To solve the given system of linear equations:
[tex]\[
\begin{array}{l}
-x - 5y = -2 \quad \text{(1)} \\
2x + 3y = 18 \quad \text{(2)}
\end{array}
\][/tex]
we can use the method of substitution or elimination. Here, let's use the elimination method to find the solution step-by-step.
First, we will eliminate one of the variables by making the coefficients of either [tex]\( x \)[/tex] or [tex]\( y \)[/tex] the same (or additive inverses) in both equations. Let's choose to eliminate [tex]\( x \)[/tex].
We have:
[tex]\[
-x - 5y = -2 \quad \text{(Equation 1)}
\][/tex]
[tex]\[
2x + 3y = 18 \quad \text{(Equation 2)}
\][/tex]
To eliminate [tex]\( x \)[/tex], we can multiply the first equation by 2, so that the coefficients of [tex]\( x \)[/tex] in both equations are opposites:
[tex]\[
2(-x - 5y) = 2(-2)
\][/tex]
This simplifies to:
[tex]\[
-2x - 10y = -4 \quad \text{(Equation 3)}
\][/tex]
Now, we can add Equation 3 to Equation 2 to eliminate [tex]\( x \)[/tex]:
[tex]\[
-2x - 10y = -4 \quad \text{(Equation 3)}
\][/tex]
[tex]\[
2x + 3y = 18 \quad \text{(Equation 2)}
\][/tex]
Adding these two equations together, we get:
[tex]\[
(-2x + 2x) + (-10y + 3y) = -4 + 18
\][/tex]
This simplifies to:
[tex]\[
0x - 7y = 14
\][/tex]
So, we have:
[tex]\[
-7y = 14
\][/tex]
Solving for [tex]\( y \)[/tex]:
[tex]\[
y = \frac{14}{-7}
\][/tex]
[tex]\[
y = -2
\][/tex]
Now that we have the value of [tex]\( y \)[/tex], we can substitute it back into one of the original equations to find [tex]\( x \)[/tex]. Let's substitute [tex]\( y = -2 \)[/tex] into Equation 1:
[tex]\[
-x - 5(-2) = -2
\][/tex]
This simplifies to:
[tex]\[
-x + 10 = -2
\][/tex]
Subtracting 10 from both sides, we get:
[tex]\[
-x = -2 - 10
\][/tex]
[tex]\[
-x = -12
\][/tex]
Multiplying both sides by -1, we get:
[tex]\[
x = 12
\][/tex]
Therefore, the solution to the system of equations is:
[tex]\[
x = 12
\][/tex]
[tex]\[
y = -2
\][/tex]
So, the solution is [tex]\( (12, -2) \)[/tex].
