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Question
$3|-8x|+8=80$
$|-4+5x|=16$
$x^2-12x+11=0$
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First Equation: $3|-8x| + 8 = 80$
Step1: Isolate the absolute value term
Subtract 8 from both sides:
$3|-8x| = 80 - 8$
$3|-8x| = 72$
Step2: Solve for the absolute value
Divide both sides by 3:
$|-8x| = \frac{72}{3}$
$|-8x| = 24$
Step3: Split into two linear equations
Since $|A|=B$ means $A=B$ or $A=-B$:
$-8x = 24$ or $-8x = -24$
Step4: Solve for x in each case
For $-8x=24$: $x = \frac{24}{-8} = -3$
For $-8x=-24$: $x = \frac{-24}{-8} = 3$
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Second Equation: $|-4 + 5x| = 16$
Step1: Split into two linear equations
Apply absolute value rule:
$-4 + 5x = 16$ or $-4 + 5x = -16$
Step2: Solve first linear equation
Add 4 to both sides, then divide by 5:
$5x = 16 + 4$
$5x = 20$
$x = \frac{20}{5} = 4$
Step3: Solve second linear equation
Add 4 to both sides, then divide by 5:
$5x = -16 + 4$
$5x = -12$
$x = \frac{-12}{5} = -2.4$
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Third Equation: $x^2 - 12x + 11 = 0$
Step1: Factor the quadratic
Find two numbers that multiply to 11 and add to -12:
$(x - 1)(x - 11) = 0$
Step2: Solve for x using zero product rule
Set each factor equal to 0:
$x - 1 = 0$ or $x - 11 = 0$
$x = 1$ or $x = 11$
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- For $3|-8x| + 8 = 80$: $x = 3$ or $x = -3$
- For $|-4 + 5x| = 16$: $x = 4$ or $x = -\frac{12}{5}$
- For $x^2 - 12x + 11 = 0$: $x = 1$ or $x = 11$