Question

The solution to the system of equations shown is (2, 0).

3x − 2y = 6
x + 4y = 2

When the first equation is multiplied by 2, the sum of the two equations is equivalent to 7x = 14
.

Which system of equations will also have a solution of (2, 0)?

6x - 4y = 6
x + 4y = 2
6x − 4y = 6
2x + 8y = 2
x + 4y = 2
7x = 14
6x − 4y = 6
7x = 14

Answer 1

To find systems of equations that also have the solution (2, 0), we can start by substituting \(x = 2\) and \(y = 0\) into each equation in the systems you provided.

1. **First system:**
- \(6x - 4y = 6\)
- \(x + 4y = 2\)

Substitute (2, 0) into the first equation:
\[
6(2) - 4(0) = 12 - 0 = 12 \quad (\text{not true, }\text{so this system does not have the solution.})
\]

Now for the second equation:
\[
2 + 0 = 2 \quad (\text{true, so the second equation is satisfied, but the first isn't, so this system does not work overall.})
\]

2. **Second system:**
- \(6x - 4y = 6\)
- \(2x + 8y = 2\)

Substitute (2, 0) into the first equation:
\[
6(2) - 4(0) = 12 - 0 = 12 \quad (\text{not true, so this system does not have the solution.})
\]

Now for the second equation:
\[
2(2) + 8(0) = 4 + 0 = 4 \quad (\text{not true, so this system does not work overall.})
\]

3. **Third system:**
- \(x + 4y = 2\)
- \(7x = 14\)

Substitute (2, 0) into the first equation:
\[
2 + 0 = 2 \quad (\text{true})
\]

Now for the second equation:
\[
7(2) = 14 \quad (\text{true})
\]

Thus, this system has the solution (2, 0).

4. **Fourth system:**
- \(6x - 4y = 6\)
- \(7x = 14\)

Substitute (2, 0) into the first equation:
\[
6(2) - 4(0) = 12 - 0 = 12 \quad (\text{not true, so this system does not have the solution.})
\]

Now for the second equation:
\[
7(2) = 14 \quad (\text{true, but we need the first equation to be true as well, so this system does not work overall.})
\]

So, the only system provided that has the solution (2, 0) is:

- \(x + 4y = 2\)
- \(7x = 14\)