Solve this system of equations by using elimination.
[tex]\[
\begin{array}{c}
\left\{
\begin{array}{l}
3x + 5y = -2 \\
-x + 2y = 8
\end{array}
\right.
\end{array}
\][/tex]



Answer :

To solve the system of equations using the elimination method, follow these detailed steps:

The given system of equations is:
[tex]\[ \begin{cases} 3x + 5y = -2 \\ -x + 2y = 8 \end{cases} \][/tex]

1. Eliminate one variable:

To eliminate [tex]\(x\)[/tex], we want to make the coefficients of [tex]\(x\)[/tex] in both equations equal in magnitude but opposite in sign so that they cancel out when we add the equations. To do this, we can multiply the second equation by [tex]\(3\)[/tex]:
[tex]\[ 3(-x + 2y) = 3(8) \][/tex]
Which simplifies to:
[tex]\[ -3x + 6y = 24 \][/tex]

2. Add the two equations:

Now add the new equation [tex]\(-3x + 6y = 24\)[/tex] to the first original equation [tex]\(3x + 5y = -2\)[/tex]:
[tex]\[ (3x + 5y) + (-3x + 6y) = -2 + 24 \][/tex]
Simplifying this, we get:
[tex]\[ 3x - 3x + 5y + 6y = -2 + 24 \][/tex]
[tex]\[ 0x + 11y = 22 \][/tex]
[tex]\[ 11y = 22 \][/tex]

3. Solve for [tex]\(y\)[/tex]:

Divide both sides of the equation [tex]\(11y = 22\)[/tex] by [tex]\(11\)[/tex] to isolate [tex]\(y\)[/tex]:
[tex]\[ y = \frac{22}{11} \][/tex]
[tex]\[ y = 2 \][/tex]

4. Substitute [tex]\(y\)[/tex] back into one of the original equations to find [tex]\(x\)[/tex]:

Now substitute [tex]\(y = 2\)[/tex] back into the original first equation [tex]\(3x + 5y = -2\)[/tex]:
[tex]\[ 3x + 5(2) = -2 \][/tex]
Simplifying this, we get:
[tex]\[ 3x + 10 = -2 \][/tex]
[tex]\[ 3x = -2 - 10 \][/tex]
[tex]\[ 3x = -12 \][/tex]

5. Solve for [tex]\(x\)[/tex]:

Divide both sides of the equation [tex]\(3x = -12\)[/tex] by [tex]\(3\)[/tex] to isolate [tex]\(x\)[/tex]:
[tex]\[ x = \frac{-12}{3} \][/tex]
[tex]\[ x = -4 \][/tex]

Thus, the solution to the system of equations is:
[tex]\[ (x, y) = (-4, 2) \][/tex]