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?

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.