Solve the system of linear equations below.

[tex]\[
\begin{array}{l}
x+y=4 \\
2x+3y=0
\end{array}
\][/tex]

A. [tex][tex]$x=-6, y=2$[/tex][/tex]
B. [tex][tex]$x=-1, y=5$[/tex][/tex]
C. [tex][tex]$x=\frac{11}{5}, y=\frac{9}{5}$[/tex][/tex]
D. [tex][tex]$x=12, y=-8$[/tex][/tex]



Answer :

To solve the system of linear equations:
[tex]\[ \begin{cases} x + y = 4 \\ 2x + 3y = 0 \end{cases} \][/tex]

we will follow a step-by-step approach.

### Step 1: Express one variable in terms of the other.
From the first equation, [tex]\(x + y = 4\)[/tex], we solve for [tex]\(x\)[/tex]:
[tex]\[ x = 4 - y \][/tex]

### Step 2: Substitute this expression into the second equation.
We substitute [tex]\(x = 4 - y\)[/tex] into the second equation, [tex]\(2x + 3y = 0\)[/tex]:
[tex]\[ 2(4 - y) + 3y = 0 \][/tex]

### Step 3: Simplify the equation and solve for [tex]\(y\)[/tex].
First, distribute the 2:
[tex]\[ 8 - 2y + 3y = 0 \][/tex]

Combine like terms:
[tex]\[ 8 + y = 0 \][/tex]

Solve for [tex]\(y\)[/tex]:
[tex]\[ y = -8 \][/tex]

### Step 4: Substitute the value of [tex]\(y\)[/tex] back into the expression for [tex]\(x\)[/tex].
Now, substitute [tex]\(y = -8\)[/tex] back into the equation [tex]\(x = 4 - y\)[/tex]:

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

### Conclusion
The solution to the system of equations is [tex]\(x = 12\)[/tex] and [tex]\(y = -8\)[/tex].

Thus, the correct answer is:
[tex]\[ \boxed{D. \, x = 12, \, y = -8} \][/tex]

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