Question

for each value of x, determine whether it is a solution to -56=-8(x+4).

is it a solution?	yes	no
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Explanation:

Step1: Solve the equation \(-56 = -8(x + 4)\)

First, divide both sides by \(-8\): \(\frac{-56}{-8}=\frac{-8(x + 4)}{-8}\), which simplifies to \(7=x + 4\). Then subtract 4 from both sides: \(x=7 - 4=3\). Wait, but maybe we can test each value directly. Let's test \(x=-8\):
Substitute \(x = -8\) into \(-8(x + 4)\): \(-8(-8 + 4)=-8(-4)=32
eq - 56\), so No.

Step2: Test \(x = 2\)

Substitute \(x = 2\) into \(-8(x + 4)\): \(-8(2 + 4)=-8(6)=-48
eq - 56\), so No.

Step3: Test \(x = 10\)

Substitute \(x = 10\) into \(-8(x + 4)\): \(-8(10 + 4)=-8(14)=-112
eq - 56\), so No.

Step4: Test \(x=-5\)

Substitute \(x=-5\) into \(-8(x + 4)\): \(-8(-5 + 4)=-8(-1)=8
eq - 56\), so No. Wait, maybe I made a mistake in solving. Wait, original equation: \(-56=-8(x + 4)\). Let's solve again: divide both sides by -8: \(7=x + 4\), so \(x = 3\). But 3 is not in the list? Wait, maybe the equation is \(-56=-8x+4\)? No, the problem says \(-56=-8(x + 4)\). Wait, maybe the user's equation is different? Wait, the table has x values -8,2,10,-5. Let's re - check each:

For \(x=-8\): RHS: \(-8(-8 + 4)=-8\times(-4)=32\). LHS is -56. 32≠-56→No.

For \(x = 2\): RHS: \(-8(2 + 4)=-8\times6=-48\). -48≠-56→No.

For \(x = 10\): RHS: \(-8(10 + 4)=-8\times14=-112\). -112≠-56→No.

For \(x=-5\): RHS: \(-8(-5 + 4)=-8\times(-1)=8\). 8≠-56→No. Wait, maybe the equation is \(-56=-8x + 4\)? Let's try that. Then \(-8x= - 60\), \(x=\frac{60}{8}=7.5\), still not in the list. Maybe the equation is \(-56=8(x + 4)\)? Then divide by 8: -7=x + 4, x=-11. Not in list. Wait, maybe the original problem's equation is miswritten? But according to the given x values, we test each:

So for \(x=-8\): No; \(x = 2\): No; \(x = 10\): No; \(x=-5\): No.

Answer:

For \(x=-8\): No; \(x = 2\): No; \(x = 10\): No; \(x=-5\): No. (Mark "No" for all given x values in the table)