EXERCISE 5 (a) Solve for \( x \) and \( y \) by using the method of substitution: (1) \( x-y=8 \) and \( 2 x+y=10 \) (2) \( 3 x-y=2 \) and \( 7 x-2 y=8 \) (3) \( 3 x+5 y=8 \) and \( x-2 y=-1 \) (4) \( 7 x-3 y=41 \) and \( 3 x-y=17 \) (b) Solve for \( x \) and \( y \) using either the elimination or substitution method: (1) \( x+y=1 \) and \( x-2 y=1 \) (2) \( 3 x+2 y=2 \) and \( 5 x-2 y=-18 \) (3) \( x+4 y=14 \) and \( 3 x+2 y=12 \) (4) \( 2 y-3 x=7 \) and \( 4 y-5 x=21 \) (5) \( 3 x+2 y=6 \) and \( 5 x+3 y=11 \) (6) \( \frac{y-2}{3}-\frac{x-1}{2}=1 \) and \( \frac{y-4}{5}-\frac{x-3}{4}=1 \) (c) At ABC shop, 3 hamburgers and 2 cooldrinks cost R89, whereas 2 hamburgers and 3 cooldrinks cost R73,50. Determine the cost of 1 hamburger. 55

Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}3x-y=2\\7x-2y=8\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}y=-2+3x\\7x-2y=8\end{array}\right.\) - step2: Substitute the value of \(y:\) \(7x-2\left(-2+3x\right)=8\) - step3: Simplify: \(x+4=8\) - step4: Move the constant to the right side: \(x=8-4\) - step5: Subtract the numbers: \(x=4\) - step6: Substitute the value of \(x:\) \(y=-2+3\times 4\) - step7: Calculate: \(y=10\) - step8: Calculate: \(\left\{ \begin{array}{l}x=4\\y=10\end{array}\right.\) - step9: Check the solution: \(\left\{ \begin{array}{l}x=4\\y=10\end{array}\right.\) - step10: Rewrite: \(\left(x,y\right) = \left(4,10\right)\) Solve the system of equations \( 3 h+2 c=89;2 h+3 c=73.5 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}3h+2c=89\\2h+3c=73.5\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}c=\frac{89-3h}{2}\\2h+3c=73.5\end{array}\right.\) - step2: Substitute the value of \(c:\) \(2h+3\times \frac{89-3h}{2}=73.5\) - step3: Simplify: \(2h+\frac{3\left(89-3h\right)}{2}=73.5\) - step4: Convert the expressions: \(2h+\frac{3\left(89-3h\right)}{2}=\frac{147}{2}\) - step5: Multiply both sides of the equation by LCD: \(\left(2h+\frac{3\left(89-3h\right)}{2}\right)\times 2=\frac{147}{2}\times 2\) - step6: Simplify the equation: \(-5h+267=147\) - step7: Move the constant to the right side: \(-5h=147-267\) - step8: Subtract the numbers: \(-5h=-120\) - step9: Change the signs: \(5h=120\) - step10: Divide both sides: \(\frac{5h}{5}=\frac{120}{5}\) - step11: Divide the numbers: \(h=24\) - step12: Substitute the value of \(h:\) \(c=\frac{89-3\times 24}{2}\) - step13: Calculate: \(c=\frac{17}{2}\) - step14: Calculate: \(\left\{ \begin{array}{l}c=\frac{17}{2}\\h=24\end{array}\right.\) - step15: Check the solution: \(\left\{ \begin{array}{l}c=\frac{17}{2}\\h=24\end{array}\right.\) - step16: Rewrite: \(\left(c,h\right) = \left(\frac{17}{2},24\right)\) Solve the system of equations \( x-y=8;2 x+y=10 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x-y=8\\2x+y=10\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=8+y\\2x+y=10\end{array}\right.\) - step2: Substitute the value of \(x:\) \(2\left(8+y\right)+y=10\) - step3: Simplify: \(16+3y=10\) - step4: Move the constant to the right side: \(3y=10-16\) - step5: Subtract the numbers: \(3y=-6\) - step6: Divide both sides: \(\frac{3y}{3}=\frac{-6}{3}\) - step7: Divide the numbers: \(y=-2\) - step8: Substitute the value of \(y:\) \(x=8-2\) - step9: Calculate: \(x=6\) - step10: Calculate: \(\left\{ \begin{array}{l}x=6\\y=-2\end{array}\right.\) - step11: Check the solution: \(\left\{ \begin{array}{l}x=6\\y=-2\end{array}\right.\) - step12: Rewrite: \(\left(x,y\right) = \left(6,-2\right)\) Solve the system of equations \( 3 x+5 y=8;x-2 y=-1 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}3x+5y=8\\x-2y=-1\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}3x+5y=8\\x=-1+2y\end{array}\right.\) - step2: Substitute the value of \(x:\) \(3\left(-1+2y\right)+5y=8\) - step3: Simplify: \(-3+11y=8\) - step4: Move the constant to the right side: \(11y=8+3\) - step5: Add the numbers: \(11y=11\) - step6: Divide both sides: \(\frac{11y}{11}=\frac{11}{11}\) - step7: Divide the numbers: \(y=1\) - step8: Substitute the value of \(y:\) \(x=-1+2\times 1\) - step9: Substitute back: \(x=-1+2\) - step10: Calculate: \(x=1\) - step11: Calculate: \(\left\{ \begin{array}{l}x=1\\y=1\end{array}\right.\) - step12: Check the solution: \(\left\{ \begin{array}{l}x=1\\y=1\end{array}\right.\) - step13: Rewrite: \(\left(x,y\right) = \left(1,1\right)\) Solve the system of equations \( 7 x-3 y=41;3 x-y=17 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}7x-3y=41\\3x-y=17\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}7x-3y=41\\y=-17+3x\end{array}\right.\) - step2: Substitute the value of \(y:\) \(7x-3\left(-17+3x\right)=41\) - step3: Simplify: \(-2x+51=41\) - step4: Move the constant to the right side: \(-2x=41-51\) - step5: Subtract the numbers: \(-2x=-10\) - step6: Change the signs: \(2x=10\) - step7: Divide both sides: \(\frac{2x}{2}=\frac{10}{2}\) - step8: Divide the numbers: \(x=5\) - step9: Substitute the value of \(x:\) \(y=-17+3\times 5\) - step10: Calculate: \(y=-2\) - step11: Calculate: \(\left\{ \begin{array}{l}x=5\\y=-2\end{array}\right.\) - step12: Check the solution: \(\left\{ \begin{array}{l}x=5\\y=-2\end{array}\right.\) - step13: Rewrite: \(\left(x,y\right) = \left(5,-2\right)\) Solve the system of equations \( x+y=1;x-2 y=1 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x+y=1\\x-2y=1\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=1-y\\x-2y=1\end{array}\right.\) - step2: Substitute the value of \(x:\) \(1-y-2y=1\) - step3: Subtract the terms: \(1-3y=1\) - step4: Move the constant to the right side: \(-3y=1-1\) - step5: Subtract the terms: \(-3y=0\) - step6: Change the signs: \(3y=0\) - step7: Rewrite the expression: \(y=0\) - step8: Substitute the value of \(y:\) \(x=1-0\) - step9: Substitute back: \(x=1\) - step10: Calculate: \(\left\{ \begin{array}{l}x=1\\y=0\end{array}\right.\) - step11: Check the solution: \(\left\{ \begin{array}{l}x=1\\y=0\end{array}\right.\) - step12: Rewrite: \(\left(x,y\right) = \left(1,0\right)\) Solve the system of equations \( x+4 y=14;3 x+2 y=12 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x+4y=14\\3x+2y=12\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=14-4y\\3x+2y=12\end{array}\right.\) - step2: Substitute the value of \(x:\) \(3\left(14-4y\right)+2y=12\) - step3: Simplify: \(42-10y=12\) - step4: Move the constant to the right side: \(-10y=12-42\) - step5: Subtract the numbers: \(-10y=-30\) - step6: Change the signs: \(10y=30\) - step7: Divide both sides: \(\frac{10y}{10}=\frac{30}{10}\) - step8: Divide the numbers: \(y=3\) - step9: Substitute the value of \(y:\) \(x=14-4\times 3\) - step10: Calculate: \(x=2\) - step11: Calculate: \(\left\{ \begin{array}{l}x=2\\y=3\end{array}\right.\) - step12: Check the solution: \(\left\{ \begin{array}{l}x=2\\y=3\end{array}\right.\) - step13: Rewrite: \(\left(x,y\right) = \left(2,3\right)\) Solve the system of equations \( 3 x+2 y=2;5 x-2 y=-18 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}3x+2y=2\\5x-2y=-18\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=\frac{2-2y}{3}\\5x-2y=-18\end{array}\right.\) - step2: Substitute the value of \(x:\) \(5\times \frac{2-2y}{3}-2y=-18\) - step3: Simplify: \(\frac{5\left(2-2y\right)}{3}-2y=-18\) - step4: Multiply both sides of the equation by LCD: \(\left(\frac{5\left(2-2y\right)}{3}-2y\right)\times 3=-18\times 3\) - step5: Simplify the equation: \(10-16y=-54\) - step6: Move the constant to the right side: \(-16y=-54-10\) - step7: Subtract the numbers: \(-16y=-64\) - step8: Change the signs: \(16y=64\) - step9: Divide both sides: \(\frac{16y}{16}=\frac{64}{16}\) - step10: Divide the numbers: \(y=4\) - step11: Substitute the value of \(y:\) \(x=\frac{2-2\times 4}{3}\) - step12: Calculate: \(x=-2\) - step13: Calculate: \(\left\{ \begin{array}{l}x=-2\\y=4\end{array}\right.\) - step14: Check the solution: \(\left\{ \begin{array}{l}x=-2\\y=4\end{array}\right.\) - step15: Rewrite: \(\left(x,y\right) = \left(-2,4\right)\) Solve the system of equations \( 2 y-3 x=7;4 y-5 x=21 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}2y-3x=7\\4y-5x=21\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=\frac{-7+2y}{3}\\4y-5x=21\end{array}\right.\) - step2: Substitute the value of \(x:\) \(4y-5\times \frac{-7+2y}{3}=21\) - step3: Simplify: \(4y-\frac{5\left(-7+2y\right)}{3}=21\) - step4: Multiply both sides of the equation by LCD: \(\left(4y-\frac{5\left(-7+2y\right)}{3}\right)\times 3=21\times 3\) - step5: Simplify the equation: \(2y+35=63\) - step6: Move the constant to the right side: \(2y=63-35\) - step7: Subtract the numbers: \(2y=28\) - step8: Divide both sides: \(\frac{2y}{2}=\frac{28}{2}\) - step9: Divide the numbers: \(y=14\) - step10: Substitute the value of \(y:\) \(x=\frac{-7+2\times 14}{3}\) - step11: Calculate: \(x=7\) - step12: Calculate: \(\left\{ \begin{array}{l}x=7\\y=14\end{array}\right.\) - step13: Check the solution: \(\left\{ \begin{array}{l}x=7\\y=14\end{array}\right.\) - step14: Rewrite: \(\left(x,y\right) = \left(7,14\right)\) Solve the system of equations \( 3 x+2 y=6;5 x+3 y=11 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}3x+2y=6\\5x+3y=11\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=\frac{6-2y}{3}\\5x+3y=11\end{array}\right.\) - step2: Substitute the value of \(x:\) \(5\times \frac{6-2y}{3}+3y=11\) - step3: Simplify: \(\frac{5\left(6-2y\right)}{3}+3y=11\) - step4: Multiply both sides of the equation by LCD: \(\left(\frac{5\left(6-2y\right)}{3}+3y\right)\times 3=11\times 3\) - step5: Simplify the equation: \(30-y=33\) - step6: Move the constant to the right side: \(-y=33-30\) - step7: Subtract the numbers: \(-y=3\) - step8: Change the signs: \(y=-3\) - step9: Substitute the value of \(y:\) \(x=\frac{6-2\left(-3\right)}{3}\) - step10: Calculate: \(x=4\) - step11: Calculate: \(\left\{ \begin{array}{l}x=4\\y=-3\end{array}\right.\) - step12: Check the solution: \(\left\{ \begin{array}{l}x=4\\y=-3\end{array}\right.\) - step13: Rewrite: \(\left(x,y\right) = \left(4,-3\right)\) Solve the system of equations \( (y-2)/3-(x-1)/2=1;(y-4)/5-(x-3)/4=1 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}\frac{\left(y-2\right)}{3}-\frac{\left(x-1\right)}{2}=1\\\frac{\left(y-4\right)}{5}-\frac{\left(x-3\right)}{4}=1\end{array}\right.\) - step1: Remove the parentheses: \(\left\{ \begin{array}{l}\frac{y-2}{3}-\frac{x-1}{2}=1\\\frac{y-4}{5}-\frac{x-3}{4}=1\end{array}\right.\) - step2: Solve the equation: \(\left\{ \begin{array}{l}x=\frac{-7+2y}{3}\\\frac{y-4}{5}-\frac{x-3}{4}=1\end{array}\right.\) - step3: Substitute the value of \(x:\) \(\frac{y-4}{5}-\frac{\frac{-7+2y}{3}-3}{4}=1\) - step4: Simplify: \(\frac{y-4}{5}+\frac{8-y}{6}=1\) - step5: Multiply both sides of the equation by LCD: \(\left(\frac{y-4}{5}+\frac{8-y}{6}\right)\times 30=1\times 30\) - step6: Simplify the equation: \(y+16=30\) - step7: Move the constant to the right side: \(y=30-16\) - step8: Subtract the numbers: \(y=14\) - step9: Substitute the value of \(y:\) \(x=\frac{-7+2\times 14}{3}\) - step10: Calculate: \(x=7\) - step11: Calculate: \(\left\{ \begin{array}{l}x=7\\y=14\end{array}\right.\) - step12: Check the solution: \(\left\{ \begin{array}{l}x=7\\y=14\end{array}\right.\) - step13: Rewrite: \(\left(x,y\right) = \left(7,14\right)\) ### (a) Solve for \( x \) and \( y \) using the method of substitution: 1. **Equations:** \[ x - y = 8 \quad (1) \] \[ 2x + y = 10 \quad (2) \] **Solution:** From equation (1), we can express \( y \) in terms of \( x \): \[ y = x - 8 \] Substituting into equation (2): \[ 2x + (x - 8) = 10 \] \[ 3x - 8 = 10 \implies 3x = 18 \implies x = 6 \] Now substituting \( x \) back to find \( y \): \[ y = 6 - 8 = -2 \] **Result:** \( (x, y) = (6, -2) \) 2. **Equations:** \[ 3x - y = 2 \quad (1) \] \[ 7x - 2y = 8 \quad (2) \] **Solution:** From equation (1): \[ y = 3x - 2 \] Substituting into equation (2): \[ 7x - 2(3x - 2) = 8 \] \[ 7x - 6x + 4 = 8 \implies x + 4 = 8 \implies x = 4 \] Now substituting \( x \) back to find \( y \): \[ y = 3(4) - 2 = 12 - 2 = 10 \] **Result:** \( (x, y) = (4, 10) \) 3. **Equations:** \[ 3x + 5y = 8 \quad (1) \] \[ x - 2y = -1 \quad (2) \] **Solution:** From equation (2): \[ x = 2y - 1 \] Substituting into equation (1): \[ 3(2y - 1) + 5y = 8 \] \[ 6y - 3 + 5y = 8 \implies 11y - 3 = 8 \implies 11y = 11 \implies y = 1 \] Now substituting \( y \) back to find \( x \): \[ x = 2(1) - 1 = 1 \] **Result:** \( (x, y) = (1, 1) \) 4. **Equations:** \[ 7x - 3y = 41 \quad (1) \] \[ 3x - y = 17 \quad (2) \] **Solution:** From equation (2): \[ y = 3x - 17 \] Substituting into equation (1): \[ 7x - 3(3x - 17) = 41 \] \[ 7x - 9x + 51 = 41 \implies -2x + 51 = 41 \implies -2x = -10 \implies x = 5 \] Now substituting \( x \) back to find \( y \): \[ y = 3(5) - 17 = 15 - 17 = -2 \] **Result:** \( (x, y) = (5, -2) \) ### (b) Solve for \( x \) and \( y \) using either the elimination or substitution method: 1. **Equations:** \[ x + y = 1 \quad (1) \] \[ x - 2y = 1 \quad (2) \] **Solution:** From equation (1): \[ y = 1 - x \] Substituting into equation (2): \[ x - 2(1 - x) = 1 \] \[ x - 2 + 2x = 1 \implies 3x - 2 = 1 \implies 3x = 3 \implies x = 1 \] Now substituting \( x \) back to find \( y \): \[ y = 1 - 1 = 0 \] **Result:** \( (x, y) = (1, 0) \) 2. **Equations:** \[ 3x + 2y = 2 \quad (1) \] \[ 5x - 2y = -18 \quad (2) \] **Solution:** Adding both equations: \[ 3x + 2y + 5x - 2y = 2 - 18 \] \[ 8x = -16 \implies x = -2 \] Now substituting \( x \) back to find \( y \): \[ 3(-2) + 2y = 2 \implies -6 + 2y = 2 \implies 2y = 8 \implies y = 4 \] **Result:** \( (x, y) = (-2, 4) \) 3. **Equations:** \[ x + 4y = 14 \quad (1) \] \[ 3x + 2y = 12 \quad (2) \] **Solution:** From equation (1): \[ x = 14 - 4y \] Substituting into equation (2): \[ 3(14 - 4y) + 2y = 12 \] \[ 42 - 12y + 2y = 12 \implies -10y = -30 \implies y = 3 \] Now substituting \( y \) back to find \( x \): \[ x = 14 - 4(3) = 14 - 12 = 2 \] **Result:** \( (x, y) = (2, 3) \) 4. **Equations:** \[