Sure, let's solve the system of equations step-by-step using the elimination method.
The system of equations is:
[tex]\[
\begin{array}{l}
15 q - 4 r = 62 \quad (1) \\
5 q + 8 r = 86 \quad (2)
\end{array}
\][/tex]
First, we need to eliminate one of the variables. To do this, we will align the coefficients of either [tex]\( q \)[/tex] or [tex]\( r \)[/tex]. Let’s align the coefficients of [tex]\( r \)[/tex] by making the coefficients in both equations equal but opposite in sign.
Notice that the second equation already has an [tex]\( 8r \)[/tex]. If we multiply the first equation by 2, we will get [tex]\(-8r\)[/tex] in the first equation:
[tex]\[
2 \times (15q - 4r) = 2 \times 62
\][/tex]
which simplifies to:
[tex]\[
30q - 8r = 124 \quad (3)
\][/tex]
Now the system of equations becomes:
[tex]\[
\begin{array}{l}
30 q - 8 r = 124 \quad (3) \\
5 q + 8 r = 86 \quad (2)
\end{array}
\][/tex]
Next, we add these two equations to eliminate [tex]\( r \)[/tex]:
[tex]\[
(30 q - 8 r) + (5 q + 8 r) = 124 + 86
\][/tex]
This simplifies to:
[tex]\[
35 q = 210
\][/tex]
Solve for [tex]\( q \)[/tex]:
[tex]\[
q = \frac{210}{35} = 6
\][/tex]
Now that we have [tex]\( q = 6 \)[/tex], we will substitute this value back into one of the original equations to find [tex]\( r \)[/tex]. Using the second equation:
[tex]\[
5 q + 8 r = 86
\][/tex]
Substitute [tex]\( q = 6 \)[/tex] into the equation:
[tex]\[
5(6) + 8 r = 86
\][/tex]
This simplifies to:
[tex]\[
30 + 8 r = 86
\][/tex]
Solve for [tex]\( r \)[/tex]:
[tex]\[
8 r = 86 - 30
\][/tex]
[tex]\[
8 r = 56
\][/tex]
[tex]\[
r = \frac{56}{8} = 7
\][/tex]
Thus, the solution to the system of equations is:
[tex]\[
q = 6, \quad r = 7
\][/tex]
So the correct answer is [tex]\( q = 6, r = 7 \)[/tex].
