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]
### 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]
