To solve the system of equations given by:
- \( x - 5y = 7 \) (Equation 1)
- \( 3x - 2y = -4 \) (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 \( y = mx + b \):
-
For \( x - 5y = 7 \): \[ -5y = -x + 7 \quad \Rightarrow \quad y = \frac{1}{5}x - \frac{7}{5} \]
-
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.