Systems of Linear Equations: math test solutions, Grade 8 – download the PDF

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Test solutions with the detailed point scale. Add up your points and spot what to review.

Suggested time: 45 minutes. Out of 20 points. Calculator allowed only when the problem says so.

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$ 3(2) + 2(-1) = 4 $ true; $ 2 - 3(-1) = 5 $ true. Yes, it is a solution. (1 pt)
$ 1 + 1 = 2 $ true; $ 2(1) - 1 = 1 \neq 3 $. No, it is not a solution. (1 pt)

Line 1: $ (0, 6), (1, 5), (2, 4), (3, 3) $. Line 2: $ (0, -3), (1, -1), (2, 1), (3, 3) $. (1 pt)

The lines are drawn below. (1 pt)

They cross at $ (3, 3) $. Check: $ -3 + 6 = 3 $ and $ 2(3) - 3 = 3 $. (1 pt)

Replace $ x $: $ 3(2y + 1) - 4y = 9 $. (1 pt) Then $ 6y + 3 - 4y = 9 $, $ 2y = 6 $, $ y = 3 $. (1 pt)

So $ x = 2(3) + 1 = 7 $. Check: $ 3(7) - 4(3) = 9 $. The solution is $ (7, 3) $. (1 pt)

Multiply the second equation by $ -2 $: $ -4x + 10y = 28 $. (1 pt) Add it to the first: $ 13y = 26 $, so $ y = 2 $. (1 pt)

Then $ 4x + 6 = -2 $, so $ x = -2 $. Check: $ 2(-2) - 5(2) = -14 $. The solution is $ (-2, 2) $. (1 pt)

The second equation gives $ 2y = -8x + 6 $, so $ y = -4x + 3 $. Same line: infinitely many solutions. (1 pt)
Subtracting gives $ 0 = -5 $, which is false (also $ y = 3x - 2 $ and $ y = 3x - 7 $ are parallel). No solution. (1 pt)
Slopes $ 5 $ and $ -5 $ differ: exactly one solution, at $ (0, -1) $. (1 pt)