To find the solution to the system of equations \( y = 3x + 2 \) and \( y = -2x + 12 \), we will first set the two equations equal to each other to find the intersection point (solution).
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Set the equations equal: \[ 3x + 2 = -2x + 12 \]
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Solve for \( x \): \[ 3x + 2x = 12 - 2 \] \[ 5x = 10 \] \[ x = 2 \]
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Substitute \( x \) back into one of the original equations to find \( y \): \[ y = 3(2) + 2 = 6 + 2 = 8 \]
So, the solution to the system of equations is \( (2, 8) \).
Now, let’s complete the table for the given equations for the values of \( x \) provided: -2, -1, 0, 1, and 2.
| x values | \( y = 3x + 2 \) | \( y = -2x + 12 \) | |----------|-------------------|---------------------| | -2 | \( 3(-2) + 2 = -6 + 2 = -4 \) | \( -2(-2) + 12 = 4 + 12 = 16 \) | | -1 | \( 3(-1) + 2 = -3 + 2 = -1 \) | \( -2(-1) + 12 = 2 + 12 = 14 \) | | 0 | \( 3(0) + 2 = 0 + 2 = 2 \) | \( -2(0) + 12 = 0 + 12 = 12 \) | | 1 | \( 3(1) + 2 = 3 + 2 = 5 \) | \( -2(1) + 12 = -2 + 12 = 10 \) | | 2 | \( 3(2) + 2 = 6 + 2 = 8 \) | \( -2(2) + 12 = -4 + 12 = 8 \) |
The ordered pair that is a solution to the system of equations is \( (2, 8) \).