Certainly! Let's solve the system of equations using the elimination method.
Given the system of equations:
[tex]\[
\begin{array}{l}
x + 3y = 2 \quad \text{(Equation 1)} \\
-x - 6y = -14 \quad \text{(Equation 2)}
\end{array}
\][/tex]
Step 1: Add the two equations together to eliminate [tex]\( x \)[/tex].
[tex]\[
(x + 3y) + (-x - 6y) = 2 + (-14)
\][/tex]
Simplifying this, we get:
[tex]\[
(x - x) + (3y - 6y) = 2 - 14
\][/tex]
[tex]\[
0 - 3y = -12
\][/tex]
[tex]\[
-3y = -12
\][/tex]
Step 2: Solve for [tex]\( y \)[/tex].
Divide both sides of the equation by -3:
[tex]\[
y = \frac{-12}{-3} = 4
\][/tex]
So, we have found that [tex]\( y = 4 \)[/tex].
Step 3: Substitute [tex]\( y = 4 \)[/tex] back into one of the original equations to solve for [tex]\( x \)[/tex]. We'll use Equation 1:
[tex]\[
x + 3y = 2
\][/tex]
Substitute [tex]\( y = 4 \)[/tex]:
[tex]\[
x + 3(4) = 2
\][/tex]
[tex]\[
x + 12 = 2
\][/tex]
Subtract 12 from both sides:
[tex]\[
x = 2 - 12
\][/tex]
[tex]\[
x = -10
\][/tex]
So, the solution to the system is [tex]\( x = -10 \)[/tex] and [tex]\( y = 4 \)[/tex].
In summary:
[tex]\[
x = -10, \quad y = 4
\][/tex]
We've also confirmed through our elimination step that the resulting equations agree with our calculations:
[tex]\[
\text{Adding the equations: } -3y = -12
\][/tex]
and simplifying it gives:
[tex]\[
-3y = -12
\][/tex]
Thus, the solution to the system of equations is [tex]\( x = -10 \)[/tex] and [tex]\( y = 4 \)[/tex].
