Linear relationships - Year 10 Mathematics

What you will learn

rearrange and solve linear equations that arise from formulas,
solve and graph linear inequalities on a number line,
graph inequalities in two variables and shade the correct region,
determine whether two lines are parallel (same gradient) or perpendicular ( $m_1 m_2 = -1$ ),
write the equation of a line parallel or perpendicular to a given line through a specified point.

Worked example 0 Real-world example: budget constraint

A school has $600 to spend on pens at $2 each and notebooks at $5 each. Write an inequality and shade the feasible region.

Let $x$ = number of pens, $y$ = number of notebooks.
Constraint: $2x + 5y \leq 600$ .
Boundary line: $2x + 5y = 600$ , or $y = -\tfrac{2}{5}x + 120$ .
Test $(0, 0)$ : $2(0) + 5(0) = 0 \leq 600$ — true, so shade the side containing the origin.
Also $x \geq 0$ and $y \geq 0$ (cannot buy negative items), so the feasible region is a triangle in the first quadrant.

Key idea: the inequality defines a region of all affordable combinations, not just a single answer.

1. Solving linear equations from formulas

Many formulas in science and finance are linear in one variable. To solve for that variable, rearrange using inverse operations.

Worked example 1 Rearranging a formula

The formula for the perimeter of a rectangle is $P = 2l + 2w$ . Find $l$ when $P = 54$ and $w = 9$ .

Substitute: $54 = 2l + 2(9) = 2l + 18$ .
$2l = 54 - 18 = 36$ .
$l = 18$ .

Worked example 2 Solving a formula for a pronumeral

Make $t$ the subject of $v = u + at$ .

$v - u = at$ .
$t = \dfrac{v - u}{a}$ , provided $a \neq 0$ .

2. Linear inequalities and number lines

A linear inequality looks like a linear equation but uses $<$ , $>$ , $\leq$ , or $\geq$ . Solve it exactly as you would an equation, with one critical rule: if you multiply or divide both sides by a negative number, reverse the inequality sign.

Formula reference

\text{If } -a x < b \text{ then } x > -\frac{b}{a} \quad (\text{sign flips})

Worked example 3 Solving and graphing a linear inequality

Solve $3 - 2x \geq 7$ and show the solution on a number line.

$-2x \geq 7 - 3 = 4$ .
Divide by $-2$ (flip the sign): $x \leq -2$ .
On the number line, shade everything to the left of $-2$ with a closed circle at $-2$ (since $\leq$ includes equality).

Number line showing x is less than or equal to -2. The closed dot at -2 means -2 is included.

3. Graphing inequalities in two variables

To graph an inequality like $y < 2x + 1$ :

Draw the boundary line $y = 2x + 1$ . Use a dashed line for $<$ or $>$ (boundary not included) and a solid line for $\leq$ or $\geq$ .
Choose a test point not on the line (often $(0, 0)$ ).
If the test point satisfies the inequality, shade the side containing it; otherwise shade the other side.

Worked example 4 Graphing a two-variable inequality

Graph $y \geq -x + 3$ .

Boundary line: $y = -x + 3$ (solid, since $\geq$ ). Intercepts: $(0, 3)$ and $(3, 0)$ .
Test $(0, 0)$ : $0 \geq -0 + 3 = 3$ ? No. Shade the side away from the origin.
The solution region is above and including the line $y = -x + 3$ .

4. Parallel and perpendicular lines

Formula reference

Two lines are parallel if and only if they have the same gradient:

m_1 = m_2

Two lines are perpendicular if and only if the product of their gradients is $-1$ :

m_1 \times m_2 = -1 \quad \Longleftrightarrow \quad m_2 = -\frac{1}{m_1}

Worked example 5 Finding a parallel line

Find the equation of the line parallel to $y = 3x - 2$ that passes through $(1, 7)$ .

Parallel means same gradient: $m = 3$ .
Use point-gradient form: $y - 7 = 3(x - 1)$ .
$y = 3x + 4$ .

Worked example 6 Finding a perpendicular line

Find the equation of the line perpendicular to $y = \tfrac{2}{3}x + 5$ that passes through $(4, -1)$ .

Gradient of given line: $m_1 = \tfrac{2}{3}$ .
Perpendicular gradient: $m_2 = -\tfrac{3}{2}$ .
$y - (-1) = -\tfrac{3}{2}(x - 4)$ .
$y + 1 = -\tfrac{3}{2}x + 6$ , so $y = -\tfrac{3}{2}x + 5$ .

Practice

Fluency

Tier 1: basic skills

Solve $5x + 3 = 28$ .
Make $w$ the subject of $P = 2l + 2w$ .
Solve $4x - 7 > 9$ and state the solution as an inequality.
Solve $-3x \leq 12$ and show the solution on a number line.
State whether $y = 2x + 1$ and $y = 2x - 5$ are parallel, perpendicular, or neither.
State whether $y = 4x + 3$ and $y = -\tfrac{1}{4}x + 7$ are parallel, perpendicular, or neither.
Find the gradient of a line perpendicular to a line with gradient $5$ .
For the inequality $y > x + 2$ , state whether the boundary line is solid or dashed, and whether you shade above or below.

Reasoning

Tier 2: mixed practice

The formula $C = \tfrac{5}{9}(F - 32)$ converts Fahrenheit to Celsius. Find $F$ when $C = 20$ .
Solve $7 - 3x \geq 2x + 22$ and graph the solution on a number line.
Find the equation of the line parallel to $y = -2x + 6$ passing through $(3, 1)$ .
Find the equation of the line perpendicular to $y = \tfrac{1}{2}x - 3$ passing through $(6, 4)$ .
Graph the region satisfying $2x + y \leq 8$ in the first quadrant (where $x \geq 0$ , $y \geq 0$ ).
A rectangle has perimeter $P = 2l + 2w$ . If $P = 40$ and $l$ must be at least twice $w$ , write two inequalities and find the range of possible values for $w$ .

Reasoning

Tier 3: explain and apply

Prove that the triangle with vertices $A(0, 0)$ , $B(4, 2)$ , and $C(2, -4)$ contains a right angle by checking perpendicular gradients. State which angle is $90°$ .
The lines $y = kx + 3$ and $y = (2k - 5)x + 1$ are parallel. Find $k$ .
Find the equation of the perpendicular bisector of the segment from $A(2, 6)$ to $B(8, 2)$ .
A factory produces $x$ standard items and $y$ premium items. Each standard item needs $2$ hours; each premium item needs $5$ hours. The factory has at most $100$ hours. Write and graph the inequality, then find three integer combinations that use all $100$ hours.

Challenge

Reasoning

Harder reasoning

Show that the quadrilateral with vertices $P(1, 1)$ , $Q(5, 3)$ , $R(4, 5)$ , and $S(0, 3)$ is a parallelogram by proving both pairs of opposite sides are parallel. Is it also a rectangle? Justify using gradients.
Two inequality constraints are $x + 2y \leq 10$ and $3x + y \leq 12$ , with $x \geq 0$ and $y \geq 0$ . Find the vertices of the feasible region and determine which vertex maximises $P = 4x + 3y$ .
The line $L_1$ passes through $(a, 2a)$ and $(3, 7)$ , and the line $L_2$ passes through $(1, -1)$ and $(4, 5)$ . If $L_1$ is perpendicular to $L_2$ , find $a$ .

Answers

Answer key

Attempt the practice first. When you're ready to check, expand the answers below.

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Year 10 core - answers

Fluency

Tier 1: basic skills

$5x = 25$ , so $x = 5$ .
$2w = P - 2l$ , so $w = \dfrac{P - 2l}{2}$ .
$4x > 16$ , so $x > 4$ .
Divide by $-3$ and flip: $x \geq -4$ . Closed circle at $-4$ , shade right.
Both have gradient $2$ , so parallel.
$m_1 = 4$ , $m_2 = -\tfrac{1}{4}$ . Product $= 4 \times (-\tfrac{1}{4}) = -1$ , so perpendicular.
$m = -\tfrac{1}{5}$ .
Dashed line (strict inequality). Shade above (since $y >$ ).

Reasoning

Tier 2: mixed practice

$20 = \tfrac{5}{9}(F - 32)$ . Multiply both sides by $\tfrac{9}{5}$ : $36 = F - 32$ . $F = 68$ .
$7 - 3x \geq 2x + 22$ . $-5x \geq 15$ . Divide by $-5$ , flip: $x \leq -3$ . Closed circle at $-3$ , shade left.
Gradient $= -2$ . Through $(3, 1)$ : $y - 1 = -2(x - 3)$ , so $y = -2x + 7$ .
Given gradient $\tfrac{1}{2}$ , perpendicular gradient $= -2$ . Through $(6, 4)$ : $y - 4 = -2(x - 6)$ , so $y = -2x + 16$ .
Boundary: $y = -2x + 8$ . Solid line. Intercepts $(4, 0)$ and $(0, 8)$ . Test $(0,0)$ : $0 \leq 8$ true, shade below (toward origin). Region is the triangle with vertices $(0, 0)$ , $(4, 0)$ , $(0, 8)$ .
$2l + 2w = 40$ gives $l = 20 - w$ . Also $l \geq 2w$ : $20 - w \geq 2w$ , so $20 \geq 3w$ , $w \leq \tfrac{20}{3} \approx 6.67$ . Since $w > 0$ and $l > 0$ (so $w < 20$ ), the range is $0 < w \leq \tfrac{20}{3}$ .

Reasoning

Tier 3: explain and apply

$m_{AB} = \tfrac{2}{4} = \tfrac{1}{2}$ . $m_{AC} = \tfrac{-4}{2} = -2$ . Product: $\tfrac{1}{2} \times (-2) = -1$ . So $AB \perp AC$ , meaning the right angle is at $A$ .
Parallel means equal gradients: $k = 2k - 5$ , so $k = 5$ .
Midpoint: $\left(\tfrac{2+8}{2}, \tfrac{6+2}{2}\right) = (5, 4)$ . Gradient of $AB$ : $\tfrac{2-6}{8-2} = -\tfrac{2}{3}$ . Perpendicular gradient: $\tfrac{3}{2}$ . Equation: $y - 4 = \tfrac{3}{2}(x - 5)$ , so $y = \tfrac{3}{2}x - \tfrac{7}{2}$ .
$2x + 5y \leq 100$ , with $x \geq 0$ , $y \geq 0$ . Boundary intercepts: $(50, 0)$ and $(0, 20)$ . Three integer solutions using exactly $100$ hours: $(0, 20)$ , $(25, 10)$ , $(50, 0)$ .

Reasoning

Challenge

$m_{PQ} = \tfrac{3-1}{5-1} = \tfrac{1}{2}$ . $m_{SR} = \tfrac{5-3}{4-0} = \tfrac{1}{2}$ . So $PQ \parallel SR$ . $m_{QR} = \tfrac{5-3}{4-5} = -2$ . $m_{PS} = \tfrac{3-1}{0-1} = -2$ . So $QR \parallel PS$ . Both pairs of opposite sides are parallel, confirming a parallelogram. Check rectangle: $m_{PQ} \times m_{QR} = \tfrac{1}{2} \times (-2) = -1$ . Adjacent sides are perpendicular, so it is a rectangle.
Vertices of feasible region: $(0, 0)$ , $(4, 0)$ , $(\tfrac{14}{5}, \tfrac{18}{5})$ , $(0, 5)$ . Intersection of $x + 2y = 10$ and $3x + y = 12$ : solve to get $x = \tfrac{14}{5}$ , $y = \tfrac{18}{5}$ . Evaluate $P$ : at $(0,0)$ : $0$ ; at $(4,0)$ : $16$ ; at $(\tfrac{14}{5}, \tfrac{18}{5})$ : $\tfrac{56+54}{5} = \tfrac{110}{5} = 22$ ; at $(0,5)$ : $15$ . Maximum $P = 22$ at $(\tfrac{14}{5}, \tfrac{18}{5})$ .
$m_{L_2} = \tfrac{5-(-1)}{4-1} = 2$ . Perpendicular: $m_{L_1} = -\tfrac{1}{2}$ . $m_{L_1} = \tfrac{7-2a}{3-a}$ . Set equal: $\tfrac{7-2a}{3-a} = -\tfrac{1}{2}$ . Cross-multiply: $2(7-2a) = -(3-a)$ . $14 - 4a = -3 + a$ . $17 = 5a$ . $a = \tfrac{17}{5}$ .

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