review the graphs of a system of two linear equations in two variables: x-5y=7 and 3x-2y=-4

solve both equations

1 answer

To solve the system of equations given by:

  1. \( x - 5y = 7 \) (Equation 1)
  2. \( 3x - 2y = -4 \) (Equation 2)

we can use either the substitution method or the elimination method. Here, we will use the substitution method first.

Step 1: Solve for one variable

From Equation 1, we can solve for \( x \):

\[ x = 5y + 7 \]

Step 2: Substitute into the other equation

Next, we substitute \( x \) in Equation 2 with \( 5y + 7 \):

\[ 3(5y + 7) - 2y = -4 \]

Expanding this gives:

\[ 15y + 21 - 2y = -4 \]

Combining like terms:

\[ 13y + 21 = -4 \]

Step 3: Isolate \( y \)

Now, isolate \( y \):

\[ 13y = -4 - 21 \] \[ 13y = -25 \] \[ y = -\frac{25}{13} \]

Step 4: Substitute \( y \) back to find \( x \)

Now, substitute \( y \) back into the equation for \( x \):

\[ x = 5\left(-\frac{25}{13}\right) + 7 \] \[ x = -\frac{125}{13} + 7 \] \[ x = -\frac{125}{13} + \frac{91}{13} \] \[ x = -\frac{34}{13} \]

Final Solution

The solution to the system of equations is:

\[ \left( x, y \right) = \left( -\frac{34}{13}, -\frac{25}{13} \right) \]

Optional: Graphical Representation

To graph these equations, you would convert each equation into slope-intercept form \( y = mx + b \):

  1. For \( x - 5y = 7 \): \[ -5y = -x + 7 \quad \Rightarrow \quad y = \frac{1}{5}x - \frac{7}{5} \]

  2. For \( 3x - 2y = -4 \): \[ -2y = -3x - 4 \quad \Rightarrow \quad y = \frac{3}{2}x + 2 \]

You would plot these lines on a graph, and the intersection point would be \(\left( -\frac{34}{13}, -\frac{25}{13} \right)\). This confirms that the solution found algebraically is the point where both lines intersect.