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?

Answer 1

To solve the system of two linear equations:

1. **Equations given**:
- \( x - 5y = 7 \)
- \( 3x - 2y = -4 \)

2. **Graphing the equations**:
To find the solution graphically, we need to plot both equations on a coordinate plane.

**For the first equation \( x - 5y = 7 \)**:
- Rearranging the equation to slope-intercept form (y = mx + b):
\[
-5y = -x + 7 \quad \Rightarrow \quad y = \frac{1}{5}x - \frac{7}{5}
\]
- This line has a slope of \( \frac{1}{5} \) and a y-intercept of \( -\frac{7}{5} \).

**For the second equation \( 3x - 2y = -4 \)**:
- Rearranging to slope-intercept form:
\[
-2y = -3x - 4 \quad \Rightarrow \quad y = \frac{3}{2}x + 2
\]
- This line has a slope of \( \frac{3}{2} \) and a y-intercept of \( 2 \).

3. **Finding intersections/solutions**:
- On the graph, the solution to the system of equations is the point where the two lines intersect.

4. **Algebraic method**:
If preferred, we can solve the equations algebraically using either substitution or elimination.

**Using substitution**:
From the first equation \( x = 5y + 7 \), substitute \( x \) into the second equation:
\[
3(5y + 7) - 2y = -4
\]
Expand and simplify:
\[
15y + 21 - 2y = -4
\]
\[
13y + 21 = -4
\]
\[
13y = -25 \quad \Rightarrow \quad y = -\frac{25}{13}
\]
Substitute \( y \) back into \( x = 5y + 7 \):
\[
x = 5\left(-\frac{25}{13}\right) + 7 = -\frac{125}{13} + \frac{91}{13} = -\frac{34}{13}
\]

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

You may now either graph the equations and check their intersection or verify the solution by substituting these values back into the original equations to confirm they satisfy both equations.