Solve the system by substitution. Check your solution.

[tex]\[
\begin{array}{l}
0.5x + 0.25y = 36 \\
y + 18 = 16x
\end{array}
\][/tex]

A. [tex]\((36, 72)\)[/tex]

B. [tex]\((9, 126)\)[/tex]

C. [tex]\((49, 81)\)[/tex]

D. [tex]\((21, 9)\)[/tex]



Answer :

To solve the system of equations by substitution, we will start by isolating one variable in one of the equations and then substituting it in the other equation.

Given the system:
[tex]\[ \begin{array}{l} 0.5x + 0.25y = 36 \\ y + 18 = 16x \end{array} \][/tex]

1. Isolate [tex]\( y \)[/tex] in the second equation:

[tex]\[ y + 18 = 16x \][/tex]

Subtract 18 from both sides:

[tex]\[ y = 16x - 18 \][/tex]

2. Substitute [tex]\( y = 16x - 18 \)[/tex] into the first equation:

[tex]\[ 0.5x + 0.25(16x - 18) = 36 \][/tex]

Distribute [tex]\( 0.25 \)[/tex] in the equation:

[tex]\[ 0.5x + 4x - 4.5 = 36 \][/tex]

Combine like terms:

[tex]\[ 4.5x - 4.5 = 36 \][/tex]

Add 4.5 to both sides:

[tex]\[ 4.5x = 40.5 \][/tex]

Solve for [tex]\( x \)[/tex]:

[tex]\[ x = \frac{40.5}{4.5} = 9 \][/tex]

3. Use [tex]\( x = 9 \)[/tex] to find [tex]\( y \)[/tex] using the isolated expression [tex]\( y = 16x - 18 \)[/tex]:

[tex]\[ y = 16(9) - 18 \][/tex]

Calculate the value:

[tex]\[ y = 144 - 18 = 126 \][/tex]

So, the solution of the system is [tex]\((x, y) = (9, 126)\)[/tex].

4. Check the solution:

Substitute [tex]\( x = 9 \)[/tex] and [tex]\( y = 126 \)[/tex] back into the original equations to verify:

First equation:
[tex]\[ 0.5(9) + 0.25(126) = 4.5 + 31.5 = 36 \][/tex]
True.

Second equation:
[tex]\[ 126 + 18 = 16(9) \\ 144 = 144 \][/tex]
True.

Since both equations are satisfied, the solution [tex]\((9, 126)\)[/tex] is correct.

So, the correct answer is [tex]\((9, 126)\)[/tex], which corresponds to option [tex]\((b)\)[/tex].