Simultaneous equations: one linear, one non-linear

You have solved simultaneous equations before, back when both equations were linear, two straight lines, and you found where they cross. In Add Math the picture changes in one important way: one equation is still linear, but the other is non-linear, it might contain $x^{2}$, $y^{2}$, $xy$, or a fraction.

Geometrically, you are now finding where a straight line meets a curve.

That single change has two consequences. First, elimination by adding or subtracting the equations no longer works cleanly, so substitution becomes the reliable method.

Second, because a line can cut a curve in two places, touch it at one, or miss it entirely, you can expect two solutions, one, or none, and the answer usually comes as pairs of $x$ and $y$ values.

There is one route that works almost every time, and doing the steps in this order avoids most of the trouble.

1. 1

   Make a subject from the linear equation

   Always rearrange the linear equation (never the non-linear one) to get $x$ or $y$ on its own. It is simpler and keeps the algebra clean.

2. 2

   Substitute into the non-linear equation

   Replace that variable in the non-linear equation. You now have one equation in one unknown.

3. 3

   Simplify to a quadratic and solve

   Expand carefully, collect everything on one side to get $=0$, then factorise or use the quadratic formula.

4. 4

   Back-substitute into the linear equation

   Put each value back into the linear equation, not the curve, to find its partner. It is faster and avoids introducing extra false answers.

5. 5

   Write the answers as matched pairs

   State them as coordinate pairs so each $x$ stays with the correct $y$.

Solve the simultaneous equations below.

$$x + y = 6 \qquad\text{and}\qquad x^{2} + y^{2} = 20$$

Step 1, make $y$ the subject of the linear equation:

$$y = 6 - x$$

Step 2, substitute this into the non-linear equation:

$$x^{2} + (6 - x)^{2} = 20$$

Step 3, expand and simplify to a quadratic. Remember that $(6-x)^{2} = 36 - 12x + x^{2}$:

$$x^{2} + 36 - 12x + x^{2} = 20 \;\Rightarrow\; 2x^{2} - 12x + 16 = 0$$

Divide through by $2$, then factorise:

$$x^{2} - 6x + 8 = 0 \;\Rightarrow\; (x-2)(x-4) = 0 \;\Rightarrow\; x = 2 \text{ or } x = 4$$

Step 4, back-substitute each into $y = 6 - x$. When $x = 2$, $y = 4$; when $x = 4$, $y = 2$.

Step 5, write the matched pairs: the solutions are $(2, 4)$ and $(4, 2)$.

A quick check confirms it: $2^{2} + 4^{2} = 4 + 16 = 20$, and $2 + 4 = 6$. Both original equations hold.

That final check costs a few seconds and catches almost every arithmetic slip.

The method is short, so most lost marks come from a handful of slips in the algebra rather than from not knowing what to do.

• Expanding a square wrongly: $(6-x)^{2}$ is $36 - 12x + x^{2}$, never $36 - x^{2}$. The middle term is where marks vanish.
• Rearranging the non-linear equation instead of the linear one, which usually creates a much messier substitution.
• Back-substituting into the curve, which can appear to give extra answers, put values back into the linear equation.
• Losing the pairing, so an $x$ is matched to the wrong $y$. Write pairs, not two separate lists.
• Stopping at $x$ and forgetting to find $y$, the question asks for both.

This is a topic where the idea is quick to grasp but the execution is easy to fumble, a squared bracket expanded wrongly, a sign dropped, a pair mismatched. Working one-to-one, a teacher can watch your algebra line by line, catch the exact step where the slip happens, and drill the safe order until substitution feels routine rather than risky.

Our teachers are experienced; lessons are online and taught in English, while the SPM papers are set bilingually in Bahasa Melayu and English.

If simultaneous equations are costing you marks you should be keeping, you are welcome to begin with a one-hour paid trial class at the teacher's own rate (from RM50 per hour, depending on experience). We won't promise a grade, but we can help you make this one of the steadiest few marks on your paper.