42 linear equations

Linear Equations I
Back to Algebra–Ready Review Content.

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

Example A.
For x pizzas it would cost 3 * x = $3x.
Linear Equations I

Example A.
To have them delivered, it would cost 3x + 10 ($) in total.
Linear Equations I

Example A.
Linear Equations I
b. Suppose the total is $34, how many pizzas did we order?

Example A.
Linear Equations I
We backtrack the calculation by subtracting the $10 for delivery
so the cost for the pizzas is $24,

Example A.
Linear Equations I
so the cost for the pizzas is $24, each pizza is $3 so we must
have ordered 8 pizzas.

Example A.
Linear Equations I
In symbols, we've the equation 3x + 10 = 34,

Example A.
Linear Equations I
backtrack-calculation: 3x + 10 = 34 subtract 10
–10 –10

Example A.
Linear Equations I
–10 –10
so 3x = 24

Example A.
Linear Equations I
–10 –10
so 3x = 24 divide by 3
so x = 8 (pizzas)

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

An equation is two expressions set equal to each other.
Equations look like:
left expression = right expression
or
LHS = RHS
Linear Equations I

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.

or
LHS = RHS
Linear Equations I
equal. Such a value is called a solution of the equation.

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

or
LHS = RHS
Linear Equations I
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.

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.

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.
3x + 10 = 34, or
8 = 4x – 6.

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
of the x2.
3x + 10 = 34, or
8 = 4x – 6.

to reveal what x is. The easiest linear equations to solve are
the single–step equations such as the following ones,
Linear Equations I
of the x2.
3x + 10 = 34, or
8 = 4x – 6.

x – 3 = 12,
Linear Equations I
of the x2.
3x + 10 = 34, or
8 = 4x – 6.

x – 3 = 12,
12 = x + 3,
Linear Equations I
of the x2.
3x + 10 = 34, or
8 = 4x – 6.

x – 3 = 12,
12 = x + 3,
3*x = 12,
12 =
all four equation are one-step equations.
x
3
Linear Equations I
of the x2.
3x + 10 = 34, or
8 = 4x – 6.

x – 3 = 12,
12 = x + 3,
3*x = 12,
12 =
all four equation are one-step equations.
x
3
Linear Equations I
of the x2.
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.

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

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

Example B. Solve for x
a. x – 3 = 12
b. x + 3 = –12
c. 3x = 15
Linear Equations I

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.

a. x – 3 = 12 Add 3 to both sides
+ 3 + 3
b. x + 3 = –12
c. 3x = 15
Linear Equations I
This says

+ 3 + 3
x = 15
b. x + 3 = –12
c. 3x = 15
Linear Equations I
This says

+ 3 + 3
x = 15 check: 15 – 3 = 12
b. x + 3 = –12
c. 3x = 15
Linear Equations I
?
This says

+ 3 + 3
x = 15 check: 15 – 3 = 12
b. x + 3 = –12
c. 3x = 15
Linear Equations I
12 = 12 (yes)
?
This says

+ 3 + 3
x = 15 check: 15 – 3 = 12
b. x + 3 = –12
c. 3x = 15
Linear Equations I
12 = 12 (yes)
?
This says
This says
“3 added to x gives –12”,
hence subtract 3 to get
back to x.

+ 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
This says
back to x.

+ 3 + 3
x = 15 check: 15 – 3 = 12
–3 –3
x = –15
c. 3x = 15
Linear Equations I
12 = 12 (yes)
?
This says
This says
back to x.

+ 3 + 3
x = 15 check: 15 – 3 = 12
–3 –3
x = –15 check: –15 + 3 = –12
c. 3x = 15
Linear Equations I
12 = 12 (yes)
?
?
This says
This says
back to x.

+ 3 + 3
x = 15 check: 15 – 3 = 12
–3 –3
x = –15 check: –15 + 3 = –12
c. 3x = 15
Linear Equations I
12 = 12 (yes)
?
–12 = –12 (yes)
?
This says
This says
back to x.

+ 3 + 3
x = 15 check: 15 – 3 = 12
–3 –3
x = –15 check: –15 + 3 = –12
c. 3x = 15
Linear Equations I
12 = 12 (yes)
?
–12 = –12 (yes)
?
This says
This says
back to x.
This says
“triple the x gives 15”,
hence divide by 3 to get
back to x.

+ 3 + 3
x = 15 check: 15 – 3 = 12
–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
This says
back to x.
This says
back to x.

+ 3 + 3
x = 15 check: 15 – 3 = 12
–3 –3
x = –15 check: –15 + 3 = –12
3x
3
15
3
=
x = 5
Linear Equations I
12 = 12 (yes)
?
–12 = –12 (yes)
?
This says
This says
back to x.
This says
back to x.

+ 3 + 3
x = 15 check: 15 – 3 = 12
–3 –3
x = –15 check: –15 + 3 = –12
3x
3
15
3
=
x = 5 check: 3(5) = 15
Linear Equations I
12 = 12 (yes)
?
–12 = –12 (yes)
?
?
This says
This says
back to x.
This says
back to x.

+ 3 + 3
x = 15 check: 15 – 3 = 12
–3 –3
x = –15 check: –15 + 3 = –12
3x
3
15
3
=
x = 5 check: 3(5) = 15
Linear Equations I
12 = 12 (yes)
?
–12 = –12 (yes)
?
15 = 15 (yes)
?
This says
This says
back to x.
This says
back to x.

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

x
3
–12=d. Multiply both sides by 3
x
3
–12=( (3))
Linear Equations I
This says
back to x.

x
3
x
3
–12=( (3))
x = –36
Linear Equations I
This says
back to x.

x
3
x
3
–12=( (3))
x = –36 Check:
3
–12=– 36
Linear Equations I
This says
back to x.

x
3
x
3
–12=( (3))
x = –36 Check:
3
–12=– 36
Linear Equations I
This says
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.

x
3
x
3
–12=( (3))
x = –36 Check:
3
–12=– 36
Linear Equations I
This says
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.
solution.

x
3
x
3
–12=( (3))
x = –36 Check:
3
–12=– 36
Linear Equations I
This says
back to x.
Example C. Solve for x
a. 4x – 6 = 30
solution.

x
3
x
3
–12=( (3))
x = –36 Check:
3
–12=– 36
Linear Equations I
This says
back to x.
a. 4x – 6 = 30 Collect the numbers by adding 6 to both sides
solution.
+6+6

x
3
x
3
–12=( (3))
x = –36 Check:
3
–12=– 36
Linear Equations I
This says
back to x.
solution.
+6+6
4x = 36

x
3
x
3
–12=( (3))
x = –36 Check:
3
–12=– 36
Linear Equations I
This says
back to x.
solution.
+6+6
4x = 36
4 4
Divide both sides by 4

x
3
x
3
–12=( (3))
x = –36 Check:
3
–12=– 36
Linear Equations I
This says
back to x.
solution.
+6+6
x = 9
4x = 36
(Check this is the right answer.)
4 4
Divide both sides by 4

Linear Equations I
b. x – 6 = 3x

Linear Equations I
b. x – 6 = 3x Collect the x's by subtracting x from both sides
–x –x

Linear Equations I
–x –x
–6 2x=

Linear Equations I
–x –x
–3 = x
–6 2x=
2 2 Divide by 2

Linear Equations I
–x –x
–3 = x
–6 2x=
2 2
In real–life, we encounter linear equations often.
Divide by 2

Linear Equations I
–x –x
–3 = x
–6 2x=
2 2
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

Linear Equations I
–x –x
–3 = x
–6 2x=
2 2
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

Linear Equations I
–x –x
–3 = x
–6 2x=
2 2
cheese?
b. What is the expression that calculate the number of calories
of a sandwich with x slices of cheese?
Divide by 2

Linear Equations I
–x –x
–3 = x
–6 2x=
2 2
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

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

Linear Equations I
The total calories 14 + 90x is 500,

Linear Equations I
The total calories 14 + 90x is 500, i.e.
140 + 90x = 500

Linear Equations I
Subtract 140 from
both sides–140 –140
140 + 90x = 500

Linear Equations I
Subtract 140 from
90x = 360
140 + 90x = 500

Linear Equations I
Subtract 140 from
90x = 360 Divide both sides by 90
90 90
x = 4
140 + 90x = 500

Linear Equations I
Subtract 140 from
90 90
x = 4
So there are 4 slices of cheese in a 500–cal sandwich.
140 + 90x = 500

Linear Equations I
Subtract 140 from
90 90
x = 4
The more general linear equations have the form
#x ± # = #x ± #,
where # can be any number.
140 + 90x = 500

Linear Equations I
Subtract 140 from
90 90
x = 4
#x ± # = #x ± #,
where # can be any number. We solve it by following steps:
140 + 90x = 500

Linear Equations I
Subtract 140 from
90 90
x = 4
#x ± # = #x ± #,
1. Add or subtract to move the x-term to one side of the
equation and get: #x ± # = # or # = #x ± #
140 + 90x = 500

Linear Equations I
Subtract 140 from
90 90
x = 4
#x ± # = #x ± #,
2. Add or subtract the # to separate the number-term from the
x-term to get: #x = # or # = #x.
140 + 90x = 500

Linear Equations I
Subtract 140 from
90 90
x = 4
#x ± # = #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
140 + 90x = 500

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

Example E.
Solve 3x – 4 = 5x + 2
–3x –3x
subtract 3x to remove
the x from one side.
Linear Equations I

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

Example E.
Solve 3x – 4 = 5x + 2
–3x –3x
– 4 = 2x + 2
–2 –2
subtract 2 to move the 2 to
the other side.
Linear Equations I

Example E.
Solve 3x – 4 = 5x + 2
–3x –3x
– 4 = 2x + 2
–2 –2
– 6 = 2x
the other side.
Linear Equations I

Example E.
Solve 3x – 4 = 5x + 2
–3x –3x
– 4 = 2x + 2
–2 –2
– 6 = 2x
2
–6
2
2x
=
the other side.
divide by 2 get x.
Linear Equations I

Example E.
Solve 3x – 4 = 5x + 2
–3x –3x
– 4 = 2x + 2
–2 –2
– 6 = 2x
2
–6
2
2x
=
–3 = x
the other side.
divide by 2 get x.
Linear Equations I

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

42 linear equations

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42 linear equations