Solve the system of equations using the elimination method.

[tex]\[
\left\{\begin{array}{l}
4x + 10y = -30 \\
5x + 11y = -27
\end{array}\right.
\][/tex]

A. [tex]\((-7, -7)\)[/tex]
B. [tex]\((-10, -7)\)[/tex]
C. [tex]\((7, -7)\)[/tex]
D. [tex]\((10, -7)\)[/tex]



Answer :

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

The system of equations is:
1. [tex]\( 4x + 10y = -30 \)[/tex]
2. [tex]\( 5x + 11y = -27 \)[/tex]

### Step 1: Align the coefficients for one of the variables.
Let's eliminate [tex]\(x\)[/tex] by aligning the coefficients. We'll multiply the first equation by 5 and the second equation by 4, so that the coefficients of [tex]\(x\)[/tex] will be the same.

1. [tex]\( 4x + 10y = -30 \)[/tex] [tex]\( \times 5 \)[/tex] becomes [tex]\( 20x + 50y = -150 \)[/tex]
2. [tex]\( 5x + 11y = -27 \)[/tex] [tex]\( \times 4 \)[/tex] becomes [tex]\( 20x + 44y = -108 \)[/tex]

Now we have:
[tex]\[ 20x + 50y = -150 \][/tex]
[tex]\[ 20x + 44y = -108 \][/tex]

### Step 2: Subtract one equation from the other to eliminate [tex]\(x\)[/tex].
Subtract the second equation from the first:
[tex]\[ (20x + 50y) - (20x + 44y) = -150 - (-108) \][/tex]
[tex]\[ 20x + 50y - 20x - 44y = -150 + 108 \][/tex]
[tex]\[ 6y = -42 \][/tex]

### Step 3: Solve for [tex]\(y\)[/tex].
[tex]\[ 6y = -42 \][/tex]
[tex]\[ y = \frac{-42}{6} \][/tex]
[tex]\[ y = -7 \][/tex]

### Step 4: Substitute [tex]\(y\)[/tex] back into one of the original equations to solve for [tex]\(x\)[/tex].
Substitute [tex]\( y = -7 \)[/tex] into the first equation:
[tex]\[ 4x + 10(-7) = -30 \][/tex]
[tex]\[ 4x - 70 = -30 \][/tex]
[tex]\[ 4x = -30 + 70 \][/tex]
[tex]\[ 4x = 40 \][/tex]
[tex]\[ x = \frac{40}{4} \][/tex]
[tex]\[ x = 10 \][/tex]

### Conclusion
The solution to the system of equations is [tex]\( (x, y) = (10, -7) \)[/tex].

Thus, the correct answer is:
[tex]\[ (10, -7) \][/tex]