To solve this system of equations by elimination, we follow these steps:
Given the system:
[tex]\[
\begin{cases}
-4x + y = -7 \quad &\text{(Equation 1)} \\
-2x + 3y = -1 \quad &\text{(Equation 2)}
\end{cases}
\][/tex]
Step 1: Make the coefficients of one of the variables in both equations the same in magnitude, so they can cancel each other out when added or subtracted.
Let's manipulate the equations to eliminate [tex]\(x\)[/tex].
Equation 1: [tex]\(-4x + y = -7\)[/tex]
Equation 2: [tex]\(-2x + 3y = -1\)[/tex]
Equation 2 can be multiplied by 2 to make the coefficient of [tex]\(x\)[/tex] in both equations equal in magnitude:
[tex]\[
2 \times (-2x + 3y) = 2 \times -1
\][/tex]
Thus, we get:
[tex]\[
-4x + 6y = -2 \quad \text{(New Equation 2)}
\][/tex]
Now we have:
[tex]\[
\begin{cases}
-4x + y = -7 \quad &\text{(Equation 1)} \\
-4x + 6y = -2 \quad &\text{(New Equation 2)}
\end{cases}
\][/tex]
Step 2: Subtract Equation 1 from New Equation 2 to eliminate [tex]\(x\)[/tex]:
[tex]\[
(-4x + 6y) - (-4x + y) = -2 - (-7)
\][/tex]
This simplifies to:
[tex]\[
-4x + 6y + 4x - y = -2 + 7
\][/tex]
Combining like terms:
[tex]\[
5y = 5
\][/tex]
Step 3: Solve for [tex]\(y\)[/tex]:
[tex]\[
y = \frac{5}{5} = 1
\][/tex]
We have found [tex]\(y = 1\)[/tex].
Step 4: Substitute [tex]\(y = 1\)[/tex] back into one of the original equations to find [tex]\(x\)[/tex]. Let's use Equation 1:
[tex]\[
-4x + y = -7
\][/tex]
Substitute [tex]\(y = 1\)[/tex]:
[tex]\[
-4x + 1 = -7
\][/tex]
Subtract 1 from both sides:
[tex]\[
-4x = -7 - 1
\][/tex]
[tex]\[
-4x = -8
\][/tex]
Divide both sides by -4:
[tex]\[
x = \frac{-8}{-4} = 2
\][/tex]
Thus, we have found [tex]\(x = 2\)[/tex].
Therefore, the solution to the system of equations is:
[tex]\[
(x, y) = (2, 1)
\][/tex]
This is the single solution to the system.
