Answer :
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].
[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].