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Solution - Absolute value equations

Exact form: x=-53,1
x=-\frac{5}{3} , 1
Mixed number form: x=-123,1
x=-1\frac{2}{3} , 1
Decimal form: x=1.667,1
x=-1.667 , 1

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
|7x+1|=|x9|
without the absolute value bars:

|x|=|y||7x+1|=|x9|
x=+y(7x+1)=(x9)
x=y(7x+1)=(x9)
+x=y(7x+1)=(x9)
x=y(7x+1)=(x9)

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||7x+1|=|x9|
x=+y , +x=y(7x+1)=(x9)
x=y , x=y(7x+1)=(x9)

2. Solve the two equations for x

11 additional steps

(7x+1)=(x-9)

Subtract from both sides:

(7x+1)-x=(x-9)-x

Group like terms:

(7x-x)+1=(x-9)-x

Simplify the arithmetic:

6x+1=(x-9)-x

Group like terms:

6x+1=(x-x)-9

Simplify the arithmetic:

6x+1=9

Subtract from both sides:

(6x+1)-1=-9-1

Simplify the arithmetic:

6x=91

Simplify the arithmetic:

6x=10

Divide both sides by :

(6x)6=-106

Simplify the fraction:

x=-106

Find the greatest common factor of the numerator and denominator:

x=(-5·2)(3·2)

Factor out and cancel the greatest common factor:

x=-53

11 additional steps

(7x+1)=-(x-9)

Expand the parentheses:

(7x+1)=-x+9

Add to both sides:

(7x+1)+x=(-x+9)+x

Group like terms:

(7x+x)+1=(-x+9)+x

Simplify the arithmetic:

8x+1=(-x+9)+x

Group like terms:

8x+1=(-x+x)+9

Simplify the arithmetic:

8x+1=9

Subtract from both sides:

(8x+1)-1=9-1

Simplify the arithmetic:

8x=91

Simplify the arithmetic:

8x=8

Divide both sides by :

(8x)8=88

Simplify the fraction:

x=88

Simplify the fraction:

x=1

3. List the solutions

x=-53,1
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|7x+1|
y=|x9|
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.