KBAT: Simultaneous Equations in Context

Q: Which equation should I substitute from, the linear or the non-linear one?

Always rearrange the linear equation and substitute into the other. Here making y=17−xy=17-xy=17−x the subject and putting it into xy=66xy=66xy=66 gives one clean quadratic. Substituting from the non-linear equation usually produces fractions or extra terms and invites mistakes.

Q: Both roots gave the same rectangle, is that a mistake?

No. x=11, y=6x=11,\,y=6x=11,y=6 and x=6, y=11x=6,\,y=11x=6,y=11 describe one rectangle measuring 11 m by 6 m; the two roots simply swap which side you call length and width. The guideline that length must exceed width by more than 4 m then tells you to take length 11 m and width 6 m.

Q: How does the discriminant show something is impossible?

The discriminant b2−4acb^{2}-4acb2−4ac tells you whether a quadratic has real solutions. For area 80 with perimeter 34, it is 289−320=−31289-320=-31289−320=−31, which is negative, so x2−17x+80=0x^{2}-17x+80=0x2−17x+80=0 has no real root, and no such rectangle exists. A negative discriminant is a clean way to prove a scenario cannot happen.

Q: Can I still score if I only find one dimension?

Yes. Add Math Paper 2 is 2 hours 30 minutes and 100 marks, and marking is analytic, method marks are awarded line by line. Correct equations, a valid substitution and one solved value can earn most of the marks even before you finish pairing and interpreting.

What makes this a KBAT question

A routine systems question gives you a linear and a non-linear equation and says 'solve simultaneously'. A KBAT question gives you a design brief, a perimeter, an area, a rule about dimensions, and expects you to write the two equations yourself, solve them, then judge the answers against the situation.

That is Kemahiran Berfikir Aras Tinggi, higher-order thinking: substitution and the quadratic formula are ordinary Form 4 work, but choosing the variables, forming the pair of equations, and deciding which root is meaningful (or whether any solution exists) are steps no one spells out. In Add Math this rewards students who can connect a system of equations to the shape or scenario behind it, and who read the discriminant as a statement about whether something is possible at all, not just as a number in a formula.

One worked problem, in the style of Paper 2

This is an original question written in the style of SPM Paper 2. Try it yourself before reading the solution.

Q1[8 marks]

A landscape designer plans a rectangular reflecting pool. To fit the courtyard the perimeter must be 34 m, and the water surface must cover 66 m $^{2}$ .

(a) Taking the length as $x$ m and the width as $y$ m, form two equations and solve them simultaneously to find the possible dimensions. (b) A guideline requires the length to exceed the width by more than 4 m.

State the dimensions that satisfy this, with justification. (c) The designer wonders whether a pool with the same 34 m perimeter could instead enclose 80 m $^{2}$ .

By reasoning about the discriminant, determine whether this is possible.

Show worked solution

Understand. Perimeter gives a linear equation in $x$ and $y$ ; area gives a non-linear one.

We solve them together, choose the dimensions that fit the guideline, then test whether 80 m $^{2}$ is achievable with the same perimeter.

Plan. From the perimeter, $2(x+y)=34\Rightarrow x+y=17$ .

From the area, $xy=66$ . Substitute $y=17-x$ into the area equation to get one quadratic, then solve.

Execute and check. (a) The two equations are:

x+y=17 \qquad\text{and}\qquad xy=66

Substitute $y=17-x$ into $xy=66$ :

x(17-x)=66 \;\Rightarrow\; 17x-x^{2}=66 \;\Rightarrow\; x^{2}-17x+66=0

Factorise: $(x-11)(x-6)=0$ , so $x=11$ or $x=6$ . The matching widths are $y=17-11=6$ or $y=17-6=11$ .

Either way the pool measures 11 m by 6 m.

(b) The guideline requires length $-$ width $>4$ . With length 11 m and width 6 m, $11-6=5>4$ ✓, so the length is 11 m and the width is 6 m.

(The assignment length 6, width 11 would make the length shorter than the width, so it is rejected as a labelling of length and width.)

x^{2}-17x+80=0

Test the discriminant $b^{2}-4ac$ :

(-17)^{2}-4(1)(80)=289-320=-31<0

The discriminant is negative, so the equation has no real solution: no rectangle with perimeter 34 m can enclose 80 m $^{2}$ . It is impossible.

Check. For a fixed perimeter the largest area is a square: side $17\div 2=8.5$ m gives area $8.5^{2}=72.25$ m $^{2}$ .

Since $66<72.25$ , the pool in (a) is comfortably possible, while $80>72.25$ exceeds the maximum, consistent with the negative discriminant.

Finding a sensible first step

When a systems question hides behind a scenario, the reliable first step is to name the two unknowns clearly and turn each piece of information into one equation. Here 'length $x$ , width $y$ ' is the choice; the perimeter becomes $x+y=17$ and the area becomes $xy=66$ .

Notice which equation is linear, that is the one to rearrange, because making $y$ the subject and substituting always collapses the pair into a single quadratic. Resist starting with the non-linear equation; substituting from it is messier and invites errors.

Once you have one quadratic in $x$ , the rest is standard: factorise or use the formula, then match each $x$ to its $y$ . Naming variables and choosing the linear equation to substitute from turns an unfamiliar brief into a routine you have practised many times.

What markers reward

Marking is analytic, so method marks are awarded line by line. On a simultaneous-equations context question a marker looks for:

Two correct equations formed from the words, $x+y=17$ (linear) and $xy=66$ (non-linear).
Substitution from the linear equation to reach a single quadratic, $x^{2}-17x+66=0$ .
A valid solving method, factorising $(x-11)(x-6)=0$ or the quadratic formula.
Both roots paired with their matching value, and the dimensions stated as 11 m by 6 m.
The guideline applied with a reason, length exceeds width by 5 m, which is more than 4 m.
A discriminant argument for part (c): $289-320=-31<0$ , therefore no real solution and 80 m $^{2}$ is impossible.

How a teacher helps

Context systems questions improve fastest when a student practises translating words into equations and then defending each answer. In a one-to-one lesson our teachers ask you to define the two variables before writing anything, to substitute from the linear equation as a habit, and to say in words why a root is kept or rejected.

We spend time on the discriminant as a yes-or-no test for feasibility, because that reasoning is where KBAT marks live. Teachers at spmaddmath.com.my are experienced, and lessons are online and taught in English.

Message us on WhatsApp to arrange a one-hour paid class from RM50/hr at the teacher's rate.

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Frequently asked questions

Which equation should I substitute from, the linear or the non-linear one?

Always rearrange the linear equation and substitute into the other. Here making $y=17-x$ the subject and putting it into $xy=66$ gives one clean quadratic.

Substituting from the non-linear equation usually produces fractions or extra terms and invites mistakes.

Both roots gave the same rectangle, is that a mistake?

No. $x=11,\,y=6$ and $x=6,\,y=11$ describe one rectangle measuring 11 m by 6 m; the two roots simply swap which side you call length and width.

The guideline that length must exceed width by more than 4 m then tells you to take length 11 m and width 6 m.

How does the discriminant show something is impossible?

The discriminant $b^{2}-4ac$ tells you whether a quadratic has real solutions. For area 80 with perimeter 34, it is $289-320=-31$ , which is negative, so $x^{2}-17x+80=0$ has no real root, and no such rectangle exists.

A negative discriminant is a clean way to prove a scenario cannot happen.

Can I still score if I only find one dimension?

Yes. Add Math Paper 2 is 2 hours 30 minutes and 100 marks, and marking is analytic, method marks are awarded line by line.

Correct equations, a valid substitution and one solved value can earn most of the marks even before you finish pairing and interpreting.

Source:SRC-DSKP-EN SRC-FORMAT

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