Complete the tables below. Note the inclusion of decimal and fractional values as per senior phase requirements. Using the skill of algebraic substitution, calculate the output values for the following linear equations.

• For $x = -2$: $y = \frac{1}{2}(-2) + 2 = -1 + 2 = 1$
• For $x = -1$: $y = \frac{1}{2}(-1) + 2 = -0.5 + 2 = 1.5$
• For $x = 0$: $y = \frac{1}{2}(0) + 2 = 0 + 2 = 2$
• For $x = 0.5$: $y = \frac{1}{2}(0.5) + 2 = 0.25 + 2 = 2.25$
• For $x = 2$: $y = \frac{1}{2}(2) + 2 = 1 + 2 = 3$

Show your substitution calculations for Equation 1 when:

Equation 2: $y = 2x - 4$

• For $x = -2$: $y = 2(-2) - 4 = -4 - 4 = -8$
• For $x = -1$: $y = 2(-1) - 4 = -2 - 4 = -6$
• For $x = 0$: $y = 2(0) - 4 = 0 - 4 = -4$
• For $x = 0.5$: $y = 2(0.5) - 4 = 1 - 4 = -3$
• For $x = 2$: $y = 2(2) - 4 = 4 - 4 = 0$