + 6 – x = –4 + 2x + x
2x + 6 = –4 + 3x
4 + 6 = 3x – 2x
10 = x
simplify each side
move 2x and –4 to the other
sides, change their signs
Linear Equations II
c. + x =
4
3
2
1
Example G. Solve.
a. –5 = x + 6
–6 –5 = x
–11 = x
move +6 to the left,
it turns into –6
119. b. 3(x + 2) – x = 2(–2 + x) + x
3x + 6 – x = –4 + 2x + x
2x + 6 = –4 + 3x
4 + 6 = 3x – 2x
10 = x
simplify each side
move 2x and –4 to the other
sides, change their signs
Linear Equations II
c. + x =
4
3 move to the other side,
change its sign
3
42
1
Example G. Solve.
a. –5 = x + 6
–6 –5 = x
–11 = x
move +6 to the left,
it turns into –6
120. b. 3(x + 2) – x = 2(–2 + x) + x
3x + 6 – x = –4 + 2x + x
2x + 6 = –4 + 3x
4 + 6 = 3x – 2x
10 = x
simplify each side
move 2x and –4 to the other
sides, change their signs
Linear Equations II
c. + x =
4
3 move to the other side,
change its sign
3
42
1
x = 2
1
4
3–
Example G. Solve.
a. –5 = x + 6
–6 –5 = x
–11 = x
move +6 to the left,
it turns into –6
121. b. 3(x + 2) – x = 2(–2 + x) + x
3x + 6 – x = –4 + 2x + x
2x + 6 = –4 + 3x
4 + 6 = 3x – 2x
10 = x
simplify each side
move 2x and –4 to the other
sides, change their signs
Linear Equations II
c. + x =
4
3 move to the other side,
change its sign
3
42
1
x = 2
1
4
3
x =
–
4
2
4
3–
Example G. Solve.
a. –5 = x + 6
–6 –5 = x
–11 = x
move +6 to the left,
it turns into –6
122. b. 3(x + 2) – x = 2(–2 + x) + x
3x + 6 – x = –4 + 2x + x
2x + 6 = –4 + 3x
4 + 6 = 3x – 2x
10 = x
simplify each side
move 2x and –4 to the other
sides, change their signs
Linear Equations II
c. + x =
4
3 move to the other side,
change its sign
3
42
1
x = 2
1
4
3
x =
x =
4
–1
–
4
2
4
3–
Example G. Solve.
a. –5 = x + 6
–6 –5 = x
–11 = x
move +6 to the left,
it turns into –6
124. II. (The Opposite rule) If –x = c, then x = –c
Example H.
a. If –x = –15
Linear Equations II
125. Linear Equations II
II. (The Opposite rule) If –x = c, then x = –c
Example H.
a. If –x = –15
then x = 15
126. 4
3
Linear Equations II
II. (The Opposite rule) If –x = c, then x = –c
Example H.
a. If –x = –15
then x = 15
b. If –x =
127. 4
3
4
3
Linear Equations II
II. (The Opposite rule) If –x = c, then x = –c
Example H.
a. If –x = –15
then x = 15
b. If –x =
then x = –
128. 4
3
4
3
Linear Equations II
Recall that an equation or a relation expressed using fractions
may always be restated in a way without using fractions.
II. (The Opposite rule) If –x = c, then x = –c
Example H.
a. If –x = –15
then x = 15
b. If –x =
then x = –
129. 4
3
4
3
Linear Equations II
Recall that an equation or a relation expressed using fractions
may always be restated in a way without using fractions.
Example I. The value of 2/3 of an apple is the same as the
value of ¾ of an orange, restate this relation in whole numbers.
II. (The Opposite rule) If –x = c, then x = –c
Example H.
a. If –x = –15
then x = 15
b. If –x =
then x = –
130. 4
3
4
3
Linear Equations II
Recall that an equation or a relation expressed using fractions
may always be restated in a way without using fractions.
Example I. The value of 2/3 of an apple is the same as the
value of ¾ of an orange, restate this relation in whole numbers.
2
3 A =
3
4 G,We've
II. (The Opposite rule) If –x = c, then x = –c
Example H.
a. If –x = –15
then x = 15
b. If –x =
then x = –
131. 4
3
4
3
Linear Equations II
Recall that an equation or a relation expressed using fractions
may always be restated in a way without using fractions.
Example I. The value of 2/3 of an apple is the same as the
value of ¾ of an orange, restate this relation in whole numbers.
2
3 A =
3
4 G,We've LCD =12, multiply 12 to both sides,
II. (The Opposite rule) If –x = c, then x = –c
Example H.
a. If –x = –15
then x = 15
b. If –x =
then x = –
132. 4
3
4
3
Linear Equations II
Recall that an equation or a relation expressed using fractions
may always be restated in a way without using fractions.
Example I. The value of 2/3 of an apple is the same as the
value of ¾ of an orange, restate this relation in whole numbers.
2
3 A =
3
4 G,
( ) *12
We've LCD =12, multiply 12 to both sides,
2
3
A =
3
4
G
II. (The Opposite rule) If –x = c, then x = –c
Example H.
a. If –x = –15
then x = 15
b. If –x =
then x = –
133. 4
3
4
3
Linear Equations II
Recall that an equation or a relation expressed using fractions
may always be restated in a way without using fractions.
Example I. The value of 2/3 of an apple is the same as the
value of ¾ of an orange, restate this relation in whole numbers.
2
3 A =
3
4 G,
4
( ) *12
We've
2
3
A =
3
4
G
LCD =12, multiply 12 to both sides,
II. (The Opposite rule) If –x = c, then x = –c
Example H.
a. If –x = –15
then x = 15
b. If –x =
then x = –
134. 4
3
4
3
Linear Equations II
Recall that an equation or a relation expressed using fractions
may always be restated in a way without using fractions.
Example I. The value of 2/3 of an apple is the same as the
value of ¾ of an orange, restate this relation in whole numbers.
2
3 A =
3
4 G,
34
( ) *12
We've
2
3
A =
3
4
G
LCD =12, multiply 12 to both sides,
II. (The Opposite rule) If –x = c, then x = –c
Example H.
a. If –x = –15
then x = 15
b. If –x =
then x = –
135. 4
3
4
3
Linear Equations II
Recall that an equation or a relation expressed using fractions
may always be restated in a way without using fractions.
Example I. The value of 2/3 of an apple is the same as the
value of ¾ of an orange, restate this relation in whole numbers.
2
3 A =
3
4 G,
34
8A = 9G
( ) *12
We've
2
3
A =
3
4
G
or the value of 8 apples equal to 9 oranges.
LCD =12, multiply 12 to both sides,
II. (The Opposite rule) If –x = c, then x = –c
Example H.
a. If –x = –15
then x = 15
b. If –x =
then x = –
136. Linear Equations II
III. (Fractional Equations Rule) Multiply fractional equations
by the LCD to remove the fractions first, then solve.
137. Example J. Solve the equations
Linear Equations II
x = –6
4
3a.
III. (Fractional Equations Rule) Multiply fractional equations
by the LCD to remove the fractions first, then solve.
138. III. (Fractional Equations Rule) Multiply fractional equations
by the LCD to remove the fractions first, then solve.
Example J. Solve the equations
Linear Equations II
x = –6
4
3
x = –64
3
)( 4
a. clear the denominator,
multiply both sides by 4
139. Example J. Solve the equations
Linear Equations II
x = –6
4
3
x = –64
3
3x = –24
)( 4
a. clear the denominator,
multiply both sides by 4
III. (Fractional Equations Rule) Multiply fractional equations
by the LCD to remove the fractions first, then solve.
140. Example J. Solve the equations
Linear Equations II
x = –6
4
3
x = –64
3
3x = –24
x = –8
)( 4
a. clear the denominator,
multiply both sides by 4
div by 3
III. (Fractional Equations Rule) Multiply fractional equations
by the LCD to remove the fractions first, then solve.
141. Example J. Solve the equations
x
4 3
1 2
6
5
4
3
Linear Equations II
–+ = xb.
x = –6
4
3
x = –64
3
3x = –24
x = –8
)( 4
a. clear the denominator,
multiply both sides by 4
div by 3
III. (Fractional Equations Rule) Multiply fractional equations
by the LCD to remove the fractions first, then solve.
142. Example J. Solve the equations
x
4 3
1 2
6
5
4
3 the LCD = 12, multiply it to both
sides to clear the denominators
( ) *12
Linear Equations II
–+ = x
x
4 3
1 2
6
5
4
3–+ = x
b.
x = –6
4
3
x = –64
3
3x = –24
x = –8
)( 4
a. clear the denominator,
multiply both sides by 4
div by 3
III. (Fractional Equations Rule) Multiply fractional equations
by the LCD to remove the fractions first, then solve.
143. Example J. Solve the equations
x
4 3
1 2
6
5
4
3 the LCD = 12, multiply it to both
sides to clear the denominators
( ) *12
3
Linear Equations II
–+ = x
x
4 3
1 2
6
5
4
3–+ = x
b.
x = –6
4
3
x = –64
3
3x = –24
x = –8
)( 4
a. clear the denominator,
multiply both sides by 4
div by 3
III. (Fractional Equations Rule) Multiply fractional equations
by the LCD to remove the fractions first, then solve.
144. Example J. Solve the equations
x
4 3
1 2
6
5
4
3 the LCD = 12, multiply it to both
sides to clear the denominators
( ) *12
3 4
Linear Equations II
–+ = x
x
4 3
1 2
6
5
4
3–+ = x
b.
x = –6
4
3
x = –64
3
3x = –24
x = –8
)( 4
a. clear the denominator,
multiply both sides by 4
div by 3
III. (Fractional Equations Rule) Multiply fractional equations
by the LCD to remove the fractions first, then solve.
145. Example J. Solve the equations
x
4 3
1 2
6
5
4
3 the LCD = 12, multiply it to both
sides to clear the denominators
( ) *12
3 4 2 3
Linear Equations II
–+ = x
x
4 3
1 2
6
5
4
3–+ = x
b.
x = –6
4
3
x = –64
3
3x = –24
x = –8
)( 4
a. clear the denominator,
multiply both sides by 4
div by 3
III. (Fractional Equations Rule) Multiply fractional equations
by the LCD to remove the fractions first, then solve.
146. Example J. Solve the equations
x
4 3
1 2
6
5
4
3 the LCD = 12, multiply it to both
sides to clear the denominators
( ) *12
3 4 2 3
3x + 8 = 10x – 9
Linear Equations II
–+ = x
x
4 3
1 2
6
5
4
3–+ = x
b.
x = –6
4
3
x = –64
3
3x = –24
x = –8
)( 4
a. clear the denominator,
multiply both sides by 4
div by 3
III. (Fractional Equations Rule) Multiply fractional equations
by the LCD to remove the fractions first, then solve.
147. Example J. Solve the equations
x
4 3
1 2
6
5
4
3 the LCD = 12, multiply it to both
sides to clear the denominators
( ) *12
3 4 2 3
3x + 8 = 10x – 9
Linear Equations II
–+ = x
x
4 3
1 2
6
5
4
3–+ = x
b.
x = –6
4
3
x = –64
3
3x = –24
x = –8
)( 4
a. clear the denominator,
multiply both sides by 4
move 3x to the right and –9
to the left and switch signs.
div by 3
III. (Fractional Equations Rule) Multiply fractional equations
by the LCD to remove the fractions first, then solve.
148. Example J. Solve the equations
x
4 3
1 2
6
5
4
3 the LCD = 12, multiply it to both
sides to clear the denominators
( ) *12
3 4 2 3
3x + 8 = 10x – 9
9 + 8 = 10x – 3x
Linear Equations II
–+ = x
x
4 3
1 2
6
5
4
3–+ = x
b.
x = –6
4
3
x = –64
3
3x = –24
x = –8
)( 4
a. clear the denominator,
multiply both sides by 4
move 3x to the right and –9
to the left and switch signs.
div by 3
III. (Fractional Equations Rule) Multiply fractional equations
by the LCD to remove the fractions first, then solve.
149. Example J. Solve the equations
x
4 3
1 2
6
5
4
3 the LCD = 12, multiply it to both
sides to clear the denominators
( ) *12
3 4 2 3
3x + 8 = 10x – 9
9 + 8 = 10x – 3x
17 = 7x
Linear Equations II
–+ = x
x
4 3
1 2
6
5
4
3–+ = x
b.
x = –6
4
3
x = –64
3
3x = –24
x = –8
)( 4
a. clear the denominator,
multiply both sides by 4
move 3x to the right and –9
to the left and switch signs.
div by 3
III. (Fractional Equations Rule) Multiply fractional equations
by the LCD to remove the fractions first, then solve.
150. Example J. Solve the equations
x
4 3
1 2
6
5
4
3 the LCD = 12, multiply it to both
sides to clear the denominators
( ) *12
3 4 2 3
3x + 8 = 10x – 9
9 + 8 = 10x – 3x
17 = 7x
= x
7
17
Linear Equations II
–+ = x
x
4 3
1 2
6
5
4
3–+ = x
b.
x = –6
4
3
x = –64
3
3x = –24
x = –8
)( 4
a. clear the denominator,
multiply both sides by 4
move 3x to the right and –9
to the left and switch signs.
div by 3
div by 7
III. (Fractional Equations Rule) Multiply fractional equations
by the LCD to remove the fractions first, then solve.
151. (x – 20) = x – 27
100
15
100
45
Linear Equations II
c.
152. (x – 20) = x – 27
100
15
100
45 multiply 100 to both sides
to remove denominators
[ (x – 20) = x – 27
100
15
100
45
] * 100
Linear Equations II
c.
153. (x – 20) = x – 27
100
15
100
45 multiply 100 to both sides
to remove denominators
[ (x – 20) = x – 27
100
15
100
45
] * 100
1 1 100
Linear Equations II
c.
154. (x – 20) = x – 27
100
15
100
45 multiply 100 to both sides
to remove denominators
[ (x – 20) = x – 27
100
15
100
45
] * 100
1 1 100
15 (x – 20) = 45x – 2700
Linear Equations II
c.
155. (x – 20) = x – 27
100
15
100
45 multiply 100 to both sides
to remove denominators
[ (x – 20) = x – 27
100
15
100
45
] * 100
1 1 100
15 (x – 20) = 45x – 2700
15x – 300 = 45x – 2700
Linear Equations II
c.
156. (x – 20) = x – 27
100
15
100
45 multiply 100 to both sides
to remove denominators
[ (x – 20) = x – 27
100
15
100
45
] * 100
1 1 100
15 (x – 20) = 45x – 2700
15x – 300 = 45x – 2700
2700 – 300 = 45x –15x
Linear Equations II
c.
157. (x – 20) = x – 27
100
15
100
45 multiply 100 to both sides
to remove denominators
[ (x – 20) = x – 27
100
15
100
45
] * 100
1 1 100
15 (x – 20) = 45x – 2700
15x – 300 = 45x – 2700
2700 – 300 = 45x –15x
2400 = 30x
Linear Equations II
c.
158. (x – 20) = x – 27
100
15
100
45 multiply 100 to both sides
to remove denominators
[ (x – 20) = x – 27
100
15
100
45
] * 100
1 1 100
15 (x – 20) = 45x – 2700
15x – 300 = 45x – 2700
2700 – 300 = 45x –15x
2400 = 30x
2400 / 30 = x
80 = x
Linear Equations II
c.
159. (x – 20) = x – 27
100
15
100
45 multiply 100 to both sides
to remove denominators
[ (x – 20) = x – 27
100
15
100
45
] * 100
1 1 100
15 (x – 20) = 45x – 2700
15x – 300 = 45x – 2700
2700 – 300 = 45x –15x
2400 = 30x
2400 / 30 = x
80 = x
Linear Equations II
c.
d. 0.25(x – 100) = 0.10x – 1
160. (x – 20) = x – 27
100
15
100
45 multiply 100 to both sides
to remove denominators
[ (x – 20) = x – 27
100
15
100
45
] * 100
1 1 100
15 (x – 20) = 45x – 2700
15x – 300 = 45x – 2700
2700 – 300 = 45x –15x
2400 = 30x
2400 / 30 = x
80 = x
Linear Equations II
c.
(x – 100) = x – 1100
25
100
10
d. 0.25(x – 100) = 0.10x – 1
Change the decimal into fractions, we get
161. (x – 20) = x – 27
100
15
100
45 multiply 100 to both sides
to remove denominators
[ (x – 20) = x – 27
100
15
100
45
] * 100
1 1 100
15 (x – 20) = 45x – 2700
15x – 300 = 45x – 2700
2700 – 300 = 45x –15x
2400 = 30x
2400 / 30 = x
80 = x
Linear Equations II
c.
(x – 100) = x – 1100
25
100
10 multiply 100 to both sides
to remove denominators
d. 0.25(x – 100) = 0.10x – 1
Change the decimal into fractions, we get
You finish it…
162. (x – 20) = x – 27
100
15
100
45 multiply 100 to both sides
to remove denominators
[ (x – 20) = x – 27
100
15
100
45
] * 100
1 1 100
15 (x – 20) = 45x – 2700
15x – 300 = 45x – 2700
2700 – 300 = 45x –15x
2400 = 30x
2400 / 30 = x
80 = x
Linear Equations II
c.
(x – 100) = x – 1100
25
100
10 multiply 100 to both sides
to remove denominators
d. 0.25(x – 100) = 0.10x – 1
Change the decimal into fractions, we get
You finish it…
Ans: x =160
164. Linear Equations II
IV. (Reduction Rule) Use division to reduce equations to
simpler ones.
Specifically divide the common factor of the coefficients
of each term to make them smaller and easier to work with.
165. 14x – 49 = 70x – 98
Linear Equations II
Example. K. Simplify the equation first then solve.
IV. (Reduction Rule) Use division to reduce equations to
simpler ones.
Specifically divide the common factor of the coefficients
of each term to make them smaller and easier to work with.
166. 14x – 49 = 70x – 98 divide each term by 7
Linear Equations II
Example. K. Simplify the equation first then solve.
14x – 49 70x – 98=
7 7 7 7
IV. (Reduction Rule) Use division to reduce equations to
simpler ones.
Specifically divide the common factor of the coefficients
of each term to make them smaller and easier to work with.
167. 14x – 49 = 70x – 98 divide each term by 7
Linear Equations II
Example. K. Simplify the equation first then solve.
14x – 49 70x – 98=
7 7 7 7
2x – 7 = 10x – 14
IV. (Reduction Rule) Use division to reduce equations to
simpler ones.
Specifically divide the common factor of the coefficients
of each term to make them smaller and easier to work with.
168. 14x – 49 = 70x – 98 divide each term by 7
Linear Equations II
Example. K. Simplify the equation first then solve.
14x – 49 70x – 98=
7 7 7 7
2x – 7 = 10x – 14
14 – 7 = 10x – 2x
IV. (Reduction Rule) Use division to reduce equations to
simpler ones.
Specifically divide the common factor of the coefficients
of each term to make them smaller and easier to work with.
169. 14x – 49 = 70x – 98 divide each term by 7
Linear Equations II
Example. K. Simplify the equation first then solve.
14x – 49 70x – 98=
7 7 7 7
2x – 7 = 10x – 14
14 – 7 = 10x – 2x
7 = 8x
= x
8
7
IV. (Reduction Rule) Use division to reduce equations to
simpler ones.
Specifically divide the common factor of the coefficients
of each term to make them smaller and easier to work with.
170. 14x – 49 = 70x – 98 divide each term by 7
Linear Equations II
Example. K. Simplify the equation first then solve.
14x – 49 70x – 98=
7 7 7 7
2x – 7 = 10x – 14
14 – 7 = 10x – 2x
7 = 8x
= x
8
7
We should always reduce the equation first if it is possible.
IV. (Reduction Rule) Use division to reduce equations to
simpler ones.
Specifically divide the common factor of the coefficients
of each term to make them smaller and easier to work with.
171. There’re two type of equations that give unusual results.
The first type is referred to as identities.
Linear Equations II
172. There’re two type of equations that give unusual results.
The first type is referred to as identities. A simple identity is the
equation “x = x”.
Linear Equations II
173. There’re two type of equations that give unusual results.
The first type is referred to as identities. A simple identity is the
equation “x = x”. This corresponds to the trick question “What
number x is equal to itself?”
Linear Equations II
174. There’re two type of equations that give unusual results.
The first type is referred to as identities. A simple identity is the
equation “x = x”. This corresponds to the trick question “What
number x is equal to itself?” The answer of course is that x
can be any number or that the solutions of equation
“x = x” are all numbers.
Linear Equations II
175. There’re two type of equations that give unusual results.
The first type is referred to as identities. A simple identity is the
equation “x = x”. This corresponds to the trick question “What
number x is equal to itself?” The answer of course is that x
can be any number or that the solutions of equation
“x = x” are all numbers.
Linear Equations II
This is also the case for the any equation where both sides are
identical such as 2x + 1 = 2x + 1,1 – 4x = 1 – 4x etc…
176. There’re two type of equations that give unusual results.
The first type is referred to as identities. A simple identity is the
equation “x = x”. This corresponds to the trick question “What
number x is equal to itself?” The answer of course is that x
can be any number or that the solutions of equation
“x = x” are all numbers.
Linear Equations II
This is also the case for the any equation where both sides are
identical such as 2x + 1 = 2x + 1,1 – 4x = 1 – 4x etc…
An equation with identical expressions on both sides or can be
rearranged into identical sides has all numbers as its solutions.
Such an equation is called an identity.
177. There’re two type of equations that give unusual results.
The first type is referred to as identities. A simple identity is the
equation “x = x”. This corresponds to the trick question “What
number x is equal to itself?” The answer of course is that x
can be any number or that the solutions of equation
“x = x” are all numbers.
Example L. Solve.
Linear Equations II
2(x – 1) + 3 = x – (– x –1)
This is also the case for the any equation where both sides are
identical such as 2x + 1 = 2x + 1,1 – 4x = 1 – 4x etc…
An equation with identical expressions on both sides or can be
rearranged into identical sides has all numbers as its solutions.
Such an equation is called an identity.
178. There’re two type of equations that give unusual results.
The first type is referred to as identities. A simple identity is the
equation “x = x”. This corresponds to the trick question “What
number x is equal to itself?” The answer of course is that x
can be any number or that the solutions of equation
“x = x” are all numbers.
Example L. Solve.
Linear Equations II
2(x – 1) + 3 = x – (– x –1)
This is also the case for the any equation where both sides are
identical such as 2x + 1 = 2x + 1,1 – 4x = 1 – 4x etc…
expand
2x – 2 + 3 = x + x + 1
An equation with identical expressions on both sides or can be
rearranged into identical sides has all numbers as its solutions.
Such an equation is called an identity.
179. There’re two type of equations that give unusual results.
The first type is referred to as identities. A simple identity is the
equation “x = x”. This corresponds to the trick question “What
number x is equal to itself?” The answer of course is that x
can be any number or that the solutions of equation
“x = x” are all numbers.
Example L. Solve.
Linear Equations II
2(x – 1) + 3 = x – (– x –1)
This is also the case for the any equation where both sides are
identical such as 2x + 1 = 2x + 1,1 – 4x = 1 – 4x etc…
expand
2x – 2 + 3 = x + x + 1 simplify
2x + 1 = 2x + 1
An equation with identical expressions on both sides or can be
rearranged into identical sides has all numbers as its solutions.
Such an equation is called an identity.
180. There’re two type of equations that give unusual results.
The first type is referred to as identities. A simple identity is the
equation “x = x”. This corresponds to the trick question “What
number x is equal to itself?” The answer of course is that x
can be any number or that the solutions of equation
“x = x” are all numbers.
Example L. Solve.
Linear Equations II
2(x – 1) + 3 = x – (– x –1)
This is also the case for the any equation where both sides are
identical such as 2x + 1 = 2x + 1,1 – 4x = 1 – 4x etc…
expand
2x – 2 + 3 = x + x + 1 simplify
2x + 1 = 2x + 1
An equation with identical expressions on both sides or can be
rearranged into identical sides has all numbers as its solutions.
Such an equation is called an identity.
two sides are identical
181. There’re two type of equations that give unusual results.
The first type is referred to as identities. A simple identity is the
equation “x = x”. This corresponds to the trick question “What
number x is equal to itself?” The answer of course is that x
can be any number or that the solutions of equation
“x = x” are all numbers.
Example L. Solve.
Linear Equations II
2(x – 1) + 3 = x – (– x –1)
This is also the case for the any equation where both sides are
identical such as 2x + 1 = 2x + 1,1 – 4x = 1 – 4x etc…
expand
2x – 2 + 3 = x + x + 1 simplify
2x + 1 = 2x + 1
So this equation is an identity and every number is a solution.
An equation with identical expressions on both sides or can be
rearranged into identical sides has all numbers as its solutions.
Such an equation is called an identity.
two sides are identical
182. At the opposite end of the identities are the “impossible”
equations where there is no solution at all.
Linear Equations II
183. At the opposite end of the identities are the “impossible”
equations where there is no solution at all. An example is the
equation x = x + 1.
Linear Equations II
184. At the opposite end of the identities are the “impossible”
equations where there is no solution at all. An example is the
equation x = x + 1. This corresponds to the trick question
“What number is still the same after we add 1 to it?”
Linear Equations II
185. At the opposite end of the identities are the “impossible”
equations where there is no solution at all. An example is the
equation x = x + 1. This corresponds to the trick question
“What number is still the same after we add 1 to it?”
Of course there no such number.
Linear Equations II
186. At the opposite end of the identities are the “impossible”
equations where there is no solution at all. An example is the
equation x = x + 1. This corresponds to the trick question
“What number is still the same after we add 1 to it?”
Of course there no such number.
Linear Equations II
If we attempt to solve x = x + 1
x – x = 1
187. At the opposite end of the identities are the “impossible”
equations where there is no solution at all. An example is the
equation x = x + 1. This corresponds to the trick question
“What number is still the same after we add 1 to it?”
Of course there no such number.
Linear Equations II
If we attempt to solve x = x + 1
x – x = 1
0 = 1we get
which is an impossibility.
188. At the opposite end of the identities are the “impossible”
equations where there is no solution at all. An example is the
equation x = x + 1. This corresponds to the trick question
“What number is still the same after we add 1 to it?”
Of course there no such number.
Linear Equations II
If we attempt to solve x = x + 1
x – x = 1
0 = 1we get
which is an impossibility. These equations are called
inconsistent equations.
189. At the opposite end of the identities are the “impossible”
equations where there is no solution at all. An example is the
equation x = x + 1. This corresponds to the trick question
“What number is still the same after we add 1 to it?”
Of course there no such number.
Example M. Solve the equation
Linear Equations II
2(x – 1) + 4 = x – (– x –1)
If we attempt to solve x = x + 1
x – x = 1
0 = 1we get
which is an impossibility. These equations are called
inconsistent equations.
190. At the opposite end of the identities are the “impossible”
equations where there is no solution at all. An example is the
equation x = x + 1. This corresponds to the trick question
“What number is still the same after we add 1 to it?”
Of course there no such number.
Example M. Solve the equation
Linear Equations II
2(x – 1) + 4 = x – (– x –1)
If we attempt to solve x = x + 1
x – x = 1
0 = 1we get
expand
2x – 2 + 4 = x + x + 1
which is an impossibility. These equations are called
inconsistent equations.
191. At the opposite end of the identities are the “impossible”
equations where there is no solution at all. An example is the
equation x = x + 1. This corresponds to the trick question
“What number is still the same after we add 1 to it?”
Of course there no such number.
Example M. Solve the equation
Linear Equations II
2(x – 1) + 4 = x – (– x –1)
If we attempt to solve x = x + 1
x – x = 1
0 = 1we get
expand
2x – 2 + 4 = x + x + 1 simplify
2x + 2 = 2x + 1
which is an impossibility. These equations are called
inconsistent equations.
192. At the opposite end of the identities are the “impossible”
equations where there is no solution at all. An example is the
equation x = x + 1. This corresponds to the trick question
“What number is still the same after we add 1 to it?”
Of course there no such number.
Example M. Solve the equation
Linear Equations II
2(x – 1) + 4 = x – (– x –1)
If we attempt to solve x = x + 1
x – x = 1
0 = 1we get
expand
2x – 2 + 4 = x + x + 1 simplify
2x + 2 = 2x + 1
2 = 1
So this is an inconsistent equation and there is no solution.
which is an impossibility. These equations are called
inconsistent equations.
193. Exercise
A. Solve in one step by addition or subtraction .
Linear Equations I
1. x + 2 = 3 2. x – 1 = –3 3. –3 = x –5
4. x + 8 = –15 5. x – 2 = –1/2 6. = x –
3
2
2
1
B. Solve in one step by multiplication or division.
7. 2x = 3 8. –3x = –1 9. –3 = –5x
10. 8 x = –15 11. –4 =
2
x 12. 7 =
3
–x
13. = –4
3
–x
14. 7 = –x 15. –x = –7
C. Solve by collecting the x’s to one side first. (Remember to
keep the x’s positive.)
16. x + 2 = 5 – 2x 17. 2x – 1 = – x –7 18. –x = x – 8
19. –x = 3 – 2x 20. –5x = 6 – 3x 21. –x + 2 = 3 + 2x
22. –3x – 1= 3 – 6x 23. –x + 7 = 3 – 3x 24. –2x + 2 = 9 + x
