To solve the system of equations
[tex]\[
\begin{array}{l}
x + 3y = 42 \\
2x - y = 14
\end{array}
\][/tex]
using the elimination method, follow these steps:
1. Determine the Multiplication Factor:
To eliminate the [tex]\( y \)[/tex]-variable, the coefficients of [tex]\( y \)[/tex] in both equations must be made equal and opposite in value. If we multiply the second equation by 3, the coefficient of [tex]\( y \)[/tex] in the second equation will become -3, which is the negative of the coefficient of [tex]\( y \)[/tex] in the first equation.
[tex]\[
3 \times (2x - y) = 3 \times 14
\][/tex]
This gives:
[tex]\[
6x - 3y = 42
\][/tex]
2. Rewrite the System:
Now, the system of equations is:
[tex]\[
\begin{array}{l}
x + 3y = 42 \\
6x - 3y = 42
\end{array}
\][/tex]
3. Add the Equations:
Adding these two equations will eliminate the [tex]\( y \)[/tex]-variable:
[tex]\[
(x + 3y) + (6x - 3y) = 42 + 42
\][/tex]
Simplifying, we get:
[tex]\[
7x = 84
\][/tex]
4. Solve for [tex]\( x \)[/tex]:
[tex]\[
x = \frac{84}{7} = 12
\][/tex]
5. Substitute [tex]\( x \)[/tex] into the First Equation:
Substitute [tex]\( x = 12 \)[/tex] back into the first equation to find [tex]\( y \)[/tex]:
[tex]\[
12 + 3y = 42
\][/tex]
Solving for [tex]\( y \)[/tex]:
[tex]\[
3y = 42 - 12
\][/tex]
[tex]\[
3y = 30
\][/tex]
[tex]\[
y = \frac{30}{3} = 10
\][/tex]
Therefore, the solution to the system is [tex]\( x = 12 \)[/tex] and [tex]\( y = 10 \)[/tex]. The correct choice is:
D. Multiply the second equation by 3. The solution is [tex]\( x = 12 \)[/tex], [tex]\( y = 10 \)[/tex].
