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Solution - Absolute value equations

Exact form: a=13,4
a=-13 , 4

Other Ways to Solve

Absolute value equations

Step-by-step explanation

1. Rewrite the equation without absolute value bars

Use the rules:
|x|=|y|x=±y and |x|=|y|±x=y
to write all four options of the equation
|5a3|=|3a29|
without the absolute value bars:

|x|=|y||5a3|=|3a29|
x=+y(5a3)=(3a29)
x=y(5a3)=(3a29)
+x=y(5a3)=(3a29)
x=y(5a3)=(3a29)

When simplified, equations x=+y and +x=y are the same and equations x=y and x=y are the same, so we end up with only 2 equations:

|x|=|y||5a3|=|3a29|
x=+y , +x=y(5a3)=(3a29)
x=y , x=y(5a3)=(3a29)

2. Solve the two equations for a

11 additional steps

(5a-3)=(3a-29)

Subtract from both sides:

(5a-3)-3a=(3a-29)-3a

Group like terms:

(5a-3a)-3=(3a-29)-3a

Simplify the arithmetic:

2a-3=(3a-29)-3a

Group like terms:

2a-3=(3a-3a)-29

Simplify the arithmetic:

2a3=29

Add to both sides:

(2a-3)+3=-29+3

Simplify the arithmetic:

2a=29+3

Simplify the arithmetic:

2a=26

Divide both sides by :

(2a)2=-262

Simplify the fraction:

a=-262

Find the greatest common factor of the numerator and denominator:

a=(-13·2)(1·2)

Factor out and cancel the greatest common factor:

a=13

12 additional steps

(5a-3)=-(3a-29)

Expand the parentheses:

(5a-3)=-3a+29

Add to both sides:

(5a-3)+3a=(-3a+29)+3a

Group like terms:

(5a+3a)-3=(-3a+29)+3a

Simplify the arithmetic:

8a-3=(-3a+29)+3a

Group like terms:

8a-3=(-3a+3a)+29

Simplify the arithmetic:

8a3=29

Add to both sides:

(8a-3)+3=29+3

Simplify the arithmetic:

8a=29+3

Simplify the arithmetic:

8a=32

Divide both sides by :

(8a)8=328

Simplify the fraction:

a=328

Find the greatest common factor of the numerator and denominator:

a=(4·8)(1·8)

Factor out and cancel the greatest common factor:

a=4

3. List the solutions

a=13,4
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|5a3|
y=|3a29|
The equation is true where the two lines cross.

Why learn this

We encounter absolute values almost daily. For example: If you walk 3 miles to school, do you also walk minus 3 miles when you go back home? The answer is no because distances use absolute value. The absolute value of the distance between home and school is 3 miles, there or back.
In short, absolute values help us deal with concepts like distance, ranges of possible values, and deviation from a set value.