Download 1 Linear Equations with One Variable An equation is a statement of

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Linear Equations with One Variable
An equation is a statement of two algebraic expressions separated by an equal
sign. When solving equations, there are a few steps to follow:
1. Clear fractions and decimals. Multiply each term by the L.C.D. of the
equation to get rid of fractions or multiply each term by powers of 10 to
get rid of decimals.
2. Get rid of any parentheses using distribution.
3. Use the Addition Property of Equalities (Add or Subtract) to get all
variable terms on one side and all constant terms on the opposite side.
4. Use the Multiplication Property of Equalities (Multiply or Divide) on
both sides to eliminate the coefficient of the variable term. (THIS IS THE
LAST STEP)
Example:
Solve
−9y − 18 = 6 − 2y − 10y
The left hand side of the equations (LHS) is already simplified. Simplify
the right hand side (RHS)
-9y -18 = 6 -12y
Use Addition Property of Equalities (At this step, you can move any term)
Let’s move -12y from the RHS
-9y – 18 = 6 – 12y
+12y
+ 12y
3y – 18 = 6
The constant term on the RHS is isolated. Don’t move the isolated term.
Move the constant term from the LHS to the RHS
3y – 18 = 6
+18 +18
3y
= 24
Use the Multiplication Property of Equalities and divide both sides by the
coefficient of y.
3y = 24
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Linear Equations with One Variable
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3
3
y=8
You can always check to see if your answer is right by plugging the
solution into the original equation.
Check:
-9(8) – 18 = 6 – 2(8) – 10 (8)
-90 = -90
This is a True statement; therefore, the solution to the equation is y = 8
Example:
Solve
3(5x + 2) − 4(2x) = −8
Get rid of the parentheses by using distribution
15x + 6 – 8x = -8
Simplify LHS
7x + 6 = -8
Use Addition Property of Equalities to move the constant term from the
LHS to the RHS
7x + 6 = -8
-6
7x
-6
= -14
Use Multiplication Property of Equalities and divide both sides by the
coefficient of x
7x = -14
7
7
x = -2
Check.
7(-2) + 6 = -8
-14 + 6 = -8
-8 = -8
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Linear Equations with One Variable
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Example
Solve
1
5 1
5
x  x
3
6 2
3
Get rid of the fractions by finding the LCD of the equation and then
multiplying by the numerator of each term by the L.C.D.
LCD = 6
(6)1
(6)5 (6)1
(6)5
x

x
3
6
2
3
6
30 6
30
x
 x
3
6 2
3
Simplify each term
2x + 5 = 3x -10
The expressions on LHS and RHS are simplified, so use the Addition
Property of Equalities
2x + 5 = 3x – 10
-5
2x
-5
= 3x – 15
The variable term on the LHS is isolated. Move the variable term from the
RHS
2x = 3x -15
-3x -3x
-x =
-15
Use the Multiplication Property of Equalities and divide both sides by the
coefficient of x
-x = -15
-1 -1
x = 15
Don’t forget to check
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Linear Equations with One Variable
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Example:
Solve
0.4(x + 5) − 0.1(2x + 6) = −0.6
Get rid of the decimal by multiplying each term by powers of 10. We want
to get rid of one decimal place, so multiply all the decimals by 10. Do not
use distribution on this step.
(10) 0.4(x + 5) –(10) 0.1(2x + 6) = (10) −0.6
4(x + 5) – 1(2x + 6)
= -6
Get rid of the parentheses by using distribution
4x + 20 – 2x – 6 = -6
Simplify the LHS
2x + 14 = -6
The constant term on the RHS is isolated, so use Addition Property of
Equalities to move the constant term from the LHS.
2x + 14 = -6
-14
2x
-14
= -20
Use the Multiplicative Property of Equalities and divide both sides by the
coefficient of x.
2x = -20
2
2
x = -10
Don’t forget to check
Other examples:
Solve
5
3
𝑥 + 2𝑥 = 3𝑥 +
5
2
𝑥=
15
𝑥=
3
4
Solve
2(𝑥+3)
3
= 4x - 3
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Linear Equations with One Variable
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SLC Math Lab Lake Worth
Linear Equations with One Variable