Question

for each value of x, determine whether it is a solution to $-4(x + 1) = 28$.
table with columns: $x$, is it a solution? (yes, no); rows: $x = 10$, $x = -2$, $x = -5$, $x = 1$; each row has empty circles for yes/no. two buttons: refresh (arrow) and $\times$.

Explanation:

Step1: Solve the equation \(-4(x + 1)=28\)

Divide both sides by \(-4\): \(x + 1=\frac{28}{-4}=-7\)
Subtract 1: \(x=-7 - 1=-8\)? Wait, no, wait: Wait, \(\frac{28}{-4}=-7\), so \(x + 1=-7\), then \(x=-7 - 1=-8\)? Wait, no, let's check again. Wait, original equation: \(-4(x + 1)=28\). Divide both sides by \(-4\): \(x + 1 = 28\div(-4)=-7\). Then \(x=-7 - 1=-8\)? Wait, but maybe I made a mistake. Wait, no, let's test each value:

For \(x = 10\):

Step2: Substitute \(x = 10\) into \(-4(x + 1)\)

\(-4(10 + 1)=-4\times11=-44
eq28\). So No.

For \(x=-2\):

Step3: Substitute \(x=-2\) into \(-4(x + 1)\)

\(-4(-2 + 1)=-4\times(-1)=4
eq28\). So No.

For \(x=-5\):

Step4: Substitute \(x=-5\) into \(-4(x + 1)\)

\(-4(-5 + 1)=-4\times(-4)=16
eq28\). So No.

For \(x = 1\):

Step5: Substitute \(x = 1\) into \(-4(x + 1)\)

\(-4(1 + 1)=-4\times2=-8
eq28\). So No.

Wait, but maybe I solved the equation wrong. Let's re - solve the equation \(-4(x + 1)=28\):

Divide both sides by \(-4\): \(x + 1=\frac{28}{-4}=-7\)

Subtract 1 from both sides: \(x=-7 - 1=-8\). So the solution is \(x = - 8\). So none of the given \(x\) values (10, - 2, - 5, 1) are solutions.

Answer:

For \(x = 10\): No
For \(x=-2\): No
For \(x=-5\): No
For \(x = 1\): No