Question

Systems of Linear Equations Unit Test
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Question
Solve the system of equations.
−6x+2y=−32

4x+3y=−9

Answer 1

To solve the system of equations:

1. **First equation**:
\(-6x + 2y = -32\)

2. **Second equation**:
\(4x + 3y = -9\)

Let's solve it step-by-step using the substitution or elimination method. I will use the elimination method for this solution.

### Step 1: Multiply the equations to align coefficients
We can multiply the first equation by \(2\) to make the coefficients of \(y\) in both equations compatible:

\[
2(-6x + 2y) = 2(-32)
\]
This simplifies to:
\[
-12x + 4y = -64
\]

Now our system of equations is:
1. \(-12x + 4y = -64\)
2. \(4x + 3y = -9\)

### Step 2: Eliminate \(y\)
We can eliminate \(y\) by aligning the coefficients. We can multiply the second equation by \(4\) to match the coefficient of \(y\) in the first equation.

\[
4(4x + 3y) = 4(-9)
\]
This simplifies to:
\[
16x + 12y = -36
\]

### Now our system of equations is:
1. \(-12x + 4y = -64\)
2. \(16x + 12y = -36\)

### Step 3: Solve for one variable
Next, we can multiply the first equation by \(3\) to get another equivalent equation.

\[
3(-12x + 4y) = 3(-64)
\]
This simplifies to:
\[
-36x + 12y = -192
\]

Now we have:
1. \(-36x + 12y = -192\)
2. \(16x + 12y = -36\)

### Step 4: Subtract the second from the first
Now we can eliminate \(y\):
\[
(-36x + 12y) - (16x + 12y) = -192 - (-36)
\]
This simplifies to:
\[
-36x - 16x = -192 + 36
\]
\[
-52x = -156
\]

### Step 5: Solve for \(x\)
Dividing both sides by \(-52\):
\[
x = \frac{-156}{-52}
\]
\[
x = 3
\]

### Step 6: Substitute back to find \(y\)
Now substitute \(x = 3\) back into one of the original equations, let's use the second equation:

\[
4(3) + 3y = -9
\]
\[
12 + 3y = -9
\]
Subtract \(12\) from both sides:
\[
3y = -9 - 12
\]
\[
3y = -21
\]
Now divide by \(3\):
\[
y = -7
\]

### Solution
The solution to the system of equations is:
\[
\boxed{(3, -7)}
\]