1. **Problem:** Solve the system of linear equations using the substitution method and check the answer. --- **A.** Given: $$y = x + 3$$ $$y = 2x + 4$$ Step 1: Set the two expressions for $y$ equal: $$x + 3 = 2x + 4$$ Step 2: Solve for $x$: $$x + 3 = 2x + 4$$ $$\cancel{x} + 3 = \cancel{x} + x + 4$$ $$3 = x + 4$$ $$x = 3 - 4 = -1$$ Step 3: Substitute $x = -1$ into $y = x + 3$: $$y = -1 + 3 = 2$$ Step 4: Check in second equation: $$y = 2x + 4 = 2(-1) + 4 = -2 + 4 = 2$$ **Solution:** $x = -1$, $y = 2$ --- **B.** Given: $$2x + 2y = 2$$ $$-4x + 4y = 12$$ Step 1: Solve first equation for $y$: $$2x + 2y = 2$$ $$2y = 2 - 2x$$ $$y = \frac{2 - 2x}{2} = 1 - x$$ Step 2: Substitute into second equation: $$-4x + 4(1 - x) = 12$$ $$-4x + 4 - 4x = 12$$ $$-8x + 4 = 12$$ $$-8x = 12 - 4 = 8$$ $$x = \frac{8}{-8} = -1$$ Step 3: Substitute $x = -1$ into $y = 1 - x$: $$y = 1 - (-1) = 1 + 1 = 2$$ Step 4: Check in first equation: $$2(-1) + 2(2) = -2 + 4 = 2$$ **Solution:** $x = -1$, $y = 2$ --- **C.** Given: $$\frac{1}{2}x + \frac{1}{3}y = 6$$ $$x - y - 2 = 0$$ Step 1: Solve second equation for $x$: $$x = y + 2$$ Step 2: Substitute into first equation: $$\frac{1}{2}(y + 2) + \frac{1}{3}y = 6$$ $$\frac{1}{2}y + 1 + \frac{1}{3}y = 6$$ $$\left(\frac{1}{2} + \frac{1}{3}\right)y + 1 = 6$$ $$\frac{3}{6} + \frac{2}{6} = \frac{5}{6}$$ $$\frac{5}{6}y + 1 = 6$$ $$\frac{5}{6}y = 5$$ $$y = 5 \times \frac{6}{5} = 6$$ Step 3: Substitute $y = 6$ into $x = y + 2$: $$x = 6 + 2 = 8$$ Step 4: Check in first equation: $$\frac{1}{2}(8) + \frac{1}{3}(6) = 4 + 2 = 6$$ **Solution:** $x = 8$, $y = 6$ --- **D.** Given: $$3m - n = 7$$ $$2m + 3n = 1$$ Step 1: Solve first equation for $n$: $$n = 3m - 7$$ Step 2: Substitute into second equation: $$2m + 3(3m - 7) = 1$$ $$2m + 9m - 21 = 1$$ $$11m - 21 = 1$$ $$11m = 22$$ $$m = 2$$ Step 3: Substitute $m = 2$ into $n = 3m - 7$: $$n = 3(2) - 7 = 6 - 7 = -1$$ Step 4: Check in first equation: $$3(2) - (-1) = 6 + 1 = 7$$ **Solution:** $m = 2$, $n = -1$ --- **E.** Given: $$p - q = 4$$ $$p + 3q = 12$$ Step 1: Solve first equation for $p$: $$p = q + 4$$ Step 2: Substitute into second equation: $$q + 4 + 3q = 12$$ $$4q + 4 = 12$$ $$4q = 8$$ $$q = 2$$ Step 3: Substitute $q = 2$ into $p = q + 4$: $$p = 2 + 4 = 6$$ Step 4: Check in first equation: $$6 - 2 = 4$$ **Solution:** $p = 6$, $q = 2$ --- **F.** Given: $$\frac{2}{5}u + \frac{3}{2}v = 2$$ $$\frac{7}{3}u - \frac{5}{4}v = -5$$ Step 1: Solve first equation for $u$: $$\frac{2}{5}u = 2 - \frac{3}{2}v$$ $$u = \frac{5}{2} \left(2 - \frac{3}{2}v\right) = \frac{5}{2} \times 2 - \frac{5}{2} \times \frac{3}{2}v = 5 - \frac{15}{4}v$$ Step 2: Substitute into second equation: $$\frac{7}{3}(5 - \frac{15}{4}v) - \frac{5}{4}v = -5$$ $$\frac{35}{3} - \frac{7}{3} \times \frac{15}{4}v - \frac{5}{4}v = -5$$ $$\frac{35}{3} - \frac{105}{12}v - \frac{5}{4}v = -5$$ Step 3: Find common denominator for $v$ terms (12): $$\frac{35}{3} - \frac{105}{12}v - \frac{15}{12}v = -5$$ $$\frac{35}{3} - \frac{120}{12}v = -5$$ $$\frac{35}{3} - 10v = -5$$ Step 4: Solve for $v$: $$-10v = -5 - \frac{35}{3} = -\frac{15}{3} - \frac{35}{3} = -\frac{50}{3}$$ $$v = \frac{50}{3} \times \frac{1}{10} = \frac{50}{30} = \frac{5}{3}$$ Step 5: Substitute $v = \frac{5}{3}$ into $u = 5 - \frac{15}{4}v$: $$u = 5 - \frac{15}{4} \times \frac{5}{3} = 5 - \frac{75}{12} = 5 - \frac{25}{4} = \frac{20}{4} - \frac{25}{4} = -\frac{5}{4}$$ Step 6: Check in first equation: $$\frac{2}{5} \times -\frac{5}{4} + \frac{3}{2} \times \frac{5}{3} = -\frac{2}{4} + \frac{15}{6} = -\frac{1}{2} + \frac{5}{2} = 2$$ **Solution:** $u = -\frac{5}{4}$, $v = \frac{5}{3}$ --- **Final answers:** A: $x = -1$, $y = 2$ B: $x = -1$, $y = 2$ C: $x = 8$, $y = 6$ D: $m = 2$, $n = -1$ E: $p = 6$, $q = 2$ F: $u = -\frac{5}{4}$, $v = \frac{5}{3}$