Solve the system using the addition method:

[tex]\[ \begin{cases}
4x + 3y = -18 \\
x + 9y = -21
\end{cases} \][/tex]

A. One solution: [tex]$\square$[/tex]

B. No solution

C. Infinite number of solutions



Answer :

Sure! Let's solve the system of equations using the addition (elimination) method step by step.

We have the following system of equations:
[tex]\[ \begin{cases} 4x + 3y = -18 \quad \text{(1)} \\ x + 9y = -21 \quad \text{(2)} \end{cases} \][/tex]

Step 1: Make the coefficients of one of the variables equal in both equations.

To eliminate [tex]\(x\)[/tex], we need the coefficients of [tex]\(x\)[/tex] to be equal. We can do this by multiplying the second equation by 4.

[tex]\[ 4(x + 9y = -21) \implies 4x + 36y = -84 \quad \text{(3)} \][/tex]

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

Now we have:
[tex]\[ \begin{cases} 4x + 3y = -18 \quad \text{(1)} \\ 4x + 36y = -84 \quad \text{(3)} \end{cases} \][/tex]

Subtract equation (1) from equation (3):
[tex]\[ (4x + 36y) - (4x + 3y) = -84 - (-18) \][/tex]

Simplify the left-hand side:
[tex]\[ 4x + 36y - 4x - 3y = -84 + 18 \][/tex]
[tex]\[ 33y = -66 \][/tex]

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

Divide both sides by 33:
[tex]\[ y = \frac{-66}{33} \][/tex]
[tex]\[ y = -2 \][/tex]

Step 4: Substitute [tex]\(y = -2\)[/tex] back into one of the original equations to solve for [tex]\(x\)[/tex].

Using the second equation [tex]\(x + 9y = -21\)[/tex]:
[tex]\[ x + 9(-2) = -21 \][/tex]
[tex]\[ x - 18 = -21 \][/tex]
[tex]\[ x = -21 + 18 \][/tex]
[tex]\[ x = -3 \][/tex]

Step 5: Write the solution.

The solution to the system is:
[tex]\[ (x, y) = (-3, -2) \][/tex]

So, the solution is:
[tex]\[ \boxed{(-3, -2)} \][/tex]