Answer :
To solve the system of equations using elimination, follow these steps:
Given equations:
[tex]\[ \begin{aligned} (1) & \quad -x + y = 1 \\ (2) & \quad -6x + 2y = -34 \end{aligned} \][/tex]
1. Multiply the first equation by 2 to align the coefficients of [tex]\( y \)[/tex] in both equations:
[tex]\[ 2(-x + y) = 2 \cdot 1 \implies -2x + 2y = 2 \][/tex]
Now we have the system:
[tex]\[ \begin{aligned} (1') & \quad -2x + 2y = 2 \\ (2) & \quad -6x + 2y = -34 \end{aligned} \][/tex]
2. Subtract equation [tex]\((1')\)[/tex] from equation [tex]\((2)\)[/tex]:
[tex]\[ (-6x + 2y) - (-2x + 2y) = -34 - 2 \][/tex]
[tex]\[ -6x + 2y + 2x - 2y = -34 - 2 \][/tex]
[tex]\[ -4x = -36 \][/tex]
3. Solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{-36}{-4} = 9 \][/tex]
4. Substitute [tex]\( x = 9 \)[/tex] back into the first equation to find [tex]\( y \)[/tex]:
[tex]\[ -x + y = 1 \implies -9 + y = 1 \implies y = 1 + 9 \implies y = 10 \][/tex]
Thus, the solution to the system of equations is:
[tex]\[ (x, y) = (9, 10) \][/tex]
Given equations:
[tex]\[ \begin{aligned} (1) & \quad -x + y = 1 \\ (2) & \quad -6x + 2y = -34 \end{aligned} \][/tex]
1. Multiply the first equation by 2 to align the coefficients of [tex]\( y \)[/tex] in both equations:
[tex]\[ 2(-x + y) = 2 \cdot 1 \implies -2x + 2y = 2 \][/tex]
Now we have the system:
[tex]\[ \begin{aligned} (1') & \quad -2x + 2y = 2 \\ (2) & \quad -6x + 2y = -34 \end{aligned} \][/tex]
2. Subtract equation [tex]\((1')\)[/tex] from equation [tex]\((2)\)[/tex]:
[tex]\[ (-6x + 2y) - (-2x + 2y) = -34 - 2 \][/tex]
[tex]\[ -6x + 2y + 2x - 2y = -34 - 2 \][/tex]
[tex]\[ -4x = -36 \][/tex]
3. Solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{-36}{-4} = 9 \][/tex]
4. Substitute [tex]\( x = 9 \)[/tex] back into the first equation to find [tex]\( y \)[/tex]:
[tex]\[ -x + y = 1 \implies -9 + y = 1 \implies y = 1 + 9 \implies y = 10 \][/tex]
Thus, the solution to the system of equations is:
[tex]\[ (x, y) = (9, 10) \][/tex]
