To solve the system of equations using the linear combination (elimination) method, we start with the two equations given:
- \( 5x + 3y = -10 \) (Equation 1)
- \( -20x - 7y = 15 \) (Equation 2)
First, we can simplify Equation 2 by multiplying it by -1 to make calculations easier:
\[ 20x + 7y = -15 \quad \text{(Equation 2 simplified)} \]
Now we have:
- \( 5x + 3y = -10 \)
- \( 20x + 7y = -15 \)
Next, we will eliminate \( x \). To do this, we can multiply Equation 1 by 4:
\[ 4(5x + 3y) = 4(-10) \] \[ 20x + 12y = -40 \quad \text{(Equation 3)} \]
Now we have:
- \( 20x + 12y = -40 \) (Equation 3)
- \( 20x + 7y = -15 \) (Equation 2 simplified)
Now, we can eliminate \( x \) by subtracting Equation 2 from Equation 3:
\[ (20x + 12y) - (20x + 7y) = -40 - (-15) \] \[ 12y - 7y = -40 + 15 \] \[ 5y = -25 \]
Dividing both sides by 5:
\[ y = -5 \]
Now that we have \( y \), we can substitute \( y \) back into Equation 1 to find \( x \):
\[ 5x + 3(-5) = -10 \] \[ 5x - 15 = -10 \] \[ 5x = -10 + 15 \] \[ 5x = 5 \] \[ x = 1 \]
Thus, the solution to the system of equations is:
\[ (x, y) = (1, -5) \]
So the correct answer is:
(1, –5).