Question

question 5
solve the system by elimination.
\\(\

$$\begin{cases}2x - 3y=-20\\-2x - 4y=-8\\end{cases}$$

\\)
\\((x,y)=(\square,\square)\\)
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Explanation:

Step1: Add the two equations.

\[

$$\begin{align*} (2x - 3y)+(-2x - 4y)&=-20+( - 8)\\ 2x-3y - 2x-4y&=-20 - 8\\ -7y&=-28 \end{align*}$$

\]

Step2: Solve for y.

Divide both sides of \(-7y = - 28\) by \(-7\):
\[y=\frac{-28}{-7}=4\]

Step3: Substitute y = 4 into the first equation.

Substitute \(y = 4\) into \(2x-3y=-20\), we get \(2x-3\times4=-20\), which simplifies to \(2x - 12=-20\).

Step4: Solve for x.

Add 12 to both sides of \(2x - 12=-20\): \(2x=-20 + 12=-8\). Then divide both sides by 2: \(x=\frac{-8}{2}=-4\).

Answer:

\((-4,4)\)