To solve the system of equations using elimination, follow these steps:
Given equations:
- \( 8x + 7y = 39 \) (Equation 1)
- \( 4x - 14y = -68 \) (Equation 2)
Step 1: Multiply the first equation to enable the elimination of the y-terms
To eliminate the y-term, we can multiply Equation 1 by 2 so that the coefficients of y in both equations can be the same (since \( 7 \times 2 = 14 \)).
Multiply Equation 1 by 2: \[ 2(8x + 7y) = 2(39) \] This results in: \[ 16x + 14y = 78 \quad \text{(Equation 3)} \]
Step 2: Add the equations to eliminate the y-terms
Now we have:
- Equation 3: \( 16x + 14y = 78 \)
- Equation 2: \( 4x - 14y = -68 \)
Now add Equation 3 and Equation 2: \[ (16x + 14y) + (4x - 14y) = 78 - 68 \] This simplifies to: \[ 20x + 0y = 10 \] So, we have: \[ 20x = 10 \]
Step 3: Solve the new equation for the x-value
Divide both sides by 20: \[ x = \frac{10}{20} = \frac{1}{2} \]
Step 4: Substitute the x-value back into either original equation to find the y-value
Now substitute \( x = \frac{1}{2} \) back into Equation 1: \[ 8\left(\frac{1}{2}\right) + 7y = 39 \] This simplifies to: \[ 4 + 7y = 39 \] Now, solve for \( y \) by subtracting 4 from both sides: \[ 7y = 39 - 4 \] \[ 7y = 35 \] Now divide by 7: \[ y = 5 \]
Step 5: Check the solution
We have found \( x = \frac{1}{2} \) and \( y = 5 \). Let's check this solution in both original equations.
Check Equation 1: \[ 8\left(\frac{1}{2}\right) + 7(5) = 39 \] \[ 4 + 35 = 39 \quad \text{(True)} \]
Check Equation 2: \[ 4\left(\frac{1}{2}\right) - 14(5) = -68 \] \[ 2 - 70 = -68 \quad \text{(True)} \]
Final answer
The solution to the system of equations is: \[ \left(\frac{1}{2}, 5\right) \]
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