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Given the system of equations:

[tex]\[
\begin{array}{l}
5x + 2y = 3 \\
4x - 8y = 12
\end{array}
\][/tex]

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

A. [tex]\((-7, 5)\)[/tex]

B. [tex]\((-5, -4)\)[/tex]

C. [tex]\((1, -1)\)[/tex]

D. [tex]\((3, -6)\)[/tex]


Sagot :

Certainly! To solve the system of equations using the elimination method, let's follow these steps:

Given the system of equations:
[tex]\[ \begin{array}{l} 1.\quad 5x + 2y = 3 \\ 2.\quad 4x - 8y = 12 \end{array} \][/tex]

### Step 1: Make the Coefficients of [tex]\( y \)[/tex] Equal

First, we want to make the coefficients of [tex]\( y \)[/tex] in both equations equal so that we can eliminate [tex]\( y \)[/tex]. Notice that the second equation already has [tex]\(-8y\)[/tex], which is a multiple of the coefficient of [tex]\( y \)[/tex] in the first equation ([tex]\(+2y \times -4\)[/tex]).

### Step 2: Eliminate [tex]\( y \)[/tex]

We will multiply the entire first equation by 4 to make the coefficients of [tex]\( y \)[/tex] the same (in magnitude) but opposite in sign:

[tex]\[ 4 \cdot (5x + 2y) = 4 \cdot 3 \][/tex]
[tex]\[ 20x + 8y = 12 \quad \quad \text{New equation (3)} \][/tex]

Now, our system of equations is:
[tex]\[ \begin{array}{l} 3.\quad 20x + 8y = 12 \\ 2.\quad 4x - 8y = 12 \end{array} \][/tex]

We can eliminate [tex]\( y \)[/tex] by adding equations (3) and (2):

[tex]\[ (20x + 8y) + (4x - 8y) = 12 + 12 \][/tex]
[tex]\[ 20x + 4x + 8y - 8y = 24 \][/tex]
[tex]\[ 24x = 24 \][/tex]

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

[tex]\[ x = \frac{24}{24} \][/tex]
[tex]\[ x = 1 \][/tex]

### Step 4: Substitute [tex]\( x \)[/tex] back to Find [tex]\( y \)[/tex]

We substitute [tex]\( x = 1 \)[/tex] back into the first equation to solve for [tex]\( y \)[/tex]:

[tex]\[ 5(1) + 2y = 3 \][/tex]
[tex]\[ 5 + 2y = 3 \][/tex]
[tex]\[ 2y = 3 - 5 \][/tex]
[tex]\[ 2y = -2 \][/tex]
[tex]\[ y = \frac{-2}{2} \][/tex]
[tex]\[ y = -1 \][/tex]

### Conclusion

Thus, the solution to the system of equations is
[tex]\[ (x, y) = (1, -1) \][/tex]

Among the given options, the correct answer is:
[tex]\[ (1, -1) \][/tex]
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