Solve the system of two equations using eliminations with multiplications

Question

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.