Systems of Linear Equations: practice solutions, Grade 9 – download the PDF

1 Is it a solution? ★★★

(a) Test \((2, 5)\). First equation: \(3(2) - 1 = 5\), true. Second equation: \(2 + 5 = 7\), true. So \((2, 5)\) is a solution.

(b) Test \((3, 4)\). First equation: \(3(3) - 1 = 8 \neq 4\), false. So \((3, 4)\) is not a solution (it fails the first equation, even though \(3 + 4 = 7\)).

2 Table of values ★★★

For \(y = x + 2\): 2, 3, 4, 5, 6. For \(y = -x + 6\): 6, 5, 4, 3, 2.

Both lines have \(y = 4\) when \(x = 2\). The solution is \((2, 4)\).

3 A quick substitution ★★★

Replace \(y\) by \(2x\) in the second equation: \(x + 2x = 12\), so \(3x = 12\) and \(x = 4\).

Then \(y = 2(4) = 8\). Check: \(4 + 8 = 12\). The solution is \((4, 8)\).

4 A quick elimination ★★★

Add the equations: \(2x = 18\), so \(x = 9\). The \(y\) terms cancel because they are opposites.

Substitute in the first equation: \(9 + y = 15\), so \(y = 6\). Check: \(9 - 6 = 3\). The solution is \((9, 6)\).

5 How many solutions? ★★★

(a) Both slopes are 4, the intercepts 1 and -5 differ: parallel lines, no solution.

(b) The slopes -1 and 1 differ: the lines cross once, one solution. (They share the intercept 2, so the solution is \((0, 2)\).)

(c) Solve the first equation for \(y\): \(2y = -2x + 6\), so \(y = -x + 3\). It is the same line: infinitely many solutions.

6 Chess club ★★★

Let \(b\) be the number of boys and \(g\) the number of girls. Then \(b + g = 31\) and \(b - g = 9\).

Add the equations: \(2b = 40\), so \(b = 20\). Then \(g = 31 - 20 = 11\).

The club has 20 boys and 11 girls. Check: \(20 - 11 = 9\).

7 Test points in a region ★★★

\((1, 2)\): \(2 \ge 2\) is true and \(2 \lt 4\) is true, so it is a solution.

\((2, 3)\): \(3 \ge 4\) is false, so it is not a solution.

\((0, 0)\): \(0 \ge 0\) is true and \(0 \lt 3\) is true, so it is a solution. (Points on a solid boundary count, so \((1, 2)\) and \((0, 0)\) are fine.)

8 Substitution with negatives ★★★

Replace \(y\): \(5x + 2(3x - 9) = 4\). Distribute: \(5x + 6x - 18 = 4\), so \(11x = 22\) and \(x = 2\).

Then \(y = 3(2) - 9 = -3\). Check: \(5(2) + 2(-3) = 10 - 6 = 4\). The solution is \((2, -3)\).

9 Isolate first ★★★

From the first equation, \(x = 11 - 3y\). Substitute in the second: \(2(11 - 3y) - 5y = 0\), so \(22 - 6y - 5y = 0\) and \(11y = 22\), so \(y = 2\).

Then \(x = 11 - 3(2) = 5\). Check: \(2(5) - 5(2) = 0\). The solution is \((5, 2)\).

10 Elimination, one multiplier ★★★

Multiply the second equation by 3: \(6x - 3y = 15\). Add it to the first: \(4x + 3y + 6x - 3y = 25 + 15\), so \(10x = 40\) and \(x = 4\).

From the second equation, \(y = 2x - 5 = 3\). Check in the first: \(4(4) + 3(3) = 25\). The solution is \((4, 3)\).

11 Elimination, two multipliers ★★★

To cancel \(y\), make \(12y\) and \(-12y\): multiply the first equation by 3 and the second by 2. This gives \(9x + 12y = 3\) and \(10x - 12y = 54\).

Add: \(19x = 57\), so \(x = 3\). Then \(3(3) + 4y = 1\) gives \(4y = -8\), so \(y = -2\).

Check: \(5(3) - 6(-2) = 15 + 12 = 27\). The solution is \((3, -2)\).

12 Reading a graph ★★★

(a) The lines cross at \(P\), which appears to be \((4, 3)\).

(b) Set the expressions for \(y\) equal: \(0.5x + 1 = -x + 7\), so \(1.5x = 6\) and \(x = 4\). Then \(y = 0.5(4) + 1 = 3\). The second line gives \(-4 + 7 = 3\), so the solution is \((4, 3)\), which matches the graph.

13 Pool tickets ★★★

Let \(a\) be the adult price and \(c\) the child price, in dollars: \(4a + 3c = 41\) and \(2a + 5c = 31\).

Multiply the second equation by 2: \(4a + 10c = 62\). Subtract the first equation: \(7c = 21\), so \(c = 3\). Then \(4a + 9 = 41\), so \(a = 8\).

An adult ticket costs 8 dollars and a child ticket costs 3 dollars. Check: \(2(8) + 5(3) = 31\).

14 A parameter ★★★

(a) The first line has slope 3 and intercept 2. If \(k = 3\), the second line is parallel (slope 3, intercept -5 different), so there is no solution. For every other \(k\) the slopes differ, so there is exactly one solution.

(b) With \(k = 2\): \(3x + 2 = 2x - 5\), so \(x = -7\). Then \(y = 3(-7) + 2 = -19\). Check: \(2(-7) - 5 = -19\). The solution is \((-7, -19)\).

15 Finding the missing constant ★★★

(a) Multiply the first equation by 2: \(4x + 6y = 24\). If \(c = 24\) the two equations describe the same line, so there are infinitely many solutions.

(b) If \(c = 20\), the second equation says \(4x + 6y = 20\), while the first says \(4x + 6y = 24\). The same expression cannot equal two different numbers, so no pair works. Subtracting gives \(0 = 4\), a false statement: no solution. (Both lines have slope \(-\tfrac{2}{3}\) but different intercepts, so they are parallel.)

16 Kayak trip ★★★

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

Add the equations: \(2b = 10\), so \(b = 5\). Then \(c = 6 - 5 = 1\).

The kayaker paddles at 5 mph in still water and the current flows at 1 mph. Check: \(5 + 1 = 6\) and \(5 - 1 = 4\).

17 Juice mixture ★★★

Let \(x\) be the gallons of the 15% drink and \(y\) the gallons of the 40% drink. The total volume gives \(x + y = 10\). The amount of pure juice gives \(0.15x + 0.40y = 0.25(10) = 2.5\).

Substitute \(x = 10 - y\): \(0.15(10 - y) + 0.40y = 2.5\), so \(1.5 + 0.25y = 2.5\) and \(y = 4\). Then \(x = 6\).

The shop needs 6 gallons of the 15% drink and 4 gallons of the 40% drink. Check: \(0.15(6) + 0.40(4) = 0.9 + 1.6 = 2.5\).

18 Fractions as coefficients ★★★

Multiply the first equation by 6: \(3x + 2y = 24\). From the second equation, \(x = y + 3\).

Substitute: \(3(y + 3) + 2y = 24\), so \(5y + 9 = 24\) and \(y = 3\). Then \(x = 6\).

Check in the original first equation: \(\dfrac{6}{2} + \dfrac{3}{3} = 3 + 1 = 4\). The solution is \((6, 3)\).

19 Ages ★★★

Let \(m\) be Mia’s age and \(l\) Leo’s age now. Then \(m = 3l\). In 6 years: \(m + 6 = 2(l + 6)\).

Substitute: \(3l + 6 = 2l + 12\), so \(l = 6\) and \(m = 18\).

Mia is 18 and Leo is 6. Check: in 6 years they will be 24 and 12, and \(24 = 2(12)\).

20 Shopping constraints ★★★

(a) Cost: \(2n + p \le 30\). Number of items: \(n + p \ge 5\). Plus \(n \ge 0\) and \(p \ge 0\).

(b) For \((n, p) = (8, 10)\): \(2(8) + 10 = 26 \le 30\) and \(8 + 10 = 18 \ge 5\), so it is possible. For \((12, 8)\): \(2(12) + 8 = 32 \gt 30\), so it is not possible.

(c) From \(2n + p \le 30\) we get \(n + p \le 30 - n \le 30\) because \(n \ge 0\). The value 30 is reached with \(n = 0\) and \(p = 30\). The greatest number of items is 30 (all pens).

21 A triangular region ★★★

(a) The two slanted lines meet where \(x - 2 = -x + 4\), so \(x = 3\) and \(y = 1\): corner \(C(3, 1)\). The line \(y = x - 2\) meets \(y = 0\) at \(A(2, 0)\). The line \(y = -x + 4\) meets \(y = 0\) at \(B(4, 0)\).

(b) The region is triangle \(ABC\) with base \(AB = 4 - 2 = 2\) and height 1 (the y-coordinate of \(C\)). Its area is \(\tfrac{1}{2} \times 2 \times 1 = 1\) square unit.