Sec 2 Simultaneous Equations: Methods & Worked Examples

Secondary 2 G3 Mathematics · Study guide

Connect the intersection of two straight lines to elimination and substitution, then form equations from practical conditions and recognise all three solution cases.

Concept → worked examples → practice9 min read · practice linked below

What you will learn

Solving simultaneous linear equations means finding values that make both equations true at the same time. In this Sec 2 Maths guide, you will connect a graph’s intersection to elimination and substitution, then form equations from a word problem.

Before you begin: be comfortable with signed numbers, expanding brackets, rearranging a linear equation and plotting coordinates. You should recognise $y=mx+c$ , where $m$ is the gradient and $c$ is the vertical intercept.

One equation gives a line; two equations give a shared condition

The equation $x+y=7$ has many solutions: $(1,6)$ , $(2,5)$ and $(3,4)$ , for example. Every point on its graph satisfies that equation. A second equation usually singles out one of those points.

A simultaneous solution is an ordered pair $(x,y)$ . The same values must work in every original equation. Finding only $x$ , or finding a point on only one line, does not finish the problem.

All three methods describe the same shared condition. Graphing locates it visually. Elimination combines equations to remove an unknown. Substitution replaces an unknown with an equal expression.

Graphical method: read the intersection

Worked example 1 — two lines, one solution

Solve graphically:

$y=2x-1,\qquad y=-x+5.$

Choose two convenient points on each line. For the first, use $(0,-1)$ and $(3,5)$ . For the second, use $(0,5)$ and $(5,0)$ .
Plot the points on the same labelled axes, using a consistent scale on each axis. Join each pair with a straight line, extend it across the grid, and label the lines.
Read the point lying on both lines: $(2,3)$ . Its x-coordinate is the same input for both equations, and its y-coordinate is their common output. Hence $x=2$ and $y=3$ .

Line A:

y=2x-1

. Line B:

y=-x+5

. At their intersection, both equations have the same

x

- and

y

-values.

Check: $2(2)-1=3$ and $-2+5=3$ . Both equations hold. A graph reading may be approximate; use the precision requested in the question. Algebra gives exact values when the intersection falls between grid marks.

Elimination: make one unknown cancel

Multiplying every term of an equation by the same non-zero number preserves its solutions. After matching coefficient magnitudes, subtract equal terms or add opposite terms.

Worked example 2 — multiply both equations

Solve the simultaneous equations:

$\begin{aligned}3x+2y&=16 &&(1)\\5x-3y&=14 &&(2)\end{aligned}$

The y-coefficients are $+2$ and $-3$ . Their magnitudes have lowest common multiple $6$ , so we can create $+6y$ and $-6y$ , which cancel when added. Multiply every term of (1) by $3$ , and every term of (2) by $2$ :

$\begin{aligned}9x+6y&=48 &&(3)\\10x-6y&=28 &&(4)\end{aligned}$

Add the left sides and add the right sides. Since $6y+(-6y)=0$ , this leaves $9x+10x=48+28$ . Divide the resulting equation by $19$ :

$19x=76\quad\Rightarrow\quad x=4.$

Substitute into (1) to find the other unknown:

$3(4)+2y=16\quad\Rightarrow\quad2y=4\quad\Rightarrow\quad y=2.$

Therefore $(x,y)=(4,2)$ . Check (2): $5(4)-3(2)=14$ . Equation (1) also gives $12+4=16$ .

Keep the equation labels: they make the reasoning easy to follow. When subtracting equations, put the entire second expression in brackets so every sign changes correctly.

Substitution: replace an unknown with an equal expression

Substitution is often efficient when an equation already gives $x$ or $y$ , or when a coefficient is 1 or −1. Substitute into the other equation to bring both conditions together.

Worked example 3 — keep the brackets

Solve the simultaneous equations:

$y=3x-7\quad(1),\qquad 2x+3y=1\quad(2).$

Equation (1) already makes $y$ the subject, so substitution gives an equation containing only $x$ . Replace $y$ in (2) with the whole expression $3x-7$ . The brackets ensure that the factor $3$ multiplies both terms:

$\begin{aligned}2x+3(3x-7)&=1\\2x+9x-21&=1\\11x&=22\\x&=2.\end{aligned}$

After expanding, add $21$ to both sides to obtain $11x=22$ , then divide by $11$ . Now substitute $x=2$ into (1): $y=3(2)-7=-1$ . Hence $(x,y)=(2,-1)$ . Check (2): $2(2)+3(-1)=1$ .

The negative answer is valid. Algebraic unknowns can be negative unless the problem’s context rules this out.

Choosing an efficient method
What you notice	Good first move
One unknown is already isolated	Substitute that expression.
Coefficients match or have a small common multiple	Eliminate an unknown.
The question asks for a graphical solution	Draw both lines and read their intersection.

Both algebraic methods give the same result. Choose the one that keeps the arithmetic manageable, unless a method is specified.

Word problems: translate each condition separately

Define the unknown quantities and units first. Then distinguish a count equation from a cost equation: the coefficient of a quantity often tells you the value of one item.

Worked example 4 — tickets and total revenue

A performance sells 23 tickets. Adult tickets cost $12 each and student tickets cost $8 each. Total revenue is $232. How many of each ticket were sold?

Let $a$ be the number of adult tickets and $s$ the number of student tickets. The ticket count gives $a+s=23$ . The adult tickets bring in $12a$ dollars and the student tickets bring in $8s$ dollars, giving the second equation:

$\begin{aligned}a+s&=23 &&\text{(ticket count)}\\12a+8s&=232 &&\text{(revenue in dollars)}\end{aligned}$

Multiply the count equation by $8$ to obtain $8a+8s=184$ . Both equations now contain $8s$ . Subtract the whole scaled count equation from the revenue equation to eliminate $s$ :

$(12a+8s)-(8a+8s)=232-184\quad\Rightarrow\quad4a=48.$

Divide by $4$ to get $a=12$ . Then use the count equation: $s=23-12=11$ .

Answer: 12 adult tickets and 11 student tickets were sold. Check: the count is $12+11=23$ , and the revenue is $12(12)+8(11)=232$ dollars.

Both answers are non-negative whole numbers, as ticket counts must be. A fractional count would signal an error or incompatible information.

One solution, no solution or infinitely many?

Two straight lines need not cross once. Their relationship explains what happens when elimination removes both unknowns.

From top to bottom: intersecting, parallel and coincident lines. Colour and line pattern identify each equation; overlapping equations impose the same condition.

One solution: elimination gives an unknown’s value, such as $3x=9$ . The lines intersect once.
No solution: a false statement such as $0=4$ remains. For example, $x+y=3$ and $x+y=7$ cannot both hold.
Infinitely many: a true identity such as $0=0$ remains. For example, $x+y=3$ and $2x+2y=6$ describe the same line. Every point on that line works; not every point in the plane.

Explore the number of solutions

Line A stays at $y=x+1$ . Change Line B’s gradient and vertical intercept. Predict which settings will make the lines parallel or identical before choosing a preset.

The graph below shows Line B with gradient −1 and vertical intercept 3. Compare its intersection with Line A to the three cases above.

Line B has gradient −1 and vertical intercept 3.

The red point marks a unique intersection. When the lines coincide, the dashed line lies directly on the solid line.

One solution: (1, 2). The gradients differ, so the lines meet once.

Common errors and how to catch them

Multiplying only one term: multiplying $2x+y=5$ by 3 gives $6x+3y=15$ . Every term, including the right-hand side, changes.
Losing a sign: $-2(x-4)=-2x+8$ . Expand the bracket before collecting terms.
Stopping after cancellation: $0=0$ means the two line equations are equivalent; it does not mean $x=y=0$ .
Checking only one condition: substitute the final pair into both original equations. In context, also check units and whether the values are sensible.

Practise with our topic questions

Apply the methods in this guide using our Sec 2 simultaneous equations practice questions. The topic bank brings the questions and their existing worked solutions together, so you can attempt a question before checking the method.

Choose elimination or substitution, define the unknowns in a word problem, and check both original equations using the same pair of values.

Frequently asked questions

How do I choose between elimination and substitution?

Look for an isolated unknown first. If there is one, substitution is usually short. Otherwise compare the coefficient multiples. Elimination is efficient when an unknown cancels after simple multiplication.

Can two straight lines have exactly two solutions?

No. Two distinct straight lines intersect at most once. If they share two different points, they are the same line and share infinitely many points.

Does a graph always give an exact answer?

No. A plotted intersection often gives an estimate, limited by the scale and drawing accuracy. Use the requested graphical precision; use algebra when exact fractions are required.

What if both unknowns disappear during elimination?

Read the remaining statement. A contradiction such as $0=5$ means no solution. An identity such as $0=0$ means equivalent line equations and infinitely many solutions.

Ready for more practice?

You are ready to move on when you can explain the intersection, choose an algebraic method, form two equations from words and interpret the final pair. Try the Math Academy Sec 2 simultaneous equations practice questions for further questions and worked solutions.