Answer :
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].
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].
