The document provides examples of solving linear equations. It explains that a linear equation is one where the expressions on both sides are linear, such as 3x + 10 = 34. An example problem involves calculating the cost of x pizzas including delivery. By setting up and solving the equation 3x + 10 = 34, it is determined that x = 8 pizzas. The document discusses manipulating linear equations through steps like subtraction to solve for the variable.

1. Linear Equations I

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 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.

26. Basic principle for solving one- step-equations:
Linear Equations I

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.

44. x
3
–12=d.
Linear Equations I
This says
“x divided by 3 gives –12”,
hence multiply 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

55. Linear Equations I
Example C. Solve for x
b. x – 6 = 3x

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 with
cheese where each slices 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 with
cheese where each slices 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 with
cheese where each slices 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 with
cheese where each slices 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

64. Linear Equations I
c. How many slices of cheese are there in a 500–cal sandwich?

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

76. Example E.
Solve 3x – 4 = 5x + 2
Linear Equations I

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. 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