Systems of Equations Practice

Q: How do I pair up x and y correctly?

Substitute each xxx back into the linear equation to get its own yyy. Never mix an xxx from one root with a yyy from another, that gives a point that does not lie on both graphs.

How to use these practice questions

The six questions below rise in difficulty across the whole Systems of Equations chapter, from a two-variable elimination to a linear equation solved together with a non-linear one. Give yourself roughly six to ten minutes per question and work on paper first, writing each line as you would in the real exam, a method line, the substitution or elimination, then the final answer.

Do not open the solution until you have committed a full attempt to paper.

When your working differs from ours, stop at the exact line where the two part company; that single step is almost always where the mark was lost. Because Add Math is marked analytically, a clear method line still earns credit even when the final arithmetic slips, so always show the equation you eliminate or substitute into.

A system usually has more than one solution pair, make a habit of pairing each $x$ with its matching $y$ , and of checking every pair in both equations.

Six practice questions

Q1[3 marks]

Solve the simultaneous equations $2x+3y=13$ and $x+y=5$ .

Show worked solution

Use elimination. Multiply the second equation by $2$ so that the $x$ -terms match the first equation:

2x+3y=13 \qquad 2(x+y)=2(5) \;\Rightarrow\; 2x+2y=10

Subtract the new equation from the first to eliminate $x$ , leaving one unknown:

(2x+3y)-(2x+2y)=13-10 \;\Rightarrow\; y=3

Back-substitute $y=3$ into $x+y=5$ :

x+3=5 \;\Rightarrow\; x=2

Answer

$x=2$ , $y=3$ . Check in the first equation: $2(2)+3(3)=4+9=13$ , as required.

Q2[4 marks]

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

Show worked solution

One equation is linear and one is a curve, so substitute the linear expression into the curve. Since both equal $y$ , set them equal to each other:

x^{2}=x+2

Bring everything to one side to form a quadratic, then factorise:

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

Find the matching $y$ for each $x$ using the linear equation $y=x+2$ :

x=2 \;\Rightarrow\; y=4 \qquad x=-1 \;\Rightarrow\; y=1

Answer

$(x,y)=(2,4)$ or $(-1,1)$ . Check $(2,4)$ : $4=2+2$ and $4=2^{2}$ , both hold; and $(-1,1)$ : $1=-1+2$ and $1=(-1)^{2}$ , both hold.

Q3[5 marks]

Solve the system of three linear equations: $x+y+z=6$ , $2x+y-z=1$ , and $x+2y+z=8$ .

Show worked solution

Label the equations (1), (2), (3). Subtract (1) from (3) to remove both $x$ and $z$ at once, since they appear identically in the two equations:

(x+2y+z)-(x+y+z)=8-6 \;\Rightarrow\; y=2

Put $y=2$ back into (1) and (2) to get two equations in $x$ and $z$ :

x+z=4 \qquad 2x-z=1-2=-1

Add these two to eliminate $z$ , then solve and back-substitute:

(x+z)+(2x-z)=4+(-1) \;\Rightarrow\; 3x=3 \;\Rightarrow\; x=1 \;\Rightarrow\; z=3

Answer

$x=1$ , $y=2$ , $z=3$ . Check in (2): $2(1)+2-3=1$ , correct; and in (3): $1+2(2)+3=8$ , correct.

Q4[5 marks]

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

Show worked solution

Substitute the linear equation into the non-linear one, replacing $y$ with $x+1$ :

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

Expand and simplify into standard quadratic form:

x^{2}+x^{2}+2x+1=25 \;\Rightarrow\; 2x^{2}+2x-24=0 \;\Rightarrow\; x^{2}+x-12=0

Factorise and solve for $x$ :

(x+4)(x-3)=0 \;\Rightarrow\; x=-4 \;\text{or}\; x=3

Find each matching $y$ from $y=x+1$ :

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

Answer

$(x,y)=(3,4)$ or $(-4,-3)$ . Check $(3,4)$ : $3^{2}+4^{2}=9+16=25$ ; and $(-4,-3)$ : $(-4)^{2}+(-3)^{2}=16+9=25$ .

Both pairs satisfy the circle.

Q5[5 marks]

The total cost of $3$ notebooks and $2$ pens is RM $13$ . The total cost of $5$ notebooks and $2$ pens is RM $19$ .

Find the price of one notebook and the price of one pen.

Show worked solution

Let the price of one notebook be $x$ and one pen be $y$ , both in ringgit. Turn each sentence into an equation:

3x+2y=13 \qquad 5x+2y=19

The $2y$ terms are identical, so subtract the first equation from the second to eliminate $y$ :

(5x+2y)-(3x+2y)=19-13 \;\Rightarrow\; 2x=6 \;\Rightarrow\; x=3

Back-substitute $x=3$ into the first equation:

3(3)+2y=13 \;\Rightarrow\; 9+2y=13 \;\Rightarrow\; 2y=4 \;\Rightarrow\; y=2

Answer

A notebook costs RM $3$ and a pen costs RM $2$ . Check the second equation: $5(3)+2(2)=15+4=19$ , as required.

Always state the answer in the words of the question, not just as numbers.

Q6[6 marks]

Solve the simultaneous equations $2x+y=7$ and $xy=6$ .

Show worked solution

Make $y$ the subject of the linear equation, then substitute into the product equation. From $2x+y=7$ :

y=7-2x

Substitute into $xy=6$ and expand:

x(7-2x)=6 \;\Rightarrow\; 7x-2x^{2}=6

Rearrange into standard form. This is a non-monic quadratic, so factorise carefully:

2x^{2}-7x+6=0 \;\Rightarrow\; (2x-3)(x-2)=0 \;\Rightarrow\; x=\tfrac{3}{2} \;\text{or}\; x=2

Find each matching $y$ from $y=7-2x$ :

x=2 \;\Rightarrow\; y=3 \qquad x=\tfrac{3}{2} \;\Rightarrow\; y=7-3=4

Answer

$(x,y)=(2,3)$ or $\left(\tfrac{3}{2},4\right)$ . Check the products: $2\times 3=6$ and $\tfrac{3}{2}\times 4=6$ , both correct.

A non-integer answer is not a mistake, the factor $(2x-3)$ simply gives $x=\tfrac{3}{2}$ .

How to mark yourself like an examiner

Add Math is marked analytically, which means marks are attached to steps, not only to the final numbers. When you check your own script, award yourself credit the way a marker would: look for the correct method being started, the right substitution or elimination, and complete solution pairs.

Method mark: did you write the correct move, scaling to eliminate, or making a variable the subject and substituting?
Reduction mark: is the resulting single-variable equation correct, and reduced to standard form where it is a quadratic?
Pairing mark: did you find the matching second variable for every value, so each solution is a complete pair?
Answer mark: are all solution pairs stated clearly and checked in both original equations?
If your final number is wrong but the method line is right, give yourself the method mark, that is exactly what a real marker does.

How a teacher helps

Marking yourself is powerful, but it is hard to see your own blind spots. In a one-to-one lesson our teacher watches the exact line where a mark slips away, a forgotten second solution pair, or a sign error while expanding a bracket, and corrects the habit on the spot.

Because our teachers are experienced, you work with someone who explains the why behind each step. 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.

Get 1-to-1 help.

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Frequently asked questions

When do I eliminate, and when do I substitute?

For two linear equations, eliminate a variable by scaling one equation so a pair of terms match, then add or subtract. When one equation is non-linear, make a variable the subject of the linear equation and substitute it into the other.

Why does a linear-and-non-linear system often have two solutions?

Substituting usually produces a quadratic, which can have two roots. Each root is a value of one variable, and it pairs with its own value of the other, so you typically report two complete $(x,y)$ pairs.

How do I pair up

x

and

y

correctly?

Substitute each $x$ back into the linear equation to get its own $y$ . Never mix an $x$ from one root with a $y$ from another, that gives a point that does not lie on both graphs.

Is a fraction like

x=\tfrac{3}{2}

a wrong answer?

Not at all. Systems of equations often have non-integer solutions.

As long as the pair satisfies both original equations, it is correct, so always check by substitution.

Source:SRC-DSKP-EN

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

Systems of Equations, Practice Questions

How to use these practice questions

Six practice questions

How to mark yourself like an examiner

How a teacher helps

Frequently asked questions

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