Sure, let's solve the given system B step-by-step:
[tex]\[
\begin{aligned}
1) & \quad x + 6y = 5 \\
2) & \quad -2y = -50
\end{aligned}
\][/tex]
First, isolate [tex]\( y \)[/tex] in the second equation:
[tex]\[
-2y = -50
\][/tex]
Divide both sides by -2:
[tex]\[
y = \frac{-50}{-2} \Rightarrow y = 25
\][/tex]
Now that we have [tex]\( y = 25 \)[/tex], substitute this value into the first equation [tex]\( x + 6y = 5 \)[/tex]:
[tex]\[
x + 6(25) = 5
\][/tex]
Simplify inside the parenthesis:
[tex]\[
x + 150 = 5
\][/tex]
Next, isolate [tex]\( x \)[/tex] by subtracting 150 from both sides of the equation:
[tex]\[
x + 150 - 150 = 5 - 150 \Rightarrow x = -145
\][/tex]
So, the solution to the system B is:
[tex]\[
x = -145 \quad \text{and} \quad y = 25
\][/tex]
Thus, the solution is [tex]\((-145, 25)\)[/tex].
