Year 9 - Linear equations and graphs - Free practice

Year 9 Linear Equations & Graphs Worksheets

Sketch from gradient/intercept, solve from any form.

Quick answer

How do you sketch linear graphs and solve linear equations?

A linear equation in the form y = mx + c has gradient m and y-intercept c. Plot the y-intercept, use the gradient to find a second point, then draw the line. To solve a linear equation, use inverse operations to isolate the pronumeral. For example, 2x + 3 = 11 gives x = 4.

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Worked examples

Year 9 linear equations and graphs: worked examples

Three examples, from a standard question to a worded problem. Have a go at each one before you open the solution.

Example 1 · Sketching a line from gradient-intercept form
Sketch the line $y = 2x - 3$ by identifying the gradient and $y$ -intercept.
Show worked solution
Key idea: In $y = mx + c$ , $m$ is the gradient and $c$ is the $y$ -intercept. Plot the $y$ -intercept, then use the gradient to find a second point.
1. Identify $m = 2$ and $c = -3$ , so the $y$ -intercept is $(0, -3)$ .
2. From $(0, -3)$ , apply the gradient: rise $= 2$ , run $= 1$ , giving the second point $(1, -1)$ .
3. Plot $(0, -3)$ and $(1, -1)$ and draw a straight line through them.
Answer: A straight line passing through $(0, -3)$ and $(1, -1)$ .
Example 2 · Solving a linear equation algebraically
Solve $3(2x - 4) = 2x + 8$ .
Show worked solution
Key idea: Expand brackets first, then collect like terms and isolate $x$ .
1. Expand the left side: $6x - 12 = 2x + 8$ .
2. Subtract $2x$ from both sides: $4x - 12 = 8$ .
3. Add $12$ to both sides: $4x = 20$ .
4. Divide both sides by $4$ : $x = 5$ .
5. Check: $3(2(5) - 4) = 3(6) = 18$ and $2(5) + 8 = 18$ . ✓
Answer: $x = 5$
Example 3 · Finding where two lines intersect
A phone plan from Company A costs $20 per month plus $0.10 per minute of calls. Company B charges $35 per month with no call costs. Write an equation for the total monthly cost of each plan in terms of $m$ minutes, then find the number of minutes at which both plans cost the same.
Show worked solution
Key idea: Write a linear equation for each plan, then set them equal and solve for $m$ .
1. Company A: $C = 0.10m + 20$ . Company B: $C = 35$ .
2. Set the costs equal: $0.10m + 20 = 35$ .
3. Subtract $20$ from both sides: $0.10m = 15$ .
4. Divide both sides by $0.10$ : $m = 150$ .
5. Check: $0.10(150) + 20 = 15 + 20 = 35$ . ✓
Answer: The plans cost the same at $150$ minutes per month.

Common mistakes

Where students usually lose marks

Swapping gradient and intercept

For $y = 2x - 3$ , student writes $y$ -intercept $= 2$ and gradient $= -3$ .

In $y = mx + c$ , the gradient is always the coefficient of $x$ (the number multiplying $x$ ), and the $y$ -intercept is the constant term $c$ . Check by substituting $x = 0$ : the result should equal $c$ .

Applying gradient in the wrong direction

Gradient $= 2$ is used as run $= 2$ , rise $= 1$ , giving an incorrect second point.

Gradient $= \dfrac{\text{rise}}{\text{run}}$ , so a gradient of $2 = \dfrac{2}{1}$ means move $2$ units up and $1$ unit right. Always write the gradient as a fraction to identify rise over run clearly.

Forgetting to expand brackets before solving

$3(2x - 4) = 2x + 8 \Rightarrow 6x - 4 = 2x + 8$ (only the $2x$ is multiplied by $3$ ).

Every term inside the brackets must be multiplied by the factor outside. Write out each multiplication separately: $3 \times 2x = 6x$ and $3 \times (-4) = -12$ , giving $6x - 12$ .

Adding instead of subtracting when rearranging

$4x - 12 = 8 \Rightarrow 4x = 8 - 12 = -4$ (subtracting $12$ from the right instead of adding).

To remove $-12$ from the left side, add $12$ to both sides: $4x - 12 + 12 = 8 + 12$ , so $4x = 20$ . Always perform the same operation on both sides.

Try one yourself

Your turn

Solve the equation $3x - 7 = 2x + 5$ for $x$ .

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About this worksheet

Why we built it

Year 9 students should be able to look at any linear equation - in any form - and sketch it. These worksheets practise that across the forms they'll meet (y = mx + c, ax + by = c, x = k, y = k) and tie sketching back to solving.

What's covered

Sub-skills your student will practise

✓Sketching from y = mx + c using gradient and intercept
✓Sketching from two known points
✓Sketching horizontal and vertical lines
✓Solving linear equations algebraically
✓Finding the intersection of two lines

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Frequently asked

Questions parents ask about linear equations and graphs

What forms can a linear equation take?

y = mx + c (gradient-intercept form), ax + by + c = 0 (general form), and special cases like x = k (vertical line) and y = k (horizontal line). Year 9 expects students to sketch any of them.

How do students sketch from gradient-intercept form?

Plot the y-intercept (the 'c' value) on the y-axis. Then use the gradient (rise/run) to find a second point. Draw a straight line through both points.

How do students find the intersection of two lines?

Set the equations equal to each other (or use substitution / elimination), solve for x, then substitute back to find y. The intersection point is (x, y). This is a Year 9 preview of simultaneous equations.

Are vertical and horizontal lines linear equations?

Yes - y = k is a horizontal line at height k, and x = k is a vertical line at position k. Vertical lines have an undefined gradient (rise over zero), which Year 9 introduces formally.

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