To solve the system of equations
- \( -5x - 5y = -30 \)
- \( 10x + 3y = -3 \)
we can start by simplifying each equation.
For the first equation, we can divide everything by -5:
\[ x + y = 6 \quad (1) \]
For the second equation, we can try to isolate \( y \):
\[ 10x + 3y = -3 \] \[ 3y = -3 - 10x \] \[ y = -1 - \frac{10}{3}x \quad (2) \]
Now, we can substitute \( y \) from equation (2) into equation (1):
\[ x + \left(-1 - \frac{10}{3}x\right) = 6 \] \[ x - 1 - \frac{10}{3}x = 6 \] Combining like terms: \[ x - \frac{10}{3}x = 7 \] To combine, we can use a common denominator: \[ \left(\frac{3}{3}x - \frac{10}{3}x\right) = 7 \] \[ -\frac{7}{3}x = 7 \]
Now, multiply both sides by -3/7 to isolate \( x \):
\[ x = -3 \]
Now that we have \( x \), we can substitute \( x = -3 \) back into equation (1) to find \( y \):
\[ -3 + y = 6 \] \[ y = 6 + 3 \] \[ y = 9 \]
So, we have the solution:
\[ (x, y) = (-3, 9) \]
To verify, we can substitute \( x \) and \( y \) back into both original equations:
- Substituting into \( -5x - 5y = -30 \):
\[ -5(-3) - 5(9) = 15 - 45 = -30 \quad \text{(True)} \]
- Substituting into \( 10x + 3y = -3 \):
\[ 10(-3) + 3(9) = -30 + 27 = -3 \quad \text{(True)} \]
Both equations are satisfied, confirming our solution is correct. The solution to the system of equations is:
\[ \boxed{(-3, 9)} \]