Year 10A - Linear and non-linear simultaneous equations - Free practice

Year 10A Linear and Non-Linear Simultaneous Equations Worksheets & Practice

Solve pairs of equations graphically and algebraically, with worked solutions every time, free.

Quick answer

How do you solve linear and non-linear simultaneous equations?

Substitute one equation into the other to reduce the system to a single equation, then solve. For a line and a parabola, substitute the linear expression for y into the quadratic and solve the resulting equation, e.g. y = x + 2 and y = x² gives x² - x - 2 = 0, so x = 2 or x = -1. Graphically, solutions are intersection points.

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Worked examples

Year 10A linear and non-linear simultaneous equations: 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 · Solving two linear equations by elimination
Solve the simultaneous equations $3x + 2y = 16$ and $x - 2y = 0$ .
Show worked solution
Key idea: When the coefficients of one variable are opposites, add the equations to eliminate that variable.
1. Add the two equations: $3x + 2y + x - 2y = 16 + 0 \implies 4x = 16 \implies x = 4$
2. Substitute $x = 4$ into $x - 2y = 0$ : $4 - 2y = 0 \implies y = 2$
3. Check in $3x + 2y = 16$ : $3(4) + 2(2) = 12 + 4 = 16$ ✓
Answer: $x = 4,\ y = 2$
Example 2 · Solving a linear and quadratic pair by substitution
Solve the simultaneous equations $y = x + 1$ and $y = x^2 - x - 2$ .
Show worked solution
Key idea: Substitute the linear expression for $y$ into the quadratic, then solve the resulting equation.
1. Set the expressions equal: $x + 1 = x^2 - x - 2$
2. Rearrange to standard form: $0 = x^2 - 2x - 3$
3. Factorise: $0 = (x - 3)(x + 1) \implies x = 3 \text{ or } x = -1$
4. Find $y$ for each $x$ using $y = x + 1$ : $x = 3 \implies y = 4; x = -1 \implies y = 0$
5. Check $(3, 4)$ in $y = x^2 - x - 2$ : $9 - 3 - 2 = 4$ ✓. Check $(-1, 0)$ : $1 + 1 - 2 = 0$ ✓
Answer: $(3,\ 4)$ and $(-1,\ 0)$
Example 3 · Linear and circle equations: real-life problem
A circular fountain in a park has the equation $x^2 + y^2 = 25$ , where distances are in metres and the origin is the centre of the fountain. A straight path through the park follows the line $y = x + 1$ . Find the coordinates of the two points where the path crosses the edge of the fountain.
Show worked solution
Key idea: Substitute the linear expression for $y$ into the circle equation and solve the resulting quadratic.
1. Substitute $y = x + 1$ into $x^2 + y^2 = 25$ : $x^2 + (x + 1)^2 = 25$
2. Expand and simplify: $x^2 + x^2 + 2x + 1 = 25 \implies 2x^2 + 2x - 24 = 0 \implies x^2 + x - 12 = 0$
3. Factorise: $(x + 4)(x - 3) = 0 \implies x = -4 \text{ or } x = 3$
4. Find $y$ for each $x$ using $y = x + 1$ : $x = -4 \implies y = -3; x = 3 \implies y = 4$
5. Check $(-4, -3)$ : $(-4)^2 + (-3)^2 = 16 + 9 = 25$ ✓. Check $(3, 4)$ : $9 + 16 = 25$ ✓
Answer: The path crosses the fountain edge at $(-4,\ -3)$ and $(3,\ 4)$ .

Common mistakes

Where students usually lose marks

Forgetting to find the y-values

Solving $x = 3$ or $x = -1$ and writing the answer as " $x = 3$ or $x = -1$ ".

Each $x$ -value must be substituted back into one of the original equations to find its matching $y$ -value. A solution to simultaneous equations is always a coordinate pair $(x, y)$ .

Expanding brackets incorrectly when substituting

$(x + 1)^2 = x^2 + 1$ instead of $x^2 + 2x + 1$ .

Always expand $(x + a)^2$ as $x^2 + 2ax + a^2$ . Write out every term carefully before collecting like terms.

Rearranging the quadratic incorrectly

From $x + 1 = x^2 - x - 2$ , writing $x^2 - x - 3 = 0$ by forgetting to move all terms.

Move every term to one side in one step: subtract the entire left side from both sides to get $0 = x^2 - 2x - 3$ . Then check by expanding the factorised form to verify.

Assuming two equations always give two solutions

Expecting two solutions when the line is a tangent to the curve, and missing that a repeated root means only one intersection point.

A repeated root (discriminant equal to zero) means the line just touches the curve at exactly one point. Always check the discriminant or graph the pair to confirm the number of solutions.

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Your turn

Solve the simultaneous equations $y = 2x + 1$ and $y = x^2 - 1$ by substitution. The two $x$ -values of the solutions are the answers. What are the two $x$ -values?

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About this worksheet

Why we built it

Linear and non-linear simultaneous equations appear in Year 10A as students prepare for senior Methods maths. The key idea is finding where two equations are satisfied at the same time, whether both are straight lines, or one is a parabola, circle or hyperbola paired with a line. These worksheets build from graphical methods and digital tools through to systematic algebraic approaches, helping students understand why solutions exist, when there are two of them, and what it means when there are none.

What's covered

Sub-skills your student will practise

✓Solving two linear simultaneous equations by substitution and elimination
✓Solving a linear equation paired with a quadratic (parabola) by substitution
✓Solving a linear equation paired with a circle equation
✓Interpreting the number of solutions graphically (0, 1 or 2 intersection points)
✓Using graphing technology to find or verify solutions numerically
✓Systematic guess-check-and-refine with digital tools for non-linear pairs

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Frequently asked

Questions parents ask about linear and non-linear simultaneous equations

What makes non-linear simultaneous equations different from the linear ones students already know?

With two linear equations the graphs are straight lines and there is at most one intersection point. When one equation is a parabola or circle, the graphs can cross at two points, one point (tangent), or not at all, so students need to interpret the discriminant or graph to know how many solutions to expect.

Does my child need to solve these algebraically, or is using a graphing app enough?

The curriculum explicitly allows graphing tools and guess-check-and-refine with digital technology. Algebraic substitution is also expected for standard linear-quadratic pairs, so ideally students practise both and use technology to check their working.

Why does substitution work for a line paired with a parabola but elimination usually does not?

Elimination relies on adding or subtracting equations to cancel a variable, which works neatly when both equations are linear. When one equation is quadratic, rearranging the linear equation and substituting into the quadratic gives a single quadratic equation in one variable that can then be solved directly.

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