The document discusses linear equations and how to solve them. It begins by providing an example of solving a multi-step linear equation to find the number of pizzas ordered given the total cost. It then defines linear equations as those containing only first degree terms of the variable and no higher powers. The document states that linear equations are easy to solve by manipulating the equation to isolate the variable. It provides examples of single-step linear equations and explains the basic principle is to apply the opposite operation to both sides to isolate the variable.
Algebra the way to do it | Free Sample eBook | Mathslearning.com | Mathematic...Mathslearning.com
Algebra ...the way to do it contains good, clearly - explained teaching text, worked examples and exercises (with fully - worked answers), 'building' algebra from basics up to approximately GCE Advanced Subsidiary/GCSE Additional level. It teaches the necessary skills in adding, subtracting, multiplying and dividing algebraic quantities, simplifying algebraic expressions, expanding brackets, H.C.F. and L.C.M., solving equations/ Inequations, factorization, graphs of straight lines, quadratic, exponential and reciprocal functions. It highlights key points. Published by Mathematics Publishing Company, available from www.mathslearning.com.
Synthetic Fiber Construction in lab .pptxPavel ( NSTU)
Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
Ethnobotany and Ethnopharmacology:
Ethnobotany in herbal drug evaluation,
Impact of Ethnobotany in traditional medicine,
New development in herbals,
Bio-prospecting tools for drug discovery,
Role of Ethnopharmacology in drug evaluation,
Reverse Pharmacology.
This is a presentation by Dada Robert in a Your Skill Boost masterclass organised by the Excellence Foundation for South Sudan (EFSS) on Saturday, the 25th and Sunday, the 26th of May 2024.
He discussed the concept of quality improvement, emphasizing its applicability to various aspects of life, including personal, project, and program improvements. He defined quality as doing the right thing at the right time in the right way to achieve the best possible results and discussed the concept of the "gap" between what we know and what we do, and how this gap represents the areas we need to improve. He explained the scientific approach to quality improvement, which involves systematic performance analysis, testing and learning, and implementing change ideas. He also highlighted the importance of client focus and a team approach to quality improvement.
Palestine last event orientationfvgnh .pptxRaedMohamed3
An EFL lesson about the current events in Palestine. It is intended to be for intermediate students who wish to increase their listening skills through a short lesson in power point.
The Indian economy is classified into different sectors to simplify the analysis and understanding of economic activities. For Class 10, it's essential to grasp the sectors of the Indian economy, understand their characteristics, and recognize their importance. This guide will provide detailed notes on the Sectors of the Indian Economy Class 10, using specific long-tail keywords to enhance comprehension.
For more information, visit-www.vavaclasses.com
Model Attribute Check Company Auto PropertyCeline George
In Odoo, the multi-company feature allows you to manage multiple companies within a single Odoo database instance. Each company can have its own configurations while still sharing common resources such as products, customers, and suppliers.
Read| The latest issue of The Challenger is here! We are thrilled to announce that our school paper has qualified for the NATIONAL SCHOOLS PRESS CONFERENCE (NSPC) 2024. Thank you for your unwavering support and trust. Dive into the stories that made us stand out!
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?
Expressions produce outputs. Equations recover inputs.
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.
Expressions produce outputs. Equations recover inputs.
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.
Expressions produce outputs. Equations recover inputs.
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.
Expressions produce outputs. Equations recover inputs.
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.
Expressions produce outputs. Equations recover inputs.
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.
Expressions produce outputs. Equations recover inputs.
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.
Expressions produce outputs. Equations recover inputs.
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.
Expressions produce outputs. Equations recover inputs.
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.
Expressions produce outputs. Equations recover inputs.
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.
Expressions produce outputs. Equations recover inputs.
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 solutions
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 solutions
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 solutions
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 solutions
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 solutions
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 solutions
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;
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 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.
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,
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,
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,
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.
26. 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.
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.
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
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
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
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
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
?
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.
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
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
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
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)
?
?
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.
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
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.
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
=
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
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)
?
?
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. 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.
46. 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.
47. 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.
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.
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.
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.
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
Fact: Given a linear equation if we +, –, * , /, to both sides by
the same quantity, the new equation will have the same
solution.
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
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
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
4x = 36
4 4
Divide both sides by 4
55. 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
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
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
–6 2x=
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 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.
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?
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.
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?
Divide by 2
64. 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
66. Linear Equations I
c. How many slices of cheese are there in a 500–cal sandwich?
The total calories 14 + 90x is 500,
67. 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
68. 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
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
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
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
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.
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:
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 ± #
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.
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. 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
78. Example E.
Solve 3x – 4 = 5x + 2
–3x –3x
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
subtract 3x to remove
the x from one side.
Linear Equations I
80. 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
81. 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
82. 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
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.
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
86. 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.
87. 2x – 3(1 – 3x) = 3(x – 6) – 1
Linear Equations I