Question

A coordinate plane with four quadrants shows the x- and y-axes ranging from negative 5 to 5 in increments of 1. A solid line and a dotted line intersect each other. The equation of the solid line is x minus 5 y equals 3. The equation of the dotted line is 3 x minus 2 y equals negative 4. The intersection of both lines is shown at negative 2 on the x-axis and negative 1 on the y-axis in quadrant 3."/>Review the graphs of a system of two linear equations in two variables: x−5y=7 and 3x−2y=−4. Find the solution to both equations. (1 point) The intersection point is (). Remaining Attempts : 3

Answer 1

To find the solution to the system of equations given by the lines \(x - 5y = 7\) and \(3x - 2y = -4\), we can solve this system using the substitution or elimination method. Here, I'll use the substitution method.

1. **Solve the first equation for \(x\):**
\[
x - 5y = 7
\]
Adding \(5y\) to both sides, we get:
\[
x = 5y + 7
\]

2. **Substitute into the second equation:**
Now, substitute \(x = 5y + 7\) into the second equation:
\[
3(5y + 7) - 2y = -4
\]
Expanding this, we have:
\[
15y + 21 - 2y = -4
\]
Simplifying, we combine like terms:
\[
13y + 21 = -4
\]
Now, isolate \(y\) by subtracting 21 from both sides:
\[
13y = -4 - 21
\]
\[
13y = -25
\]
\[
y = -\frac{25}{13}
\]

3. **Substitute back to find \(x\):**
Using the value of \(y\), substitute back to find \(x\):
\[
x = 5\left(-\frac{25}{13}\right) + 7
\]
\[
x = -\frac{125}{13} + \frac{91}{13}
\]
\[
x = -\frac{125 - 91}{13} = -\frac{34}{13}
\]

4. **Conclusion:**
The solution to the system of equations \(x - 5y = 7\) and \(3x - 2y = -4\) is:
\[
\left(-\frac{34}{13}, -\frac{25}{13}\right)
\]

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