Copy and complete the table of values for the relations y1 = 2x + 5 and y2 = 3 - 2x for x from -4 to 3.

Step 1: Complete the table of values for $y_1 = 2x + 5$. We substitute each $x$ value into the function $y_1 = 2x + 5$.

For $x = -3$: $y_1 = 2(-3) + 5 = -6 + 5 = \textbf{-1}$ For $x = -2$: $y_1 = 2(-2) + 5 = -4 + 5 = \textbf{1}$ For $x = 0$: $y_1 = 2(0) + 5 = 0 + 5 = \textbf{5}$ For $x = 2$: $y_1 = 2(2) + 5 = 4 + 5 = \textbf{9}$ For $x = 3$: $y_1 = 2(3) + 5 = 6 + 5 = \textbf{11}$

Step 2: Complete the table of values for $y_2 = 3 - 2x$. We substitute each $x$ value into the function $y_2 = 3 - 2x$.

For $x = -2$: $y_2 = 3 - 2(-2) = 3 + 4 = \textbf{7}$ For $x = 0$: $y_2 = 3 - 2(0) = 3 - 0 = \textbf{3}$ For $x = 1$: $y_2 = 3 - 2(1) = 3 - 2 = \textbf{1}$ For $x = 2$: $y_2 = 3 - 2(2) = 3 - 4 = \textbf{-1}$ For $x = 3$: $y_2 = 3 - 2(3) = 3 - 6 = \textbf{-3}$ Note: The value for $y_2$ at $x=3$ in the original table is given as 11, which contradicts the relation $y_2 = 3 - 2x$. The calculated value is -3.

The completed table is: | X | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | |-----|----|----|----|----|---|---|----|----| | $y_1 = 2x + 5$ | -3 | -1 | 1 | 3 | 5 | 7 | 9 | 11 | | $y_2 = 3 - 2x$ | 11 | 9 | 7 | 5 | 3 | 1 | -1 | -3 |