Solving Systems by Substitution Solve each system by substitution. 1) \[ \begin{array}{l} x=-2 y \\ x-y=9 \end{array} \] 3) \[ \begin{array}{l} -x-3 y=-18 \\ y=-4 x+6 \end{array} \] 5) \[ \begin{array}{l} y=-6 \\ 3 x-6 y=30 \end{array} \] 2) \[ \begin{array}{l} y=x-4 \\ -4 x-6 y=-16 \end{array} \] 4) \[ \begin{aligned} -2 x+2 y & =6 \\ 4 x+2 y & =-5 \end{aligned} \] G) \[ \begin{aligned} -2 x-y & =-5 \\ x-8 y & =-23 \end{aligned} \]

**1)** Given the system: \[ \begin{aligned} x &= -2y, \\ x - y &= 9. \end{aligned} \] Step 1: Substitute \(x = -2y\) into the second equation: \[ -2y - y = 9. \] Step 2: Combine like terms: \[ -3y = 9. \] Step 3: Solve for \(y\): \[ y = \frac{9}{-3} = -3. \] Step 4: Substitute \(y = -3\) into \(x = -2y\): \[ x = -2(-3) = 6. \] Solution: \(\boxed{(6,\,-3)}\). --- **3)** Given the system: \[ \begin{aligned} -x - 3y &= -18, \\ y &= -4x + 6. \end{aligned} \] Step 1: Substitute \(y = -4x + 6\) into the first equation: \[ -x - 3(-4x + 6) = -18. \] Step 2: Distribute \(-3\): \[ -x + 12x - 18 = -18. \] Step 3: Combine like terms: \[ 11x - 18 = -18. \] Step 4: Solve for \(x\): \[ 11x = 0 \quad \Rightarrow \quad x = 0. \] Step 5: Substitute \(x = 0\) into \(y = -4x + 6\): \[ y = -4(0) + 6 = 6. \] Solution: \(\boxed{(0,\,6)}\). --- **5)** Given the system: \[ \begin{aligned} y &= -6, \\ 3x - 6y &= 30. \end{aligned} \] Step 1: Substitute \(y = -6\) into the second equation: \[ 3x - 6(-6) = 30. \] Step 2: Simplify: \[ 3x + 36 = 30. \] Step 3: Solve for \(x\): \[ 3x = 30 - 36 = -6 \quad \Rightarrow \quad x = \frac{-6}{3} = -2. \] Solution: \(\boxed{(-2,\,-6)}\). --- **2)** Given the system: \[ \begin{aligned} y &= x - 4, \\ -4x - 6y &= -16. \end{aligned} \] Step 1: Substitute \(y = x - 4\) into the second equation: \[ -4x - 6(x - 4) = -16. \] Step 2: Distribute and simplify: \[ -4x - 6x + 24 = -16 \quad \Rightarrow \quad -10x + 24 = -16. \] Step 3: Solve for \(x\): \[ -10x = -16 - 24 = -40 \quad \Rightarrow \quad x = \frac{-40}{-10} = 4. \] Step 4: Substitute \(x = 4\) into \(y = x - 4\): \[ y = 4 - 4 = 0. \] Solution: \(\boxed{(4,\,0)}\). --- **4)** Given the system: \[ \begin{aligned} -2x + 2y &= 6, \\ 4x + 2y &= -5. \end{aligned} \] Step 1: Subtract the first equation from the second to eliminate \(y\): \[ (4x + 2y) - (-2x + 2y) = -5 - 6. \] Step 2: Simplify: \[ 4x + 2y + 2x - 2y = -11 \quad \Rightarrow \quad 6x = -11. \] Step 3: Solve for \(x\): \[ x = -\frac{11}{6}. \] Step 4: Substitute \(x = -\frac{11}{6}\) into the first equation: \[ -2\left(-\frac{11}{6}\right) + 2y = 6. \] Step 5: Simplify: \[ \frac{22}{6} + 2y = 6 \quad \Rightarrow \quad \frac{11}{3} + 2y = 6. \] Step 6: Solve for \(y\): \[ 2y = 6 - \frac{11}{3} = \frac{18}{3} - \frac{11}{3} = \frac{7}{3} \quad \Rightarrow \quad y = \frac{7}{6}. \] Solution: \(\boxed{\left(-\frac{11}{6},\,\frac{7}{6}\right)}\). --- **G)** Given the system: \[ \begin{aligned} -2x - y &= -5, \\ x - 8y &= -23. \end{aligned} \] Step 1: Solve the first equation for \(y\). Begin by rewriting: \[ -2x - y = -5 \quad \Rightarrow \quad y = -2x + 5. \] Step 2: Substitute \(y = 5 - 2x\) into the second equation: \[ x - 8(5 - 2x) = -23. \] Step 3: Distribute \(-8\): \[ x - 40 + 16x = -23. \] Step 4: Combine like terms: \[ 17x - 40 = -23. \] Step 5: Solve for \(x\): \[ 17x = -23 + 40 = 17 \quad \Rightarrow \quad x = 1. \] Step 6: Substitute \(x = 1\) back into \(y = 5 - 2x\): \[ y = 5 - 2(1) = 3. \] Solution: \(\boxed{(1,\,3)}\).