1. **Problem 5:** Determine which set of ordered pairs satisfies the condition that the second number is 2 more than twice the first number. 2. The formula for this condition is: $$y = 2x + 2$$ where $x$ is the first number and $y$ is the second number in each ordered pair $(x,y)$. 3. Check each ordered pair in each set to see if $y = 2x + 2$ holds. **Set A:** (0,2), (1,3), (2,4), (3,5), (4,6), (5,7) - For $x=0$, $y$ should be $2(0)+2=2$ ✓ - For $x=1$, $y$ should be $2(1)+2=4$, but given $3$ ✗ Since one pair fails, Set A does not satisfy the condition. **Set B:** (-3,8), (-2,6), (-1,4), (0,2), (1,4), (2,6) - For $x=-3$, $y=2(-3)+2=-6+2=-4$, given 8 ✗ Set B fails immediately. **Set C:** (0,2), (1,4), (2,6), (3,8), (4,10), (5,12) - For $x=0$, $y=2(0)+2=2$ ✓ - For $x=1$, $y=2(1)+2=4$ ✓ - For $x=2$, $y=2(2)+2=6$ ✓ - For $x=3$, $y=2(3)+2=8$ ✓ - For $x=4$, $y=2(4)+2=10$ ✓ - For $x=5$, $y=2(5)+2=12$ ✓ All pairs satisfy the condition, so Set C is correct. **Set D:** (-2,-4), (-1,-2), (0,0), (1,2), (2,4) - For $x=-2$, $y=2(-2)+2=-4+2=-2$, given -4 ✗ Set D fails. **Answer for Problem 5:** Set C. 4. **Problem 8:** Given two points $(-2,c)$ and $(16,12)$ on a line with slope $\frac{1}{6}$, find $c$. 5. The slope formula is: $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ Given $m=\frac{1}{6}$, $x_1=-2$, $y_1=c$, $x_2=16$, $y_2=12$. 6. Substitute values: $$\frac{1}{6} = \frac{12 - c}{16 - (-2)} = \frac{12 - c}{18}$$ 7. Multiply both sides by 18: $$18 \times \frac{1}{6} = 12 - c$$ $$\cancel{18} \times \frac{1}{\cancel{6}} = 12 - c$$ $$3 = 12 - c$$ 8. Solve for $c$: $$c = 12 - 3 = 9$$ **Answer for Problem 8:** $c = 9$ (Option C).