Answered

Solve the system of equations by elimination:

[tex]\[
\begin{array}{l}
8x + 7y = 39 \\
4x - 14y = -68
\end{array}
\][/tex]

1. Multiply the first equation to enable the elimination of the [tex]\(y\)[/tex]-term.
2. Add the equations to eliminate the [tex]\(y\)[/tex]-terms.
3. Solve the new equation for the [tex]\(x\)[/tex]-value.
4. Substitute the [tex]\(x\)[/tex]-value back into either original equation to find the [tex]\(y\)[/tex]-value.
5. Check the solution.

The solution to the system of equations is [tex]\(\boxed{\phantom{0}} \boxed{\phantom{0}}\)[/tex]



Answer :

Let's solve the system of equations step-by-step as outlined.

### Step 1: Multiply the first equation to enable the elimination of the [tex]\( y \)[/tex]-term.

Given system of equations:
[tex]\[ \begin{array}{l} 8x + 7y = 39 \quad \quad (1) \\ 4x - 14y = -68 \quad (2) \end{array} \][/tex]

We notice that in order to eliminate [tex]\( y \)[/tex], we need the coefficients of [tex]\( y \)[/tex] in both equations to be equal in magnitude and opposite in sign. To do this, let's multiply the first equation by 2. This multiplication will give the term [tex]\( 14y \)[/tex] in the first equation:

[tex]\[ 2 \times (8x + 7y) = 2 \times 39 \][/tex]

This results in:

[tex]\[ 16x + 14y = 78 \quad (3) \][/tex]

### Step 2: Add the equations to eliminate the [tex]\( y \)[/tex]-terms.

Now we have:
[tex]\[ \begin{array}{l} 16x + 14y = 78 \quad \quad (3) \\ 4x - 14y = -68 \quad (2) \end{array} \][/tex]

Add these two equations together to eliminate [tex]\( y \)[/tex]:

[tex]\[ (16x + 14y) + (4x - 14y) = 78 + (-68) \][/tex]

Simplifying this results in:

[tex]\[ 20x = 10 \][/tex]

### Step 3: Solve the new equation for the [tex]\( x \)[/tex]-value.

Now, solve for [tex]\( x \)[/tex]:

[tex]\[ x = \frac{10}{20} = \frac{1}{2} \][/tex]

So, [tex]\( x = \frac{1}{2} \)[/tex].

### Step 4: Substitute the [tex]\( x \)[/tex]-value back into either original equation to find the [tex]\( y \)[/tex]-value.

We'll substitute [tex]\( x = \frac{1}{2} \)[/tex] back into the first original equation:

[tex]\[ 8x + 7y = 39 \][/tex]

Substitute [tex]\( x = \frac{1}{2} \)[/tex]:

[tex]\[ 8 \left( \frac{1}{2} \right) + 7y = 39 \][/tex]

This simplifies to:

[tex]\[ 4 + 7y = 39 \][/tex]

Subtract 4 from both sides:

[tex]\[ 7y = 35 \][/tex]

Divide by 7:

[tex]\[ y = 5 \][/tex]

So, [tex]\( y = 5 \)[/tex].

### Step 5: Check the solution.

Let's substitute [tex]\( x = \frac{1}{2} \)[/tex] and [tex]\( y = 5 \)[/tex] back into both original equations to make sure they both hold true.

1. For the first equation:
[tex]\[ 8x + 7y = 39 \][/tex]
Substitute [tex]\( x = \frac{1}{2} \)[/tex] and [tex]\( y = 5 \)[/tex]:
[tex]\[ 8 \left( \frac{1}{2} \right) + 7(5) = 4 + 35 = 39 \][/tex]
This is correct.

2. For the second equation:
[tex]\[ 4x - 14y = -68 \][/tex]
Substitute [tex]\( x = \frac{1}{2} \)[/tex] and [tex]\( y = 5 \)[/tex]:
[tex]\[ 4 \left( \frac{1}{2} \right) - 14(5) = 2 - 70 = -68 \][/tex]
This is also correct.

Thus, the solution to the system of equations is:
[tex]\[ \left( \frac{1}{2}, 5 \right) \][/tex]