Systems of Linear Equations: practice solutions, Grade 8

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Practice solutions Grade 8 : Systems of Linear Equations — Zyro the alien explorer of Planète Maths

Written solutions to the chapter problems. Check each step, then correct yourself.

1 Is it a solution? ★★★

For $ (3, 2) $: $ 3 + 2 = 5 $ is true, and $ 2(3) - 2 = 4 $ is true. Both equations hold, so $ (3, 2) $ is a solution.
For $ (4, 1) $: $ 4 + 1 = 5 $ is true, but $ 2(4) - 1 = 7 \neq 4 $. The second equation fails, so $ (4, 1) $ is not a solution.

2 Table of values and graph ★★★

Line 1, $ y = x + 2 $: the points are $ (0, 2), (1, 3), (2, 4), (3, 5) $. Line 2, $ y = -x + 4 $: the points are $ (0, 4), (1, 3), (2, 2), (3, 1) $.

Both tables contain $ (1, 3) $, so the lines cross there. Check: $ 3 = 1 + 2 $ and $ 3 = -1 + 4 $. The solution is $ (1, 3) $.

3 Substitute y = 2x ★★★

Replace $ y $ by $ 2x $: $ x + 2x = 12 $, so $ 3x = 12 $ and $ x = 4 $. Then $ y = 2(4) = 8 $.

Check: $ 4 + 8 = 12 $. The solution is $ (4, 8) $.

4 Add the equations ★★★

The $ y $-terms are opposites, so add the equations: $ 2x = 14 $, so $ x = 7 $. Then $ 7 + y = 11 $ gives $ y = 4 $.

Check: $ 7 - 4 = 3 $. The solution is $ (7, 4) $.

5 Substitute an expression ★★★

Replace $ y $: $ 2x + (x - 3) = 15 $, so $ 3x - 3 = 15 $, $ 3x = 18 $, $ x = 6 $. Then $ y = 6 - 3 = 3 $.

Check: $ 2(6) + 3 = 15 $. The solution is $ (6, 3) $.

6 True or false? ★★★

False. Each equation is a line. Two lines are either crossing at exactly one point, parallel with no common point, or the same line with infinitely many common points. Two distinct lines can never share exactly two points, so a linear system never has exactly two solutions.

7 Predict the number of solutions ★★★

Same slope 3, different intercepts $ 1 $ and $ -5 $: parallel lines, no solution.
Slopes $ -1 $ and $ 2 $ are different: the lines cross, exactly one solution (they meet at $ (0, 4) $).
Divide the second equation by 2: $ y = \dfrac{1}{2}x + 2 $. It is the same line, so infinitely many solutions.

8 Elimination with one multiplier ★★★

Multiply the second equation by 3: $ 12x - 3y = 54 $. Add it to the first: $ 14x = 70 $, so $ x = 5 $.

Substitute in the first equation: $ 10 + 3y = 16 $, so $ y = 2 $.

Check in the second: $ 4(5) - 2 = 18 $. The solution is $ (5, 2) $.

9 Isolate first ★★★

In the second equation, $ y $ has coefficient $ -1 $: $ y = 3x - 6 $.

Substitute: $ x + 2(3x - 6) = 9 $, so $ 7x - 12 = 9 $, $ 7x = 21 $, $ x = 3 $. Then $ y = 3(3) - 6 = 3 $.

Check: $ 3 + 6 = 9 $. The solution is $ (3, 3) $.

10 Multiply both equations ★★★

Make the $ y $-coefficients equal (12): multiply the first equation by 3 and the second by 2. $ 9x + 12y = -3 $ and $ 10x + 12y = -6 $.

Subtract the first from the second: $ x = -3 $. Then $ 3(-3) + 4y = -1 $, so $ 4y = 8 $ and $ y = 2 $.

Check: $ 5(-3) + 6(2) = -3 $. The solution is $ (-3, 2) $.

11 A statement that is always true ★★★

Substitute $ y $: $ 4x - 2(2x - 3) = 6 $, so $ 4x - 4x + 6 = 6 $, which gives $ 6 = 6 $.

This is always true, so every point of the line works: the system has infinitely many solutions. Indeed $ 4x - 2y = 6 $ gives $ y = 2x - 3 $: the two equations describe the same line.

12 A statement that is never true ★★★

Multiply the first equation by 3: $ 6x + 3y = 21 $. Subtract the second equation: $ 0 = 11 $.

This is false for every $ x $ and $ y $, so there is no solution. The lines have the same slope $ -2 $ but different intercepts, so they are parallel.

13 Tickets at a school play ★★★

Let $ a $ be adult tickets and $ c $ child tickets. Then $ a + c = 36 $ and $ 8a + 5c = 243 $.

From the first, $ c = 36 - a $. So $ 8a + 5(36 - a) = 243 $, $ 3a + 180 = 243 $, $ 3a = 63 $, $ a = 21 $, and $ c = 15 $.

Check: $ 8(21) + 5(15) = 168 + 75 = 243 $. They sold 21 adult tickets and 15 child tickets.

14 Find the missing slope ★★★

No solution means parallel lines: same slope, different intercepts. The intercepts $ 4 $ and $ -2 $ already differ, so we need $ k = 3 $.

For any other value of $ k $ the slopes differ, so the lines cross exactly once: $ k \neq 3 $.

15 Kayak in a river ★★★

Downstream the speeds add: $ 12 = 2(b + c) $, so $ b + c = 6 $. Upstream they subtract: $ 12 = 3(b - c) $, so $ b - c = 4 $.

Add the equations: $ 2b = 10 $, $ b = 5 $. Then $ c = 6 - 5 = 1 $.

Check: $ 2(5 + 1) = 12 $ and $ 3(5 - 1) = 12 $. The kayaker paddles at 5 mph in still water and the current flows at 1 mph.

16 Two numbers ★★★

Let $ y $ be the smaller number and $ x $ the larger. Then $ x + y = 54 $ and $ x = 2y + 6 $.

Substitute: $ (2y + 6) + y = 54 $, $ 3y = 48 $, $ y = 16 $. Then $ x = 2(16) + 6 = 38 $.

Check: $ 38 + 16 = 54 $. The numbers are 38 and 16.

17 Trail mix ★★★

Let $ c $ be pounds of cashews and $ r $ pounds of raisins. The total weight gives $ c + r = 10 $. The total cost is $ 6 \times 10 = 60 $ dollars, so $ 9c + 4r = 60 $.

From the first, $ r = 10 - c $. So $ 9c + 4(10 - c) = 60 $, $ 5c + 40 = 60 $, $ c = 4 $, and $ r = 6 $.

Check: $ 9(4) + 4(6) = 36 + 24 = 60 $. Use 4 pounds (about 1.8 kg) of cashews and 6 pounds (about 2.7 kg) of raisins.

18 Intersection of two lines through points ★★★

Line $ p $: slope $ \dfrac{5 - 1}{2 - 0} = 2 $, y-intercept 1, so $ y = 2x + 1 $. Line $ q $: slope $ \dfrac{0 - 9}{3 - 0} = -3 $, y-intercept 9, so $ y = -3x + 9 $.

Set them equal: $ 2x + 1 = -3x + 9 $, so $ 5x = 8 $ and $ x = 1.6 $. Then $ y = 2(1.6) + 1 = 4.2 $.

Check in $ q $: $ -3(1.6) + 9 = 4.2 $. The lines meet at $ (1.6, 4.2) $. This point is not on a grid corner, so graphing alone would only give an estimate.

19 Fractions in a system ★★★

Multiply the first equation by 6: $ 3x + 2y = 24 $. From the second, $ y = x - 3 $.

Substitute: $ 3x + 2(x - 3) = 24 $, $ 5x - 6 = 24 $, $ x = 6 $, $ y = 3 $.

Check: $ \dfrac{6}{2} + \dfrac{3}{3} = 3 + 1 = 4 $ and $ 6 - 3 = 3 $. The solution is $ (6, 3) $.

20 Choosing a plan ★★★

Set $ 2x + 15 = 3x + 5 $: $ 10 = x $. Then $ y = 2(10) + 15 = 35 $. Both plans cost $35 after 10 visits, which matches the crossing point $ (10, 35) $.
For 6 visits: plan A costs $ 2(6) + 15 = 27 $ dollars and plan B costs $ 3(6) + 5 = 23 $ dollars, so plan B is cheaper. For 15 visits: plan A costs $ 45 $ dollars and plan B costs $ 50 $ dollars, so plan A is cheaper. Plan B is cheaper before 10 visits and plan A after.

21 A race with a head start ★★★

Mia catches Sam when their distances are equal: $ 200 + 3t = 5t $, so $ 200 = 2t $ and $ t = 100 $. Then $ d = 5(100) = 500 $.

Check: Sam is at $ 200 + 3(100) = 500 $ meters. Mia catches Sam after 100 seconds, when she has run 500 meters (about 1,640 feet).

22 Find the error ★★★

Subtracting gives $ 5x - 3x = 2x $, but also $ -2y - 2y = -4y $ and $ 11 - 5 = 6 $. So the result is $ 2x - 4y = 6 $: Lena forgot the $ y $-term, which did not vanish.

The $ y $-terms are opposites, so add instead: $ 8x = 16 $, $ x = 2 $. Then $ 3(2) + 2y = 5 $, so $ 2y = -1 $ and $ y = -0.5 $.

Check: $ 5(2) - 2(-0.5) = 11 $. The solution is $ (2, -0.5) $.