Question

complete the sentence based on the equation shown below.
$-4(x + 2) = h(-x - 2)$
answer attempt 1 out of 2
this equation will have one solution when $h = \square$ because you get one solution when you have $\quad$ number of xs on either side of the equation and $\quad$.

Explanation:

Step1: Rewrite right-hand side factor

Notice that $-x-2 = -(x+2)$, so substitute this into the equation:
$-4(x+2) = h(-(x+2))$

Step2: Rearrange to isolate $h$

Rewrite the equation as:
$-4(x+2) = -h(x+2)$
For the equation to have one solution, the coefficients of $(x+2)$ must not be equal (if they were equal, the equation would be true for all $x$, infinitely many solutions). We need to find $h$ such that the terms with $x$ are not equivalent. First, divide both sides by $(x+2)$ (assuming $x
eq -2$):
$-4 = -h$

Step3: Solve for $h$

Multiply both sides by $-1$:
$h = 4$
When $h=4$, substituting back gives $-4(x+2)=4(-x-2)$, which simplifies to $-4x-8=-4x-8$, which is infinitely many solutions? No, correction: Wait, we need the equation to have **one solution**, which occurs when the coefficients of $x$ are different. Let's expand both sides fully:
Left side: $-4(x+2) = -4x -8$
Right side: $h(-x-2) = -hx -2h$
Set equal: $-4x -8 = -hx -2h$
Bring all terms to left: $(-4 + h)x + (-8 + 2h) = 0$
For one solution, the coefficient of $x$ must not be 0: $-4 + h
eq 0 \implies h
eq 4$. Wait, no, the question says "one solution when $h=$". Wait, no—wait, if coefficients are equal, infinitely many solutions. If coefficients are unequal, one solution. But wait, let's check when $h
eq 4$:
$(-4 + h)x = 8 - 2h$
$x = \frac{8-2h}{h-4} = \frac{-2(h-4)}{h-4} = -2$ (when $h
eq 4$). Oh, so for all $h
eq 4$, the solution is $x=-2$, which is one solution. But the question asks for $h=$ a value. Wait, no, the only time there is NOT one solution is when $h=4$ (infinitely many) or... no, there's no case of no solution here. Wait, the question says "this equation will have one solution when $h=$ [ ] because you get one solution when you have [different] number of x's on either side of the equation and [constant terms match or not?]". Wait, no, when $h
eq 4$, the coefficients of $x$ are different ($-4$ vs $-h$), so we can solve for $x$, which gives $x=-2$, one solution. But the question asks for a specific $h$. Wait, maybe the question is phrased as "will have one solution (not infinitely many) when $h
eq 4$, but the blank asks for $h=$. Wait, no, let's re-express:
If $h=4$, both sides are $-4(x+2)=4(-x-2) \implies -4x-8=-4x-8$, infinitely many solutions.
If $h
eq 4$, then we can divide both sides by $(x+2)$ (if $x
eq -2$), but solving gives $x=-2$ as the only solution. So the equation has one solution for any $h
eq 4$, but the question asks for $h=$ [ ]. Wait, maybe the question has a typo, but no—wait, the drop-downs: "you get one solution when you have [different] number of x's on either side of the equation and [the constant terms are consistent]". Wait, no, the only time it does NOT have one solution is when $h=4$ (infinitely many). So to have one solution, $h$ can be any value except 4, but the blank asks for a specific $h$. Wait, no, let's check again:
Wait, if we rearrange:
$-4x -8 = -hx -2h$
$(h-4)x = 8-2h$
If $h
eq 4$, then $x = \frac{8-2h}{h-4} = \frac{-2(h-4)}{h-4} = -2$, which is a single solution.
If $h=4$, then $0x=0$, infinitely many solutions.
So the equation has one solution for all $h
eq 4$, but the question asks for $h=$ [ ]. Maybe the question is asking for the value where it does NOT have one solution, but no, it says "will have one solution when $h=$". Wait, maybe the question means "exactly one solution (not infinitely many)" so any $h$ except 4, but the blank expects a value. Wait, maybe I misread. Wait, the original equation: $-4(x+2)=h(-x-2)$. If $h=4$, it's infinitely many. If $h$ is any other number, like $h=3$, then $-4(x+2)=3(-x-2) \implies -4x-8=-3x-6 \implies -x=2 \implies x=-2$, one solution. So the answer is any $h
eq 4$, but the blank asks for a specific $h$. Wait, maybe the question is phrased as "will have one solution (not infinitely many) when $h$ is not 4, but the blank expects $h
eq 4$, but no, the blank is a box. Wait, no, maybe I made a mistake. Wait, the question says "you get one solution when you have [different] number of x's on either side of the equation". The number of x's: left side has $-4x$, right side has $-hx$. So different number of x's means $-4
eq -h \implies h
eq 4$. So to have one solution, $h$ can be any value except 4. But the blank asks for $h=$ [ ]. Maybe the question is asking for the value where it has one solution, but actually, all $h$ except 4 give one solution. But maybe the question expects $h
eq 4$, but let's go with the logic: when $h$ is not 4, we have one solution. But the blank is a box, so maybe the question has a mistake, but no—wait, no, when $h=4$, infinitely many solutions. When $h$ is any other value, one solution. So the answer can be any number except 4, but let's pick $h=5$? No, no, wait, no—wait, the equation simplifies to $x=-2$ for any $h
eq 4$, which is one solution. So the answer is that $h$ can be any value except 4, but the question asks for $h=$ [ ]. Wait, maybe the question is asking for the value where it does NOT have one solution, but no, it says "will have one solution when $h=$". Wait, maybe I misread the equation: is it $-4(x+2)=h(-x-2)$? Yes. Oh! Wait a second: if we factor out -1 from the right side: $-4(x+2) = -h(x+2)$. So if $x
eq -2$, we can divide both sides by $(x+2)$: $-4 = -h \implies h=4$. But if $h
eq 4$, then the only solution is $x=-2$, because when $x=-2$, both sides are 0, which is true. Wait, no—if $h
eq 4$, then the only $x$ that satisfies the equation is $x=-2$, which is one solution. If $h=4$, then all $x$ satisfy the equation, infinitely many solutions. So the equation has one solution when $h$ is any number except 4. But the question asks for $h=$ [ ]. Maybe the question is phrased incorrectly, but the intended answer is that $h
eq 4$, but the blank expects a value. Wait, no, the drop-downs: "you get one solution when you have [a different] number of x's on either side of the equation and [the constant terms form a true statement when solved]". Wait, no, let's go with the standard: linear equations have one solution when the coefficients of the variable are different. So coefficients of $x$: left is $-4$, right is $-h$. So $-4
eq -h \implies h
eq 4$. So to have one solution, $h$ can be any value except 4. But the blank asks for $h=$ [ ], so maybe the question wants a specific example, like $h=3$, but no, the question says "this equation will have one solution when $h=$". Wait, no, maybe I made a mistake. Wait, if $h=4$, infinitely many solutions. If $h$ is any other value, one solution. So the answer is that $h$ can be any real number except 4, but the blank expects a value. Wait, maybe the question is asking for the value where it has exactly one solution, but actually, all $h$ except 4 give one solution. But let's write the answer as $h
eq 4$, but the blank is a box, so maybe the intended answer is $h=4$ is wrong, so $h$ is not 4. Wait, no, the question says "will have one solution when $h=$". Oh! Wait, maybe I got it backwards. If $h=4$, infinitely many solutions. If $h
eq 4$, one solution. So the equation has one solution when $h$ is any number except 4. So the blank can be filled with any number except 4, e.g., $h=5$, but the standard answer here is that $h
eq 4$, but let's go with the step-by-step:

Wait, let's re-express:

Step1: Expand both sides

Left: $-4(x+2) = -4x -8$
Right: $h(-x-2) = -hx -2h$

Step2: Rearrange terms

$-4x -8 = -hx -2h$
$(h-4)x = 8-2h$

Step3: Analyze solution conditions

For one solution, the coefficient of $x$ must not be zero: $h-4
eq 0 \implies h
eq 4$. When $h
eq 4$, we can solve for $x$:
$x = \frac{8-2h}{h-4} = \frac{-2(h-4)}{h-4} = -2$, which is a single unique solution.

Answer:

This equation will have one solution when $\boldsymbol{h
eq 4}$ because you get one solution when you have a different number of x's on either side of the equation and the resulting equation simplifies to a single unique value for $x$.

(Note: If a specific numerical value is required for the blank, any number except 4 is valid, e.g., $h=3$.)