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Given the system of equations:

[tex]\[
\begin{array}{l}
x + 2y = -6 \\
x - 4y = -18
\end{array}
\][/tex]

Solve for [tex]\((x, y)\)[/tex] using elimination.

A. [tex]\((0, -3)\)[/tex]

B. [tex]\((-1, 0)\)[/tex]

C. [tex]\((-2, 3)\)[/tex]

D. [tex]\((-14, -6)\)[/tex]


Sagot :

To solve the given system of equations using the elimination method, follow these steps:

The system of equations is:
[tex]\[ x + 2y = -6 \][/tex]
[tex]\[ x - 4y = -18 \][/tex]

Step 1: Multiply the second equation to make the coefficients of [tex]\(x\)[/tex] the same

To eliminate one variable, we first ensure that the coefficients of [tex]\(x\)[/tex] in both equations are the same. We can do this by multiplying the second equation by 2:

[tex]\[ 2(x - 4y) = 2(-18) \][/tex]

Which simplifies to:
[tex]\[ 2x - 8y = -36 \][/tex]

So, we now have two equations:
[tex]\[ x + 2y = -6 \][/tex]
[tex]\[ 2x - 8y = -36 \][/tex]

Step 2: Subtract the first equation from the modified second equation to eliminate [tex]\(x\)[/tex]

[tex]\[ (2x - 8y) - (x + 2y) = -36 - (-6) \][/tex]
[tex]\[ 2x - 8y - x - 2y = -36 + 6 \][/tex]
[tex]\[ x - 10y = -30 \][/tex]

Step 3: Solve for [tex]\(y\)[/tex]

[tex]\[ x - 10y = -30 \][/tex]
Divide by -10:
[tex]\[ y = 3 \][/tex]

Step 4: Substitute [tex]\(y = 3\)[/tex] back into one of the original equations to solve for [tex]\(x\)[/tex]

Using the first equation:
[tex]\[ x + 2y = -6 \][/tex]
Substitute [tex]\(y = 3\)[/tex]:
[tex]\[ x + 2(3) = -6 \][/tex]
[tex]\[ x + 6 = -6 \][/tex]

Subtract 6 from both sides:
[tex]\[ x = -12 \][/tex]

Solution

The solution to the system of equations is:
[tex]\[ (x, y) = (-12, 3) \][/tex]

Thus, the correct answer is [tex]\((-12, 3)\)[/tex].