Systems of Equations Examples (Medium)

Q: What do I do when substitution produces an xy term?

Expand the bracket so the xyxyxy term becomes a plain x2x^{2}x2 or xxx term, then collect everything on one side. You will be left with an ordinary quadratic in a single variable to factorise.

What these examples cover

These medium Systems of Equations examples ask a little more of your algebra than the easy set. In the first, substitution creates an $xy$ term, so you have to expand and collect carefully before a quadratic appears.

In the second, a three-variable linear system will not fall to a single subtraction, you eliminate one letter to reach a pair of equations, then solve that pair. In the third, you translate a rectangle described in words into two equations, then solve and choose the answer that fits the context.

The numbers stay clean throughout. Cover the solution, attempt each question in full on paper, and only then check line by line.

The step where your working parts from ours is exactly where the real learning sits.

Worked examples

Work through all three. Attempt each fully before you read the solution, and watch how the same plan, reduce the system to one manageable equation, solve it, then substitute back, carries every question, even when the setup looks different.

Q1[4 marks]

Solve the simultaneous equations $2x+y=7$ and $x^{2}+xy=10$ .

Show worked solution

Make $y$ the subject of the linear equation, since its coefficient is $1$ :

y=7-2x

Substitute into $x^{2}+xy=10$ . Keep $x$ as a factor of the substituted bracket:

x^{2}+x(7-2x)=10 \;\Rightarrow\; x^{2}+7x-2x^{2}=10

Collect like terms. The result is $-x^{2}+7x-10=0$ ; multiply through by $-1$ so the leading term is positive, then factorise:

x^{2}-7x+10=0 \;\Rightarrow\; (x-2)(x-5)=0

So $x=2$ or $x=5$ . Substitute each back into $y=7-2x$ :

x=2 \Rightarrow y=3, \qquad x=5 \Rightarrow y=-3

Answer

The solutions are $(2,3)$ and $(5,-3)$ . Check the second in the non-linear equation: $(5)^{2}+(5)(-3)=25-15=10$ , as required.

Q2[5 marks]

Solve the system $x+y+z=6$ , $2x-y+z=5$ and $x+2y-z=6$ .

Show worked solution

Label the equations: $(1)\;x+y+z=6$ , $(2)\;2x-y+z=5$ , $(3)\;x+2y-z=6$ . The letter $z$ is easiest to remove.

Add $(1)$ and $(3)$ so the $+z$ and $-z$ cancel:

(1)+(3):\; 2x+3y=12 \quad (4)

Next remove $z$ a second way. Subtract $(2)$ from $(1)$ :

(1)-(2):\; -x+2y=1 \quad (5)

Equations $(4)$ and $(5)$ now involve only $x$ and $y$ . Make $x$ the subject of $(5)$ , giving $x=2y-1$ , and substitute into $(4)$ :

2(2y-1)+3y=12 \;\Rightarrow\; 7y-2=12 \;\Rightarrow\; y=2

Back-substitute $y=2$ into $x=2y-1$ , then use $(1)$ for $z$ :

x=2(2)-1=3, \qquad z=6-x-y=6-3-2=1

Answer

The solution is $x=3,\; y=2,\; z=1$ . Check $(2)$ : $2(3)-2+1=5$ , and $(3)$ : $3+2(2)-1=6$ , both correct.

Q3[4 marks]

A rectangular plot of land has a perimeter of $26$ m and an area of $40$ m $^{2}$ . Find its length and width.

Show worked solution

Let the length be $x$ m and the width be $y$ m, with $x>y$ . Turn each fact into an equation.

Perimeter $2(x+y)=26$ gives the linear equation, and area gives the non-linear one:

x+y=13, \qquad xy=40

Make $y$ the subject of the linear equation, $y=13-x$ , and substitute into $xy=40$ :

x(13-x)=40 \;\Rightarrow\; 13x-x^{2}=40

Rearrange into a quadratic and factorise:

x^{2}-13x+40=0 \;\Rightarrow\; (x-5)(x-8)=0

So $x=5$ or $x=8$ . Because the length is the longer side, take $x=8$ ; then $y=13-8=5$ .

Answer

The plot is $8$ m long and $5$ m wide. Check both facts: perimeter $2(8+5)=26$ m and area $8\times 5=40$ m $^{2}$ , as stated.

Q4[4 marks]

Two numbers have a sum of $11$ and the sum of their squares is $61$ . Find the two numbers.

Show worked solution

Rather than substituting one equation into the other, use the identity $(x+y)^{2}=x^{2}+2xy+y^{2}$ to reach the product $xy$ directly from the two sums you are given:

(11)^{2}=61+2xy \;\Rightarrow\; xy=30

With $x+y=11$ and $xy=30$ now known, $x$ and $y$ are the two roots of the quadratic $t^{2}-(\text{sum})t+(\text{product})=0$ :

t^{2}-11t+30=0 \;\Rightarrow\; (t-5)(t-6)=0

So $t=5$ or $t=6$ , and since $x+y=11$ , each root pairs with the other.

Answer

The two numbers are $5$ and $6$ . Check: $5+6=11$ and $5^{2}+6^{2}=25+36=61$ , as given.

Q5[4 marks]

Solve the simultaneous equations $x+y=5$ and $\frac{1}{x}+\frac{1}{y}=\frac{5}{6}$ .

Show worked solution

Combine the two fractions on the left over a common denominator; the numerator becomes $x+y$ , which you already know:

\frac{1}{x}+\frac{1}{y}=\frac{y+x}{xy}=\frac{5}{6}

Substitute $x+y=5$ into the numerator, then solve for $xy$ :

\frac{5}{xy}=\frac{5}{6} \;\Rightarrow\; xy=6

With $x+y=5$ and $xy=6$ , $x$ and $y$ are the roots of:

t^{2}-5t+6=0 \;\Rightarrow\; (t-2)(t-3)=0

Answer

The solutions are $(x,y)=(2,3)$ or $(3,2)$ . Check: $2+3=5$ and $\frac{1}{2}+\frac{1}{3}=\frac{3}{6}+\frac{2}{6}=\frac{5}{6}$ , as required.

Q6[4 marks]

Solve the simultaneous equations $2x+3y=12$ and $xy=6$ .

Show worked solution

Make $y$ the subject of the linear equation:

y=\frac{12-2x}{3}

Substitute into $xy=6$ and clear the fraction:

x\left(\frac{12-2x}{3}\right)=6 \;\Rightarrow\; 12x-2x^{2}=18

Rearrange and divide through by $-2$ ; the result is a perfect square:

x^{2}-6x+9=0 \;\Rightarrow\; (x-3)^{2}=0 \;\Rightarrow\; x=3

This repeated root gives only one value of $x$ . Substitute back to find $y$ :

y=\frac{12-2(3)}{3}=2

Answer

The only solution is $(3,2)$ . A repeated root means the line $2x+3y=12$ touches the curve $xy=6$ at exactly one point, rather than crossing it twice.

Check: $2(3)+3(2)=12$ and $3\times2=6$ .

Q7[3 marks]

Aiman is now $3$ times as old as his brother. In $8$ years, Aiman will be twice as old as his brother.

Find their present ages.

Show worked solution

Let Aiman's present age be $x$ and his brother's be $y$ . Turn the first sentence into an equation:

x=3y \quad (1)

In $8$ years both ages increase by $8$ ; turn the second sentence into an equation:

x+8=2(y+8) \;\Rightarrow\; x=2y+8 \quad (2)

Both $(1)$ and $(2)$ equal $x$ , so equate them and solve for $y$ , then find $x$ :

3y=2y+8 \;\Rightarrow\; y=8, \qquad x=3(8)=24

Answer

Aiman is $24$ and his brother is $8$ . Check: in $8$ years they are $32$ and $16$ , and $32=2\times16$ , as required.

Q8[3 marks]

The sum of three numbers is $28$ . The second number is twice the first, and the third number is $3$ more than the second.

Find the three numbers.

Show worked solution

Let the first number be $x$ . Write the second and third numbers in terms of $x$ :

y=2x, \qquad z=2x+3

Substitute both into $x+y+z=28$ and solve for $x$ :

x+2x+(2x+3)=28 \;\Rightarrow\; 5x+3=28 \;\Rightarrow\; x=5

Back-substitute to find $y$ and $z$ :

y=2(5)=10, \qquad z=2(5)+3=13

Answer

The three numbers are $5$ , $10$ , and $13$ . Check: $5+10+13=28$ , and $13$ is $3$ more than $10$ , which is twice $5$ .

Across all three, the shape of the work is the same. Reduce the system to a single equation you can solve, by substitution, by elimination, or by translating words into algebra, solve that equation, then substitute back and, in a word problem, choose the root the context allows.

Hold that plan and medium questions stop feeling like a step up.

Key method points

These three examples show how the same core method stretches to cover harder set-ups. Keep the following points in mind as you practise more.

When substitution creates an $xy$ term, expand fully and collect like terms before you try to factorise.
If the leading coefficient comes out negative, multiply the whole equation by $-1$ so factorising is cleaner.
In a three-variable system, eliminate the same letter twice to reach two equations in two unknowns, then solve that pair.
For a word problem, name the variables, turn each stated fact into one equation, and state any condition such as length $>$ width.
A quadratic may give two mathematically valid roots, but a real-world context often keeps only one, always check which fits.
Because marking is analytic, a correct set of equations and a clear method line earn marks even if the final arithmetic slips.

How a teacher helps

The medium questions are where good habits start to pay off, and where a rushed line quietly costs marks, a sign lost while collecting an $xy$ term, or the wrong root kept in a word problem. In a one-to-one lesson our teacher slows the moment down, names the step you are on, and shows how to check it before moving forward.

Because our teachers are experienced, you learn a method you can repeat under exam pressure, not a trick for one question. Lessons are taught in English, while SPM papers are set in both Malay and English, so the notation stays familiar either way.

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Frequently asked questions

What do I do when substitution produces an

xy

term?

Expand the bracket so the $xy$ term becomes a plain $x^{2}$ or $x$ term, then collect everything on one side. You will be left with an ordinary quadratic in a single variable to factorise.

In a three-variable system, which letter should I eliminate first?

Choose the letter that cancels most easily, often the one whose coefficients are $+1$ and $-1$ in two equations, so a single addition removes it. Eliminate it twice to reach two equations in two unknowns.

How do I turn a word problem into equations?

Name each unknown with a letter, then write one equation for every stated fact. A rectangle's perimeter and area give one linear and one non-linear equation, which you solve together.

Why do I sometimes reject a root in a word problem?

Because the algebra does not know the context. A negative length or a width larger than the length has no meaning here, so you keep only the root that fits the situation described.

Source:SRC-DSKP-EN

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

Systems of Equations, Worked Examples (medium)

What these examples cover

Worked examples

Key method points

How a teacher helps

Frequently asked questions

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