Question

the table can be used to determine the solution to the system of equations, $2y - x = 8$, and $y - 2x = -5$

original system	equivalent system	sum of equations in equivalent system	solution to system	new system using sum	solution to new system

which solution can be used to fill in both blanks in the table?

Explanation:

Step1: Solve \(3x = 18\) for \(x\)

To find the value of \(x\), we divide both sides of the equation \(3x = 18\) by 3.
\(x=\frac{18}{3}=6\)

Step2: Substitute \(x = 6\) into \(2y - x = 8\) to find \(y\)

Substitute \(x = 6\) into the equation \(2y - x = 8\), we get \(2y-6 = 8\).
First, add 6 to both sides: \(2y=8 + 6=14\).
Then, divide both sides by 2: \(y=\frac{14}{2}=7\)? Wait, no, wait. Wait, let's check the equivalent system. Wait, the equivalent system is \(2y - x=8\) and \(- 2y + 4x=10\), but when we sum them we get \(3x = 18\). But also, the new system using sum is \(2y - x = 8\) and \(3x = 18\). So first, solve \(3x=18\) gives \(x = 6\). Then substitute \(x = 6\) into \(2y-x=8\): \(2y-6=8\), \(2y=8 + 6=14\), \(y = 7\)? Wait, but let's check with the original second equation \(y-2x=-5\). If \(x = 6\) and \(y = 7\), then \(7-2\times6=7 - 12=-5\), which matches. Wait, but maybe I made a mistake earlier. Wait, the solution to the system (the equivalent system or the new system) should satisfy both equations. Let's re - do:

From \(3x = 18\), \(x = 6\). Then plug \(x = 6\) into \(2y-x=8\): \(2y-6 = 8\), \(2y=14\), \(y = 7\). Let's check with the original equation \(y - 2x=-5\): \(7-12=-5\), which is correct. Wait, but maybe the blanks are for the solution of the system (the equivalent system or the new system). The equivalent system is \(2y - x = 8\) and \(-2y + 4x=10\), and the new system using sum is \(2y - x = 8\) and \(3x = 18\). The solution to both systems (equivalent system, new system) should be \((x,y)=(6,7)\)? Wait, no, wait the sum of the equivalent system is \(3x = 18\), so first we solve \(3x = 18\) to get \(x = 6\), then substitute \(x = 6\) into one of the equations (like \(2y - x = 8\)) to get \(y\).

Wait, maybe the question is asking for the solution of the system (the one that fills both blanks, which are "Solution to System" (equivalent system) and "Solution to New System"). Since both systems (equivalent and new) should have the same solution as the original system. Let's solve the original system:

Original system:
\(\begin{cases}2y-x = 8\\y-2x=-5\end{cases}\)

From the second equation, we can express \(y=2x - 5\). Substitute into the first equation: \(2(2x - 5)-x=8\), \(4x-10 - x=8\), \(3x=18\), \(x = 6\), then \(y=2\times6-5 = 7\). So the solution is \((6,7)\)? Wait, but maybe I misread the table. The columns are:

- Original System: \(2y - x = 8\), \(y - 2x=-5\)
- Equivalent System: \(2y - x = 8\), \(-2y + 4x=10\)
- Sum of Equations in Equivalent System: \(3x = 18\)
- Solution to System: (blank)
- New System Using Sum: \(2y - x = 8\), \(3x = 18\)
- Solution to New System: (blank)

So the "Solution to System" (equivalent system) and "Solution to New System" should be the same as the solution to the original system, which is \((x,y)=(6,7)\)? Wait, no, wait. Wait, the equivalent system is obtained by multiplying the second original equation \(y - 2x=-5\) by \(- 2\): \( - 2y+4x = 10\). So the equivalent system is \(2y - x = 8\) and \(-2y + 4x = 10\). When we solve this equivalent system, we can add the two equations to get \(3x = 18\), so \(x = 6\), then substitute \(x = 6\) into \(2y - x = 8\) to get \(y = 7\). The new system using sum is \(2y - x = 8\) and \(3x = 18\). Solving this new system, we also get \(x = 6\) from \(3x = 18\), then \(y = 7\) from \(2y - x = 8\). So the solution to fill in both blanks is \((6,7)\)? Wait, but let's check again.

Wait, maybe I made a mistake in the calculation of \(y\). Let's re - calculate:

From \(2y-x=8\) and \(x = 6\):

\(2y-6 = 8\)

Add 6 to both sides: \(2y=8 + 6=14\)

Divide by 2: \(y = 7\)

And check with \(y-2x=-5\):

\(7-2\times6=7 - 12=-5\), which is correct.

Answer:

The solution to fill in both blanks is \((6,7)\) (or in the form \(x = 6,y = 7\))