To solve the system of equations:
- \( 3x + 5y = 5 \) (Equation 1)
- \( -2x - 8y = 6 \) (Equation 2)
We can use either substitution or elimination. Here, I will use the elimination method.
Step 1: Multiply Equation 1 to match the coefficients of \(y\) in both equations.
We can multiply Equation 1 by 2 to eliminate \(y\):
\[ 2(3x + 5y) = 2(5) \] \[ 6x + 10y = 10 \quad \text{(Equation 3)} \]
Step 2: Modify Equation 2.
Equation 2 can be left as is:
\[ -2x - 8y = 6 \quad \text{(Equation 2)} \]
Step 3: Add Equation 2 and Equation 3 to eliminate \(y\).
Now we add Equation 3 and Equation 2:
\[ (6x + 10y) + (-2x - 8y) = 10 + 6 \]
This simplifies to:
\[ (6x - 2x) + (10y - 8y) = 16 \] \[ 4x + 2y = 16 \]
Step 4: Solve for \(x\). From \(4x + 2y = 16\):
Divide the entire equation by 2:
\[ 2x + y = 8 \quad \text{(Equation 4)} \]
Step 5: Substitute \(y\) from Equation 4 back into Equation 1.
From Equation 4, express \(y\):
\[ y = 8 - 2x \]
Now substitute \(y\) into Equation 1:
\[ 3x + 5(8 - 2x) = 5 \] \[ 3x + 40 - 10x = 5 \] \[ -7x + 40 = 5 \]
Step 6: Solve for \(x\).
Subtract 40 from both sides:
\[ -7x = 5 - 40 \] \[ -7x = -35 \]
Now divide by -7:
\[ x = 5 \]
Step 7: Solve for \(y\) using the value of \(x\).
Substitute \(x = 5\) back into Equation 4:
\[ y = 8 - 2(5) \] \[ y = 8 - 10 \] \[ y = -2 \]
Final Solution:
The solution to the system of equations is:
\[ (x, y) = (5, -2) \]
To verify, substitute \(x = 5\) and \(y = -2\) back into the original equations to ensure they hold true.
- \(3(5) + 5(-2) = 15 - 10 = 5\) ✔️
- \(-2(5) - 8(-2) = -10 + 16 = 6\) ✔️
Both equations are satisfied. Thus, the solution is correct.