To solve the given system of equations and eliminate the [tex]\( y \)[/tex]-variable, we can use the elimination method. Let's follow a step-by-step approach to solve this system of equations:
Given system of equations:
[tex]\[ \begin{array}{l}
x + 3y = 42 \quad \text{(Equation 1)} \\
2x - y = 14 \quad \text{(Equation 2)}
\end{array} \][/tex]
1. Identify a common coefficient for [tex]\( y \)[/tex]:
In Equation 1, the coefficient of [tex]\( y \)[/tex] is 3.
In Equation 2, the coefficient of [tex]\( y \)[/tex] is -1.
To eliminate [tex]\( y \)[/tex], we need to make the coefficients of [tex]\( y \)[/tex] in both equations equal in magnitude but opposite in sign.
2. Multiply Equation 2 by the appropriate number to match the coefficient of [tex]\( y \)[/tex] in Equation 1:
Since the coefficient of [tex]\( y \)[/tex] in Equation 1 is 3, we multiply Equation 2 by 3:
[tex]\[ 3 \cdot (2x - y) = 3 \cdot 14 \][/tex]
[tex]\[ 6x - 3y = 42 \][/tex]
3. Rewrite the system with the new Equation 2:
[tex]\[ \begin{array}{l}
x + 3y = 42 \quad \text{(Equation 1)} \\
6x - 3y = 42 \quad \text{(New Equation 2)}
\end{array} \][/tex]
4. Add the two equations to eliminate [tex]\( y \)[/tex]:
[tex]\[ (x + 3y) + (6x - 3y) = 42 + 42 \][/tex]
[tex]\[ x + 6x + 3y - 3y = 84 \][/tex]
[tex]\[ 7x = 84 \][/tex]
5. Solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{84}{7} \][/tex]
[tex]\[ x = 12 \][/tex]
6. Substitute [tex]\( x = 12 \)[/tex] back into Equation 1 (or Equation 2) to solve for [tex]\( y \)[/tex]:
Using Equation 1:
[tex]\[ 12 + 3y = 42 \][/tex]
[tex]\[ 3y = 42 - 12 \][/tex]
[tex]\[ 3y = 30 \][/tex]
[tex]\[ y = \frac{30}{3} \][/tex]
[tex]\[ y = 10 \][/tex]
Therefore, the solution to the system is [tex]\( x = 12 \)[/tex] and [tex]\( y = 10 \)[/tex].
Answer D is correct: Multiply the second equation by 3. The solution is [tex]\( x = 12 \)[/tex], [tex]\( y = 10 \)[/tex].
