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semester 1 final review solve: -5(11r - 4) - 11 = -7r + 12(6 - 4r) r =

Question

semester 1 final review
solve:
-5(11r - 4) - 11 = -7r + 12(6 - 4r)
r =

Explanation:

Step1: Expand both sides

Left side: $-5(11r - 4)-11=-55r + 20 - 11=-55r + 9$
Right side: $-7r + 12(6 - 4r)=-7r + 72 - 48r=-55r + 72$

Step2: Set up the equation

We have $-55r + 9=-55r + 72$

Step3: Subtract $-55r$ from both sides

$9 = 72$ (which is a contradiction)

Wait, maybe I made a mistake in expansion. Let's re - expand the right side:
$12(6 - 4r)=72-48r$, so right side is $-7r+72 - 48r=-55r + 72$ (correct). Left side: $-5(11r-4)=-55r + 20$, then $-55r + 20-11=-55r + 9$ (correct).

Subtract $-55r$ from both sides: $9 = 72$, which is false. But maybe the original equation has a typo? Wait, let's check the original equation again: $-5(11r - 4)-11=-7r + 12(6 - 4r)$

Wait, maybe I misread the equation. Let's re - write the equation:

Left side: $-5\times11r-5\times(-4)-11=-55r + 20 - 11=-55r + 9$

Right side: $-7r+12\times6-12\times4r=-7r + 72-48r=-55r + 72$

Now, bring all terms to left side: $-55r + 9+55r - 72 = 0$

Simplify: $-63=0$, which is a contradiction. So there is no solution? But maybe I made a mistake. Wait, let's check the original problem again. If the equation is $-5(11r - 4)-11=-7r + 12(6 + 4r)$ (maybe a sign error in the original problem), let's try that.

Right side: $-7r+72 + 48r=41r + 72$

Left side: $-55r + 9$

Then $-55r+9 = 41r + 72$

$-55r-41r=72 - 9$

$-96r=63$

$r=-\frac{63}{96}=-\frac{21}{32}$

But since the original equation as given has a contradiction, maybe there is a typo. However, if we assume the original equation is correct as written, there is no solution. But maybe I made a mistake in the first expansion. Wait, let's check the original problem again: the equation is $-5(11r - 4)-11=-7r + 12(6 - 4r)$

Wait, another way: maybe the user made a typo, but according to the given equation, it's a contradiction. But let's check the arithmetic again.

Left side:
$-5(11r - 4)=-55r + 20$
$-55r + 20-11=-55r + 9$

Right side:
$12(6 - 4r)=72-48r$
$-7r+72 - 48r=-55r + 72$

Now, equation: $-55r + 9=-55r + 72$

Subtract $-55r$ from both sides: $9 = 72$, which is false. So the equation has no solution.

But maybe the original problem was written incorrectly. If we assume that the right - hand side is $-7r + 12(6 + 4r)$, then:

Right side: $-7r+72 + 48r=41r + 72$

Left side: $-55r + 9$

Equation: $-55r + 9=41r + 72$

$-55r-41r=72 - 9$

$-96r=63$

$r=-\frac{63}{96}=-\frac{21}{32}\approx - 0.65625$

But based on the given equation, there is no solution.

Wait, maybe I misread the coefficient of $r$ in the left - hand side. If the left - hand side is $-5(1r - 4)-11$ (instead of $11r$), let's try:

Left side: $-5(r - 4)-11=-5r + 20 - 11=-5r + 9$

Right side: $-7r+72 - 48r=-55r + 72$

Then $-5r + 9=-55r + 72$

$50r=63$

$r=\frac{63}{50}=1.26$

But since the original equation has $11r$, I think there is a contradiction in the given equation.

Answer:

No solution (or if there is a typo, and assuming a correction, the solution would be different. But based on the given equation, there is no solution as $9 = 72$ is false).