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Solution - Absolute value equations

Exact form: f=724
f=\frac{7}{24}
Decimal form: f=0.292
f=0.292

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
|f-34|=|f+16|
without the absolute value bars:

|x|=|y||f-34|=|f+16|
x=+y(f-34)=(f+16)
x=-y(f-34)=-(f+16)
+x=y(f-34)=(f+16)
-x=y-(f-34)=(f+16)

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||f-34|=|f+16|
x=+y , +x=y(f-34)=(f+16)
x=-y , -x=y(f-34)=-(f+16)

2. Solve the two equations for f

5 additional steps

(f+-34)=(f+16)

Subtract from both sides:

(f+-34)-f=(f+16)-f

Group like terms:

(f-f)+-34=(f+16)-f

Simplify the arithmetic:

-34=(f+16)-f

Group like terms:

-34=(f-f)+16

Simplify the arithmetic:

-34=16

The statement is false:

-34=16

The equation is false so it has no solution.

19 additional steps

(f+-34)=-(f+16)

Expand the parentheses:

(f+-34)=-f+-16

Add to both sides:

(f+-34)+f=(-f+-16)+f

Group like terms:

(f+f)+-34=(-f+-16)+f

Simplify the arithmetic:

2f+-34=(-f+-16)+f

Group like terms:

2f+-34=(-f+f)+-16

Simplify the arithmetic:

2f+-34=-16

Add to both sides:

(2f+-34)+34=(-16)+34

Combine the fractions:

2f+(-3+3)4=(-16)+34

Combine the numerators:

2f+04=(-16)+34

Reduce the zero numerator:

2f+0=(-16)+34

Simplify the arithmetic:

2f=(-16)+34

Find the lowest common denominator:

2f=(-1·2)(6·2)+(3·3)(4·3)

Multiply the denominators:

2f=(-1·2)12+(3·3)12

Multiply the numerators:

2f=-212+912

Combine the fractions:

2f=(-2+9)12

Combine the numerators:

2f=712

Divide both sides by :

(2f)2=(712)2

Simplify the fraction:

f=(712)2

Simplify the arithmetic:

f=7(12·2)

f=724

3. Graph

Each line represents the function of one side of the equation:
y=|f-34|
y=|f+16|
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.