Systems of Equations Examples (Easy)

Q: When do I use substitution and when do I use elimination?

Use substitution when one equation is non-linear, make a letter the subject of the linear equation and substitute it in. Use elimination for a system where every equation is linear, especially three-variable systems.

Q: How do I pair up the values correctly at the end?

Substitute each xxx back into the linear equation, never the non-linear one. The linear equation gives exactly one yyy for each xxx, so there is no risk of mismatching.

What these examples cover

These easy Systems of Equations examples build the two moves the whole chapter rests on. The first is substitution: when you have one linear equation and one non-linear equation, make a single letter the subject of the linear equation and substitute it into the non-linear one, leaving a quadratic to solve.

The second is elimination: for a linear system in three variables, add or subtract equations in pairs to remove one letter at a time. Every question here uses small, clean numbers so you can follow each line without a calculator getting in the way.

Cover the solution, attempt the question in full on paper, then check line by line. Where your working parts from ours is exactly where the learning is.

Worked examples

Work through all four. Attempt each fully before you read the matching solution, and notice how the same discipline, make one letter the subject, substitute carefully, keep brackets, and always substitute back, runs through every one.

Q1[3 marks]

Solve the simultaneous equations $x+y=5$ and $xy=6$ .

Show worked solution

Make $y$ the subject of the linear equation, because it is the simpler of the two to rearrange:

y=5-x

Substitute this into the non-linear equation $xy=6$ and expand:

x(5-x)=6 \;\Rightarrow\; 5x-x^{2}=6

Rearrange into a quadratic equal to zero, then factorise:

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

So $x=2$ or $x=3$ . Substitute each back into $y=5-x$ to pair up the values:

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

Answer

The solutions are $(x,y)=(2,3)$ and $(3,2)$ . Check the first: $2+3=5$ and $2\times 3=6$ , both correct.

Q2[3 marks]

Find the coordinates of the points where the line $y=x+1$ meets the curve $y=x^{2}-1$ .

Show worked solution

At a meeting point both equations give the same $y$ , so set the two right-hand sides equal:

x^{2}-1=x+1

Bring every term to one side to form a quadratic, then factorise:

x^{2}-x-2=0 \;\Rightarrow\; (x-2)(x+1)=0

So $x=2$ or $x=-1$ . Use the line $y=x+1$ to find each matching $y$ :

x=2 \Rightarrow y=3, \qquad x=-1 \Rightarrow y=0

Answer

The line meets the curve at $(2,3)$ and $(-1,0)$ . Check on the curve: $(2)^{2}-1=3$ and $(-1)^{2}-1=0$ , which match.

Q3[4 marks]

Solve the system $x+y+z=6$ , $x-y+z=2$ and $x+y-z=0$ .

Show worked solution

Label the equations to keep track: $(1)\;x+y+z=6$ , $(2)\;x-y+z=2$ , $(3)\;x+y-z=0$ . Subtract $(2)$ from $(1)$ ; the $x$ and $z$ terms cancel and leave only $y$ :

(1)-(2):\; 2y=4 \;\Rightarrow\; y=2

Now subtract $(3)$ from $(1)$ ; this time the $x$ and $y$ terms cancel and leave only $z$ :

(1)-(3):\; 2z=6 \;\Rightarrow\; z=3

Substitute $y=2$ and $z=3$ into $(1)$ to find $x$ :

x+2+3=6 \;\Rightarrow\; x=1

Answer

The solution is $x=1,\; y=2,\; z=3$ . Check the equations you did not solve into: $(2)\;1-2+3=2$ and $(3)\;1+2-3=0$ , both correct.

Q4[4 marks]

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

Show worked solution

The linear equation already has $y$ as the subject, so substitute $y=x-1$ straight into the non-linear equation:

x^{2}+(x-1)^{2}=25

Expand the bracket carefully, keeping the middle term, and collect like terms:

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

Divide through by $2$ to simplify, then factorise:

x^{2}-x-12=0 \;\Rightarrow\; (x-4)(x+3)=0

So $x=4$ or $x=-3$ . Substitute each back into $y=x-1$ :

x=4 \Rightarrow y=3, \qquad x=-3 \Rightarrow y=-4

Answer

The solutions are $(4,3)$ and $(-3,-4)$ . Check the second: $(-3)^{2}+(-4)^{2}=9+16=25$ , as required.

Q5[2 marks]

Solve the simultaneous equations $2x+y=9$ and $x-y=3$ .

Show worked solution

Both equations are linear, so add them directly to eliminate $y$ :

(2x+y)+(x-y)=9+3 \;\Rightarrow\; 3x=12

Divide to find $x$ , then substitute back into either equation to find $y$ :

x=4 \;\Rightarrow\; 4-y=3 \;\Rightarrow\; y=1

Answer

The solution is $x=4,\;y=1$ . Check the first equation: $2(4)+1=9$ , which is correct.

Q6[3 marks]

Solve the simultaneous equations $x-y=2$ and $x^{2}-y^{2}=20$ .

Show worked solution

Factorise the non-linear equation as a difference of two squares, then substitute $x-y=2$ from the linear equation:

x^{2}-y^{2}=(x-y)(x+y)=20 \;\Rightarrow\; 2(x+y)=20 \;\Rightarrow\; x+y=10

Solve $x-y=2$ and $x+y=10$ together by adding them, then find $y$ :

2x=12 \;\Rightarrow\; x=6, \quad y=4

Answer

The solution is $x=6,\;y=4$ . Check: $6^{2}-4^{2}=36-16=20$ , which is correct.

Q7[3 marks]

Solve the simultaneous equations $x=2y+3$ and $y^{2}=x$ .

Show worked solution

The first equation already gives $x$ in terms of $y$ , so substitute it into $y^{2}=x$ , then rearrange into a quadratic and factorise:

y^{2}=2y+3 \;\Rightarrow\; y^{2}-2y-3=0 \;\Rightarrow\; (y-3)(y+1)=0

So $y=3$ or $y=-1$ . Substitute each back into $x=2y+3$ to pair the values:

y=3 \Rightarrow x=9, \qquad y=-1 \Rightarrow x=1

Answer

The solutions are $(x,y)=(9,3)$ and $(1,-1)$ . Check the second: $(-1)^{2}=1$ , which matches $x=1$ .

Q8[3 marks]

Solve the system $x+y+z=12$ , $x-y=2$ and $z=4$ .

Show worked solution

The third equation already gives $z$ directly, so substitute it into the first equation:

x+y+4=12 \;\Rightarrow\; x+y=8

Solve $x+y=8$ and $x-y=2$ together by adding them, then find $y$ :

2x=10 \;\Rightarrow\; x=5, \quad y=3

Answer

The solution is $x=5,\;y=3,\;z=4$ . Check: $5+3+4=12$ and $5-3=2$ , both correct.

Look back over the four. Two used substitution and two used elimination, yet the same steadiness runs underneath: make one letter the subject or cancel one letter out, work the resulting equation carefully, and always substitute back to pair the values correctly.

That habit turns Systems of Equations into a dependable block of marks.

Key method points

These four examples rehearse the everyday skills that open almost every Systems of Equations question in Add Math. Keep the following points in mind as you practise more.

For one linear and one non-linear equation, make a single letter the subject of the linear equation, then substitute into the non-linear one.
Substitution leaves a quadratic, solve it by factorising, then read off both roots.
Always substitute each root back into the linear equation to pair $x$ with the correct $y$ .
For a linear system in three variables, add or subtract equations in pairs to eliminate one letter at a time.
Expand brackets fully and keep every middle term; a dropped $-2x$ is the most common slip here.
Because marking is analytic, a clear substitution or elimination line can still earn method marks even if the final arithmetic slips.

How a teacher helps

When a student loses a mark on questions like these, it is almost always a small, fixable habit, a bracket dropped while expanding $(x-1)^{2}$ , or pairing an $x$ with the wrong $y$ at the end. In a one-to-one lesson our teacher watches the exact line where the slip happens and corrects it on the spot, before it settles into a routine.

Because our teachers are experienced, you work with someone who explains the why behind each step, not just the what. Lessons are taught in English, while SPM papers are set in both Malay and English, so we make sure the notation reads the same to you either way.

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Frequently asked questions

When do I use substitution and when do I use elimination?

Use substitution when one equation is non-linear, make a letter the subject of the linear equation and substitute it in. Use elimination for a system where every equation is linear, especially three-variable systems.

Why does one linear and one non-linear equation usually have two solutions?

Substitution leaves a quadratic, and a quadratic can have two roots. Geometrically a line often cuts a curve at two points, so you typically report two coordinate pairs.

How do I pair up the values correctly at the end?

Substitute each $x$ back into the linear equation, never the non-linear one. The linear equation gives exactly one $y$ for each $x$ , so there is no risk of mismatching.

How can I check a three-variable answer?

Substitute all three values into an equation you did not use to solve. If that equation balances, your solution is secure, this is why labelling the equations $(1)$ , $(2)$ , $(3)$ is worth the moment it takes.

Source:SRC-DSKP-EN

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

Systems of Equations, Worked Examples (easy)

What these examples cover

Worked examples

Key method points

How a teacher helps

Frequently asked questions

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