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.
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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: \[ 2 + 0 = 2 \quad (\text{true, so the second equation is satisfied, but the first isn't, so this system does not work overall.}) \]
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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: \[ 2(2) + 8(0) = 4 + 0 = 4 \quad (\text{not true, so this system does not work overall.}) \]
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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).
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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:
- \(x + 4y = 2\)
- \(7x = 14\)