How to solve 8-|3-12|+5|4+1|=

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve 8-|3-12|+5|4+1|= | Equatix

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3(r+2s)=2t-4

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Question

$$8-|3-12|+5|4+1|=$$

Evaluate

$-217$

Solution Steps

Subtract $12$ from $3$ to get $-9$.

$$8-|-9|\times 5|4+1|$$

The absolute value of a real number $a$ is $a$ when $a\geq 0$, or $-a$ when $a<0$. The absolute value of $-9$ is $9$.

$$8-9\times 5|4+1|$$

Multiply $9$ and $5$ to get $45$.

$$8-45|4+1|$$

Add $4$ and $1$ to get $5$.

$$8-45|5|$$

The absolute value of a real number $a$ is $a$ when $a\geq 0$, or $-a$ when $a<0$. The absolute value of $5$ is $5$.

$$8-45\times 5$$

Multiply $45$ and $5$ to get $225$.

$$8-225$$

Subtract $225$ from $8$ to get $-217$.

$$-217$$

Show Solution

Factor

$-217$

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