[tex]\[
\begin{array}{l}
2x + 4y = 12 \\
y = \frac{1}{4}x - 3
\end{array}
\][/tex]

What is the solution to the system of equations?

A. [tex]\((-1, 8)\)[/tex]

B. [tex]\((8, -1)\)[/tex]

C. [tex]\(\left(5, \frac{1}{2}\right)\)[/tex]

D. [tex]\(\left(\frac{1}{2}, 5\right)\)[/tex]



Answer :

To find the solution to the system of equations:

[tex]\[ \begin{cases} 2x + 4y = 12 \\ y = \frac{1}{4}x - 3 \end{cases} \][/tex]

we need to use substitution or elimination methods. Here, we will use substitution because the second equation is already solved for [tex]\(y\)[/tex].

First, substitute [tex]\(y = \frac{1}{4}x - 3\)[/tex] into the first equation:

[tex]\[ 2x + 4\left(\frac{1}{4}x - 3\right) = 12 \][/tex]

Simplify the expression inside the parentheses:

[tex]\[ 2x + 4\left(\frac{1}{4}x\right) - 4 \cdot 3 = 12 \][/tex]

This becomes:

[tex]\[ 2x + x - 12 = 12 \][/tex]

Combine like terms:

[tex]\[ 3x - 12 = 12 \][/tex]

Add 12 to both sides to isolate the term with [tex]\(x\)[/tex]:

[tex]\[ 3x = 24 \][/tex]

Divide both sides by 3:

[tex]\[ x = 8 \][/tex]

Now that we have [tex]\(x\)[/tex], substitute [tex]\(x = 8\)[/tex] back into the second equation, [tex]\(y = \frac{1}{4}x - 3\)[/tex]:

[tex]\[ y = \frac{1}{4}(8) - 3 \][/tex]

Simplify the right side:

[tex]\[ y = 2 - 3 \][/tex]

Thus, we get:

[tex]\[ y = -1 \][/tex]

So the solution to the system is [tex]\( (8, -1) \)[/tex].

The correct answer is:
[tex]\[ (8, -1) \][/tex]