To solve the system of equations using substitution, we will first isolate one variable in one of the equations and then substitute it into the other equation.
The system of equations is:
- \(3x + 7y = 6\) (Equation 1)
- \(6x + 8y = -6\) (Equation 2)
Let's solve Equation 1 for \(x\):
\[ 3x + 7y = 6 \] Subtract \(7y\) from both sides:
\[ 3x = 6 - 7y \] Now divide by 3:
\[ x = \frac{6 - 7y}{3} \]
Now we will substitute this expression for \(x\) into Equation 2:
\[ 6x + 8y = -6 \] Substituting for \(x\):
\[ 6\left(\frac{6 - 7y}{3}\right) + 8y = -6 \]
Now simplify:
\[ \frac{36 - 42y}{3} + 8y = -6 \] Multiplying through by 3 to eliminate the fraction:
\[ 36 - 42y + 24y = -18 \] Combine like terms:
\[ 36 - 18y = -18 \] Now, subtract 36 from both sides:
\[ -18y = -18 - 36 \] \[ -18y = -54 \] Now divide by -18:
\[ y = \frac{-54}{-18} = 3 \]
Now we have \(y = 3\). We substitute \(y\) back into the expression we found for \(x\):
\[ x = \frac{6 - 7(3)}{3} \] \[ x = \frac{6 - 21}{3} \] \[ x = \frac{-15}{3} = -5 \]
Thus, the solution to the system of equations is:
\[ (x, y) = (-5, 3) \]
Let's also verify this solution by substituting \(x = -5\) and \(y = 3\) back into the original equations.
For Equation 1:
\[ 3(-5) + 7(3) = -15 + 21 = 6 \quad \text{(True)} \]
For Equation 2:
\[ 6(-5) + 8(3) = -30 + 24 = -6 \quad \text{(True)} \]
Both equations are satisfied, confirming that the solution is:
\[ \boxed{(-5, 3)} \]
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