Solve Simultaneous Equations (Linear + Non-linear)

What this method is for

When a question gives you a pair of equations where one is a straight line and the other is a curve, this method finds the exact points where the two meet. In Add Math it belongs to the Form 4 Systems of Equations chapter.

A linear equation has every variable to the power one, such as $x - y = 1$ . A non-linear equation carries a squared term or a product of variables, such as $x^{2} + y^{2} = 25$ or $xy = 12$ .

Solving the two together gives the coordinates where the line cuts the curve. Because a straight line can cross a curve in two places, you should usually expect two solution pairs rather than one.

The reliable tool for this is substitution: use the simple linear equation to replace a variable inside the harder non-linear one, so the whole problem collapses into a single quadratic you already know how to solve.

When to reach for it

You recognise this type instantly: there are two equations and two unknowns, one of which is straight (highest power one) while the other contains $x^{2}$ , $y^{2}$ , or an $xy$ term. Command words are a strong signal too.

Phrases like "solve the simultaneous equations", "find the coordinates of intersection", or "find the point(s) where the line meets the curve" all point here.

Picture it graphically: you are being asked where a line and a curve touch or cross. If both equations were straight lines you would use elimination and expect a single meeting point, but the moment a square or product appears, switch to substitution and be ready for two answers.

On our lessons we teach students to name the linear equation first, because that is always the one to rearrange.

The steps

1
Label the two equations
Call the linear one (1) and the non-linear one (2). This keeps your working tidy and easy for the marker to follow.
2
Make a variable the subject
From the linear equation (1), make either $x$ or $y$ the subject. Choose whichever avoids fractions.
3
Substitute into the curve
Replace that variable in the non-linear equation (2). Now there is only one unknown left.
4
Form a quadratic
Expand the brackets, collect like terms, and rearrange into the form $ay^{2} + by + c = 0$ .
5
Solve the quadratic
Factorise if it is neat, otherwise use the quadratic formula. You should get two values.
6
Pair the answers
Put each value back into the linear equation (1) to find its matching partner, giving two coordinate pairs.
7
Check
Substitute each pair into the original non-linear equation to confirm both sides balance.

Worked example

Q1[5 marks]

Solve the simultaneous equations $x - y = 1$ and $x^{2} + y^{2} = 25$ .

Show worked solution

Label the equations: $x - y = 1$ is the linear equation (1), and $x^{2} + y^{2} = 25$ is the non-linear equation (2).

From (1), make $x$ the subject:

x = y + 1

Substitute $x = y + 1$ into (2):

(y + 1)^{2} + y^{2} = 25

Expand and collect like terms:

y^{2} + 2y + 1 + y^{2} = 25 \;\Rightarrow\; 2y^{2} + 2y - 24 = 0

Divide every term by $2$ to simplify:

y^{2} + y - 12 = 0

Factorise (two numbers that multiply to $-12$ and add to $+1$ are $+4$ and $-3$ ):

(y + 4)(y - 3) = 0 \;\Rightarrow\; y = -4 \;\text{or}\; y = 3

Pair each $y$ with its $x$ using $x = y + 1$ : when $y = 3$ , $x = 4$ ; when $y = -4$ , $x = -3$ .

The solutions are $(x, y) = (4, 3)$ and $(x, y) = (-3, -4)$ .

Check $(4, 3)$ : $4 - 3 = 1$ and $4^{2} + 3^{2} = 16 + 9 = 25$ . Both equations hold, so the pair is correct, and the same check works for $(-3, -4)$ .

Common pitfalls

Rearranging the non-linear equation first, this often forces square roots and messy algebra. Always make the subject from the linear equation.
Losing the second solution. A line usually cuts a curve twice, so stopping after one value throws away marks.
Sign and expansion slips when squaring a bracket, for example writing $(y + 1)^{2} = y^{2} + 1$ instead of $y^{2} + 2y + 1$ .
Substituting the answers back into the non-linear equation to find the partner, that can create extra invalid roots. Use the linear equation to pair them.
Writing the numbers without pairing them, so $x$ and $y$ are not matched into the correct coordinates.

How a teacher helps

Most students can quote "use substitution", yet lose marks at one precise spot: expanding the squared bracket, or pairing the wrong $x$ with the wrong $y$ at the end. In a one-to-one lesson our teacher watches your working live, catches the slip the moment it happens, and has you redo that single line correctly before it becomes a habit.

Because marking in this subject is analytic, method marks are within reach even when a final number is off, so we train you to lay out each substitution clearly. Our teachers are experienced, and lessons are online and taught in English, with the Malay terms shown alongside so both versions of the exam paper feel familiar.

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Frequently asked questions

Which equation should I rearrange first?

Always make a variable the subject of the linear equation, then substitute into the non-linear one. Rearranging the non-linear equation first usually creates square roots and far more work.

Why do I get two answers?

A straight line can cross a curve at two points, so a linear-plus-non-linear system normally has two solution pairs. Give both, unless a check shows one does not fit the context of the question.

Do I lose marks if I list the values without pairing them?

Usually yes. Marking is analytic, so you still earn method marks along the way, but the final accuracy mark needs each $x$ matched to its correct $y$ as a coordinate pair such as $(4, 3)$ .

Can I use elimination instead?

Elimination is best when both equations are linear. For one linear and one non-linear equation, substitution is the dependable method that reduces everything to a single quadratic.

Source:SRC-DSKP-EN

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

Solving simultaneous equations (one linear, one non-linear)