Question

solve for ( x ). if there are multiple solutions, enter them as a list of values separated by commas. if there are no solutions, enter none. (\frac{2}{-9x - 4} = \frac{-3}{-6x + 2})

Explanation:

Step1: Cross - multiply the fractions

To solve the equation \(\frac{2}{-9x - 4}=\frac{-3}{-6x + 2}\), we cross - multiply. Cross - multiplying gives us \(2\times(-6x + 2)=-3\times(-9x - 4)\).

Step2: Expand both sides of the equation

First, expand the left - hand side: \(2\times(-6x)+2\times2=-12x + 4\).
Then, expand the right - hand side: \(-3\times(-9x)-3\times(-4)=27x + 12\).
So our equation becomes \(-12x + 4 = 27x+12\).

Step3: Move the \(x\) terms to one side and constants to the other side

Add \(12x\) to both sides of the equation: \(-12x+12x + 4=27x + 12x+12\), which simplifies to \(4 = 39x+12\).
Subtract 12 from both sides: \(4-12=39x+12 - 12\), so \(-8 = 39x\).

Step4: Solve for \(x\)

Divide both sides by 39: \(x=\frac{-8}{39}\).
We need to check if this solution makes the denominators zero.
For the denominator \(-9x - 4\): \(-9\times(\frac{-8}{39})-4=\frac{72}{39}-4=\frac{72 - 156}{39}=\frac{-84}{39}
eq0\).
For the denominator \(-6x + 2\): \(-6\times(\frac{-8}{39})+2=\frac{48}{39}+2=\frac{48 + 78}{39}=\frac{126}{39}
eq0\).

Answer:

\(-\frac{8}{39}\)