Question

Review the graphs of a system of two linear equations in two variables:x−5y=7 and 3x−2y=−4. How do you solve this?

The intersection point is____

Answer 1

To solve the system of equations given by \( x - 5y = 7 \) and \( 3x - 2y = -4 \), we can use either the substitution method or the elimination method. Here, I will demonstrate the substitution method:

1. **Rearranging the First Equation**:
Start with the first equation \( x - 5y = 7 \).
Rearranging gives us:
\[ x = 5y + 7 \]

2. **Substituting into the Second Equation**:
Substitute \( x \) in the second equation \( 3x - 2y = -4 \):
\[ 3(5y + 7) - 2y = -4 \]
Simplifying this, we have:
\[ 15y + 21 - 2y = -4 \]
Combining like terms:
\[ 13y + 21 = -4 \]

3. **Isolating \( y \)**:
Now, isolate \( y \):
\[ 13y = -4 - 21 \]
\[ 13y = -25 \]
\[ y = -\frac{25}{13} \]

4. **Substituting \( y \) Back to Find \( x \)**:
Now substitute \( y \) back into the expression for \( x \):
\[ x = 5\left(-\frac{25}{13}\right) + 7 \]
\[ x = -\frac{125}{13} + 7 \]
\[ x = -\frac{125}{13} + \frac{91}{13} \]
\[ x = -\frac{125 - 91}{13} \]
\[ x = -\frac{34}{13} \]

5. **Final Solution**:
Thus, the intersection point, which is the solution to the system of equations, is:
\[
\left(-\frac{34}{13}, -\frac{25}{13}\right)
\]

So, the intersection point is \(\left(-\frac{34}{13}, -\frac{25}{13}\right)\).