Solving simultaneous equations without slips

A simultaneous equations question in Add Math almost always pairs one linear equation with one non-linear equation, usually a quadratic in two variables, or an equation with a squared term such as $x^{2}+y^{2}=r^{2}$. Solving the pair means finding every value of $x$ and $y$ that satisfies both equations at the same time, not just one of them.

Graphically, that is exactly what it looks like: the points where a straight line crosses a curve.

Because one equation is linear, the standard approach is substitution rather than elimination. You rearrange the linear equation to make one variable the subject, then replace that variable everywhere it appears in the non-linear equation.

What is left is a single equation in one variable, almost always a quadratic, which you already know how to solve. The two roots of that quadratic usually correspond to two different points where the line and the curve meet.

• Linear + quadratic in one variable, e.g. $y=2x-1$ and $y=x^{2}-3x+4$.
• Linear + a squared-sum equation, e.g. $x+y=7$ and $x^{2}+y^{2}=25$.
• Occasionally two linear equations, solved the same way but ending in a linear equation, not a quadratic, in the last step.

The method has a fixed sequence of moves. Working through them in order, rather than jumping ahead, is what keeps the algebra manageable.

1. 1

   Isolate a variable

   Rearrange the linear equation to make $x$ or $y$ the subject, whichever gives the simpler expression to substitute.

2. 2

   Substitute into the non-linear equation

   Replace every occurrence of that variable in the second equation with the expression from step 1.

3. 3

   Expand and simplify

   Multiply out any brackets carefully, then collect terms into a single equation in one variable, usually in the form $ax^{2}+bx+c=0$.

4. 4

   Solve the resulting equation

   Factorise where possible, or apply the quadratic formula. Expect two roots unless the question states otherwise.

5. 5

   Substitute back for the paired value

   Put each root back into the rearranged linear equation from step 1 to find its matching value of the other variable.

Where factorising is not obvious, the quadratic formula on the formula list handles it directly:

$$x = \dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}$$

Most marks lost on this topic are not lost to not knowing the method, they are lost to small slips that happen while carrying it out. A handful of them come up again and again.

• Stopping after one pair of values. A quadratic step almost always gives two roots, and each root usually pairs with its own value of the other variable, four numbers, not two, unless the question or context rules one pair out.
• Substituting into the harder equation. Substituting the rearranged linear expression into the non-linear equation is the standard direction; going the other way around usually creates messier algebra and more chances to slip.
• Losing a sign when expanding brackets. $(7-x)^{2}$ is not $49-x^{2}$, every term needs expanding in full, and a rushed step here is where the discriminant quietly changes.
• Pairing the wrong values together at the end. Each root of the quadratic has its own matching value from the linear equation, mixing values from different roots gives a pair that does not actually satisfy either original equation.

There is a habit that catches almost every slip above before it reaches the final answer: substituting each solution pair back into both original equations, not the rearranged or substituted versions.

1. 1

   Take one pair of values

   Use the $x$ and $y$ values from one solution, exactly as found.

2. 2

   Test the first original equation

   Substitute both values in and confirm both sides match.

3. 3

   Test the second original equation

   Do the same with the second equation. If either side does not match, the error is somewhere in the working, not in the check.

4. 4

   Repeat for the other pair

   Every solution pair should be checked, not just the first one found.

Solve simultaneously: $y=x+1$ and $x^{2}+y^{2}=25$.

1. Substitute $y=x+1$ into the second equation: $x^{2}+(x+1)^{2}=25$.
2. Expand fully: $x^{2}+x^{2}+2x+1=25$, which simplifies to $2x^{2}+2x-24=0$.
3. Divide through by 2 and factorise: $x^{2}+x-12=0$ gives $(x+4)(x-3)=0$, so $x=-4$ or $x=3$.
4. Substitute each root back into $y=x+1$: when $x=-4$, $y=-3$; when $x=3$, $y=4$.

$$(x,y) = (-4,-3) \text{ or } (3,4)$$

Checking both pairs against the original equation $x^{2}+y^{2}=25$: $(-4)^{2}+(-3)^{2}=16+9=25$, and $3^{2}+4^{2}=9+16=25$. Both pairs hold.

Simultaneous equations reward a fixed routine, but the routine only becomes reliable once you have worked through enough examples to spot, in your own handwriting, where a sign or a bracket tends to slip. A teacher working through your solution line by line can point to that exact step and show a habit that prevents it, something that is harder to do from a printed answer key alone.

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

If simultaneous equations, or the wider Add Math syllabus, is a sticking point, a one-hour paid trial class at the teacher's own rate (from RM50 per hour, depending on experience) is a reasonable place to start. We will not promise a particular grade, but we can help checking your own working become as automatic as doing it in the first place.