To solve the system of equations [tex]\( 2y - x = 8 \)[/tex] and [tex]\( y - 2x = -5 \)[/tex], we follow these steps:
1. Original System:
[tex]\[
\begin{cases}
2y - x = 8 \\
y - 2x = -5 \\
\end{cases}
\][/tex]
2. Equivalent System:
The first equation remains the same:
[tex]\[
2y - x = 8
\][/tex]
Multiply the second equation by 2 to facilitate elimination:
[tex]\[
2(y - 2x) = 2(-5) \\
-2y + 4x = -10
\][/tex]
Therefore, the equivalent system is:
[tex]\[
\begin{cases}
2y - x = 8 \\
-2y + 4x = -10 \\
\end{cases}
\][/tex]
3. Sum of Equations in Equivalent System:
Add the two equations:
[tex]\[
(2y - x) + (-2y + 4x) = 8 + (-10) \\
2y - x - 2y + 4x = -2 \\
3x = -2
\][/tex]
4. Solution to System:
Solve for [tex]\(x\)[/tex]:
[tex]\[
x = \frac{-2}{3}
\][/tex]
5. Using [tex]\(x\)[/tex] to Find [tex]\(y\)[/tex]:
Substitute [tex]\( x = 6 \)[/tex] into the first equation to find [tex]\( y \)[/tex]:
[tex]\[
2y - 6 = 8 \\
2y = 14 \\
y = 7
\][/tex]
So, the solution to the system is [tex]\( \boxed{(6, 7)} \)[/tex].
6. New System Using Sum:
Using the sum [tex]\( 3x = 18 \)[/tex] to find [tex]\(x\)[/tex]:
[tex]\[
x = \frac{18}{3} = 6
\][/tex]
Substitute [tex]\( x = 6 \)[/tex] into the first equation:
[tex]\[
2y - 6 = 8 \\
2y = 14 \\
y = 7
\][/tex]
Thus, the solution to the new system is also [tex]\( \boxed{(6, 7)} \)[/tex].
Therefore, the solution that can be used to fill in both blanks in the table is [tex]\((6, 7)\)[/tex].
