Question

Follow the directions to solve the system of equations by elimination.

8x + 7y = 39

4x – 14y = –68

Multiply the first equation to enable the elimination of the y-term.
Add the equations to eliminate the y-terms.
Solve the new equation for the x-value.
Substitute the x-value back into either original equation to find the y-value.
Check the solution.
The solution to the system of equations is (

Answer 1

To solve the system of equations using elimination, follow these steps:

Given equations:
1. \( 8x + 7y = 39 \) (Equation 1)
2. \( 4x - 14y = -68 \) (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:
\[
\left(\frac{1}{2}, 5\right)
\]