To solve the system of equations using the elimination method, we have the following two equations:
- \( 14x - 4y = 40 \) (Equation 1)
- \( 7x + 8y = -10 \) (Equation 2)
Step 1: Make the coefficients of \(y\) in both equations equal.
We can multiply Equation 1 by 2 to get the coefficients of \(y\) to match:
\[ 2(14x - 4y) = 2(40) \]
This gives us:
\[ 28x - 8y = 80 \quad \text{(Equation 3)} \]
Now we have:
- \( 28x - 8y = 80 \) (Equation 3)
- \( 7x + 8y = -10 \) (Equation 2)
Step 2: Add Equation 3 and Equation 2.
We will add the two equations together to eliminate \(y\):
\[ (28x - 8y) + (7x + 8y) = 80 - 10 \]
This simplifies to:
\[ 35x + 0y = 70 \]
So we have:
\[ 35x = 70 \]
Step 3: Solve for \(x\).
Divide both sides by 35:
\[ x = \frac{70}{35} = 2 \]
Step 4: Substitute \(x = 2\) back into one of the original equations.
Let's use Equation 2:
\[ 7x + 8y = -10 \]
Substituting for \(x\):
\[ 7(2) + 8y = -10 \]
This simplifies to:
\[ 14 + 8y = -10 \]
Subtract 14 from both sides:
\[ 8y = -10 - 14 \] \[ 8y = -24 \]
Now, divide by 8:
\[ y = \frac{-24}{8} = -3 \]
Final Solution:
The solution for the system of equations is:
\[ x = 2, \quad y = -3 \]
Thus, the point of intersection is \( (2, -3) \).