Linear Equations, Inequalities and Systems: practice solutions, Grade 11 – download the PDF

1 Slope from two points ★★★

\(m = \dfrac{7 - (-1)}{6 - 2} = \dfrac{8}{4} = 2\).

Point-slope form with \(P\): \(y + 1 = 2(x - 2)\), so \(y = 2x - 5\). Check with \(Q\): \(2(6) - 5 = 7\).

The slope is \(2\) and the equation is \(y = 2x - 5\).

2 Rewrite in slope-intercept form ★★★

Subtract \(3x\): \(4y = -3x + 12\). Divide by \(4\): \(y = -\dfrac{3}{4}x + 3\).

The slope is \(-\dfrac{3}{4}\) and the y-intercept is \(3\). For the x-intercept set \(y = 0\): \(3x = 12\), so \(x = 4\).

3 Solve a linear equation ★★★

Distribute: \(5x - 10 + 3 = 2x + 14\), so \(5x - 7 = 2x + 14\).

Subtract \(2x\) and add \(7\): \(3x = 21\), so \(x = 7\).

Check: left side \(5(5) + 3 = 28\), right side \(14 + 14 = 28\).

4 A basic absolute value equation ★★★

The expression inside equals \(9\) or \(-9\).

\(x + 4 = 9\) gives \(x = 5\); \(x + 4 = -9\) gives \(x = -13\).

Check: \(|5 + 4| = 9\) and \(|-13 + 4| = |-9| = 9\). The solutions are \(5\) and \(-13\).

5 A three-part inequality ★★★

Add \(3\) to all three parts: \(-2 \le x < 7\).

Interval notation: \([-2, 7)\).

On the number line, draw a closed circle at \(-2\), an open circle at \(7\), and shade between them.

6 Is it a solution? ★★★

(a) \(3 + (-2) = 1\) and \(2(3) - (-2) = 8\). Both equations hold, so \((3, -2)\) is a solution.

(b) \(1 + 0 = 1\) holds, but \(2(1) - 0 = 2 \ne 8\). The pair fails the second equation, so \((1, 0)\) is not a solution.

7 Substitution warm-up ★★★

Replace \(y\) in the second equation: \(2x + 3x - 4 = 11\), so \(5x = 15\) and \(x = 3\).

Then \(y = 3(3) - 4 = 5\). Check: \(2(3) + 5 = 11\). The solution is \((3, 5)\).

8 A parallel line ★★★

Parallel lines have the same slope, so \(m = -\dfrac{1}{2}\).

Point-slope form: \(y + 2 = -\dfrac{1}{2}(x - 3)\), so \(y = -\dfrac{1}{2}x + \dfrac{3}{2} - 2 = -\dfrac{1}{2}x - \dfrac{1}{2}\).

Check: at \(x = 3\), \(-\dfrac{3}{2} - \dfrac{1}{2} = -2\).

9 An absolute value inequality ★★★

Rewrite as an and statement: \(-10 \le 4 - 3x \le 10\).

Subtract \(4\): \(-14 \le -3x \le 6\). Divide by \(-3\) and reverse both signs: \(\dfrac{14}{3} \ge x \ge -2\).

So \(-2 \le x \le \dfrac{14}{3}\), that is \(\left[-2, \dfrac{14}{3}\right]\). Check: \(x = 0\) gives \(4 \le 10\), while \(x = 5\) gives \(|-11| = 11 > 10\), which is outside.

10 An or compound inequality ★★★

First: \(2x < -8\), so \(x < -4\).

Second: \(-x \le -3\); dividing by \(-1\) reverses the sign, so \(x \ge 3\).

The solution is \(x < -4\) or \(x \ge 3\), that is \((-\infty, -4) \cup [3, \infty)\).

11 Elimination practice ★★★

Multiply the first equation by \(5\) and the second by \(3\): \(20x - 15y = -5\) and \(18x + 15y = 81\).

Add: \(38x = 76\), so \(x = 2\). Then \(8 - 3y = -1\), so \(y = 3\).

Check in the second equation: \(12 + 15 = 27\). The solution is \((2, 3)\).

12 Movie theater tickets ★★★

Let \(a\) be the number of adult tickets and \(c\) the number of child tickets. Then \(a + c = 150\) and \(12a + 8c = 1448\).

Substitute \(c = 150 - a\): \(12a + 1200 - 8a = 1448\), so \(4a = 248\) and \(a = 62\). Then \(c = 88\).

Check: \(12(62) + 8(88) = 744 + 704 = 1448\). The theater sold 62 adult tickets and 88 child tickets.

13 No solution or infinitely many? ★★★

(a) Multiplying the second equation by \(3\) gives \(6x - 9y = 12\), the first equation. The lines coincide: infinitely many solutions.

(b) Both lines have slope \(2\) but different y-intercepts (\(1\) and \(-5\)). They are parallel and distinct: no solution. Algebraically, \(2x + 1 = 2x - 5\) simplifies to \(1 = -5\), which is false.

14 Inverse of a 2 by 2 matrix ★★★

(a) \(\det A = 2 \cdot 3 - 5 \cdot 1 = 1\), so \(A^{-1} = \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}\).

(b) \(X = A^{-1}B = \begin{bmatrix} 3 \cdot 9 - 5 \cdot 5 \\ -9 + 2 \cdot 5 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}\).

Check: \(2(2) + 5(1) = 9\) and \(2 + 3(1) = 5\). So \(x = 2\) and \(y = 1\).

15 A system with three unknowns ★★★

Add the first and third equations to remove \(z\): \(5x + 3y = 7\).

Double the first equation and add the second: \(4x + 2y - 2z + x - y + 2z = 9\), so \(5x + y = 9\).

Subtract: \(2y = -2\), so \(y = -1\). Then \(5x - 1 = 9\) gives \(x = 2\). From the first equation, \(z = 2x + y = 3\).

Check the second equation: \(2 + 1 + 6 = 9\); the third: \(6 - 2 + 3 = 7\). The solution is \((2, -1, 3)\).

16 The coin jar ★★★

Let \(n\), \(d\), \(q\) be the numbers of nickels, dimes and quarters. In cents: \(n + d + q = 30\), \(5n + 10d + 25q = 330\), and \(d = 2q\).

From the first equation, \(n = 30 - 3q\). Substitute into the value equation: \(5(30 - 3q) + 20q + 25q = 330\), so \(150 + 30q = 330\) and \(q = 6\).

Then \(d = 12\) and \(n = 12\). Check: \(60 + 120 + 150 = 330\) cents. The jar has 12 nickels, 12 dimes and 6 quarters.

17 Planting plan ★★★

The axes intersections: \((0, 0)\); \((40, 0)\), since \(2(40) = 80 \le 90\); and \((0, 30)\), since \(3(30) = 90\) and \(30 \le 40\).

The two boundary lines meet where \(y = 40 - x\) and \(2x + 3(40 - x) = 90\), so \(120 - x = 90\), \(x = 30\), \(y = 10\): the vertex \((30, 10)\).

Profit: \((0, 0)\) gives \(0\); \((40, 0)\) gives \(4800\); \((30, 10)\) gives \(3600 + 1500 = 5100\); \((0, 30)\) gives \(4500\).

The maximum profit is 5,100 dollars, with 30 acres of corn and 10 acres of soybeans.

18 Minimizing a cost ★★★

Corner points: on the y-axis, \(3x + y = 15\) gives \((0, 15)\); on the x-axis, \(x + 2y = 10\) gives \((10, 0)\).

Intersection of the two lines: \(y = 15 - 3x\), so \(x + 30 - 6x = 10\), \(x = 4\), \(y = 3\).

Costs: \((0, 15)\) gives \(45\); \((4, 3)\) gives \(16 + 9 = 25\); \((10, 0)\) gives \(40\).

The minimum cost is \(25\), reached with 4 kg of A and 3 kg of B.

19 Tolerance with absolute value ★★★

(a) \(|v - 500| \le 6\), so \(-6 \le v - 500 \le 6\) and \(494 \le v \le 506\). Only 505.9 mL is accepted; 493.5 is too low and 506.2 is too high.

(b) The midpoint is \(\dfrac{68 + 76}{2} = 72\) and the distance from the midpoint to each end is \(4\). So \(|T - 72| \le 4\).

20 Boat and current ★★★

Let \(b\) be the boat speed and \(c\) the current speed, in miles per hour. Upstream: \(b - c = \dfrac{36}{3} = 12\). Downstream: \(b + c = \dfrac{36}{2} = 18\).

Add the equations: \(2b = 30\), so \(b = 15\). Then \(c = 3\).

The boat goes 15 mph in still water and the current flows at 3 mph.

21 Parameter in a system ★★★

(a) \(\det = 2 \cdot 6 - k \cdot 3 = 12 - 3k\). It equals \(0\) when \(k = 4\). For every other value, the system has exactly one solution.

(b) With \(k = 4\), the first equation is \(2x + 4y = 8\). Dividing the second equation by \(3\) gives \(x + 2y = 4\), and dividing the first by \(2\) gives the same equation. The two equations describe the same line, so there are infinitely many solutions.