Question

Solve the system of two equations using eliminations with multiplications
4x - 6y = -48
-2x + 4y = 30

Answer 1

To solve the system of equations using elimination, we start with the two given equations:

1. \( 4x - 6y = -48 \) \quad (Equation 1)
2. \( -2x + 4y = 30 \) \quad (Equation 2)

### Step 1: Multiply the equations to align the coefficients for elimination.
To eliminate \(x\), we can manipulate the equations such that their coefficients are opposites. We can multiply Equation 2 by 2:

\[
2 \times (-2x + 4y) = 2 \times 30
\]

This gives us:

\[
-4x + 8y = 60 \quad (Equation 3)
\]

### Step 2: Now, we can add Equation 1 and Equation 3.
The updated equations are:

1. \( 4x - 6y = -48 \)
2. \( -4x + 8y = 60 \)

Now, we add both equations:

\[
(4x - 6y) + (-4x + 8y) = -48 + 60
\]

This simplifies to:

\[
0x + 2y = 12
\]

So, we have:

\[
2y = 12
\]

### Step 3: Solve for \(y\).
Dividing both sides by 2, we get:

\[
y = 6
\]

### Step 4: Substitute \(y\) back into one of the original equations to solve for \(x\).
We can use Equation 1:

\[
4x - 6(6) = -48
\]

This simplifies to:

\[
4x - 36 = -48
\]

Adding 36 to both sides gives:

\[
4x = -48 + 36
\]
\[
4x = -12
\]

Now we divide by 4:

\[
x = -3
\]

### Step 5: Solution
The solution to the system of equations is:

\[
\boxed{( -3, 6 )}
\]

### Verification
Let's verify by substituting \(x = -3\) and \(y = 6\) back into the original equations:

1. Substituting into Equation 1:
\[
4(-3) - 6(6) = -12 - 36 = -48 \quad \text{(True)}
\]

2. Substituting into Equation 2:
\[
-2(-3) + 4(6) = 6 + 24 = 30 \quad \text{(True)}
\]

Both equations check out, confirming that the solution \( ( -3, 6 ) \) is correct.