Simultaneous Equations Word Problems: Worked Answers

Simultaneous equations word problems ask you to find two unknown quantities that satisfy two conditions at the same time. The difficult part is often deciding what each letter means and turning the story into equations. Once the equations are correct, substitution or elimination can do the calculation.

This lesson provides original examples for junior-school and KCSE algebra revision. It covers a pair of linear equations in two unknowns, with no squared variables. Use it when these methods are part of the work your teacher has assigned. The examples include prices, ticket numbers, lengths and an impossible result that algebra helps you detect.

Define the unknowns, including their units

Begin with a sentence for each letter. “Let b be the price of one banana in shillings” is precise. “Let b be bananas” is not: it could mean the number of bananas, their total price or the price of one banana. Your equations will mean different things in each case.

Keep the definitions unchanged throughout the solution. If b is a unit price, then 3b is the cost of three bananas. If b is a number of bananas, then 3b is three times that number. Neither expression is automatically a price.

Units help expose mistakes. You can add a cost in shillings to another cost in shillings. You cannot add “five tickets” directly to “two hundred shillings” and call the result a total cost. Convert metres and centimetres to the same unit before forming a length equation.

Find two genuinely different conditions

One equation such as x + y = 20 allows many pairs: 6 and 14, 9 and 11, or 12 and 8. A second independent condition can select one pair. For example, x − y = 4, together with x + y = 20, gives x = 12 and y = 8.

For a unique solution in these two-variable linear problems, the second condition must add information. Doubling x + y = 20 to obtain 2x + 2y = 40 does not create a new condition. Both equations describe exactly the same relationship.

Translating common statements into equations
Statement and definitions	Equation	What to check
There are 20 items altogether; x and y count the two types	x + y = 20	Every item belongs to exactly one of the two types.
x is 4 more than y	x = y + 4	x is the larger quantity.
A length L is 5 cm more than twice width W	L = 2W + 5	Both measurements are in centimetres.
Three items at p shillings and two at q shillings cost 190 shillings	3p + 2q = 190	p and q are prices of single items.

Do not choose an operation by matching one word mechanically. Read the whole relationship. “Five more than twice the width” means 2W + 5, whereas “twice the sum of the width and five” means 2(W + 5).

Worked example 1: two purchases, solved by elimination

At a fruit stall, 3 bananas and 2 oranges cost KSh 190. At the same unit prices, 2 bananas and 5 oranges cost KSh 310. Find the price of one banana and one orange.

Define: b is the price of one banana in shillings, and o is the price of one orange in shillings. The phrase “same unit prices” allows the same two letters to be used in both purchases.

Form the equations:
3b + 2o = 190 … equation 1
2b + 5o = 310 … equation 2

Elimination combines the equations so that one unknown cancels. Here, make the banana coefficients equal by multiplying every term in equation 1 by 2 and every term in equation 2 by 3:

6b + 4o = 380
6b + 15o = 930

Subtract the first new equation from the second: 11o = 550. Therefore o = 50. Substitute this value into the original equation 1:

3b + 2(50) = 190
3b + 100 = 190
3b = 90
b = 30

Answer: one banana costs KSh 30 and one orange costs KSh 50.

Check both purchases: 3(30) + 2(50) = 90 + 100 = 190, and 2(30) + 5(50) = 60 + 250 = 310. Both prices are positive and reproduce the two stated totals. A correct first total alone would not verify the solution.

Worked example 2: quantities and a total, solved by substitution

A school performance sells 36 tickets. Standard tickets cost KSh 150 each and concession tickets cost KSh 90 each. The total received is KSh 4,140. How many tickets of each type were sold?

Define: s is the number of standard tickets and c is the number of concession tickets. Here the unknowns are counts, not prices. Every ticket is in exactly one category.

The count condition gives s + c = 36. The money condition gives 150s + 90c = 4140. Dividing the money equation by 30 simplifies it to 5s + 3c = 138.

From the count equation, c = 36 − s. Substitute that entire expression for c in the other equation:

5s + 3(36 − s) = 138
5s + 108 − 3s = 138
2s = 30
s = 15

Then c = 36 − 15 = 21. Therefore 15 standard tickets and 21 concession tickets were sold.

Check the story: 15 + 21 = 36 tickets. The revenue is 15(150) + 21(90) = 2250 + 1890 = KSh 4,140. Both counts are whole numbers between 0 and 36, as the context requires.

An early estimate can also help: if all 36 tickets were concession tickets, the revenue would be KSh 3,240. The actual revenue is KSh 900 more. Each standard ticket adds KSh 60 above the concession price, so 900 ÷ 60 = 15 standard tickets. This independent check agrees with the algebra.

Worked example 3: measurements and brackets

A rectangle has a perimeter of 82 cm. Its length is 5 cm more than twice its width. Find its length and width.

Define: L is the length in centimetres and W is the width in centimetres. A rectangle has two lengths and two widths, so the perimeter condition is 2L + 2W = 82. Divide by 2 to obtain L + W = 41.

The second condition is L = 2W + 5. Substitute into L + W = 41:

(2W + 5) + W = 41
3W + 5 = 41
3W = 36
W = 12

Then L = 2(12) + 5 = 29. The rectangle measures 29 cm by 12 cm.

Check: 2(29) + 2(12) = 58 + 24 = 82 cm. Also, 29 is exactly 5 more than twice 12. Writing L + W = 82 at the start would confuse the perimeter with the sum of just one length and one width.

Choose substitution or elimination deliberately

Substitution is convenient when one equation already gives a letter on its own, as in L = 2W + 5, or when isolating one unknown is easy, as in c = 36 − s. Use brackets when replacing a letter with an expression containing addition or subtraction.

Elimination is convenient when both equations have the unknowns in the same order and you can make one pair of coefficients equal or opposite. Equal coefficients can cancel by subtraction; opposite coefficients can cancel by addition. Multiply the whole equation, including its right-hand side.

Both methods should produce the same solution for a system with a unique solution. If a question requires a named method, follow that instruction. Otherwise, choose the method that keeps the arithmetic clear rather than the one that looks shortest at first glance.

Worked example 4: a negative answer that exposes an impossible story

A record claims that 12 full containers, each holding either 2 litres or 5 litres, contain 18 litres altogether. Could that record be correct?

Let a be the number of 2-litre containers and b the number of 5-litre containers. The equations would be a + b = 12 and 2a + 5b = 18.

Multiply the first equation by 2: 2a + 2b = 24. Subtract it from the second equation: 3b = −6, so b = −2. Then a = 14.

These numbers satisfy the equations algebraically: 14 + (−2) = 12 and 2(14) + 5(−2) = 18. However, a count of negative two containers is impossible. The conditions have an algebraic solution but no solution that fits the stated physical situation.

There is an even quicker contextual check. Twelve full containers holding at least 2 litres each must contain at least 24 litres. The stated 18 litres is too small. Check the original data; do not silently change the negative answer to a positive number.

A negative value is not always invalid. It can describe a temperature below zero or a position on the negative side of an axis. Decide using the definition of the unknown, rather than rejecting every negative solution automatically.

Common errors to catch before your final answer

Mixing counts and prices: define each variable once and check the units of every term.
Reversing a comparison: if x is greater than y by 8, write x = y + 8.
Multiplying only one term: multiplying 3b + 2o = 190 by 2 gives 6b + 4o = 380.
Losing the subtraction sign: subtract all terms in the chosen equation, including its total.
Ignoring brackets: 3(36 − s) equals 108 − 3s, not 108 − s.
Checking only the rearranged equation: return to both original statements.
Accepting an impossible count: whole objects require whole-number, non-negative answers.

Also distinguish an impossible contextual answer from a system with no algebraic solution. The equations x + y = 12 and 2x + 2y = 30 are inconsistent: doubling the first requires 2x + 2y = 24. By contrast, x + y = 12 and 2x + 2y = 24 repeat one condition and do not determine a unique pair.

Practice questions

For each question, write the two variable definitions, form both equations, choose a method and check your answer in the original statements before reading the solution.

Four pencils and three erasers cost KSh 165. Two pencils and five erasers cost KSh 135 at the same unit prices. Find the price of each item.
A shop sells 28 envelopes. Large envelopes cost KSh 9 each and small envelopes cost KSh 6 each. The total received is KSh 216. How many of each size were sold?
A rectangle has perimeter 96 cm. Its length is 8 cm greater than its width. Find both dimensions.

Practice solution 1: pencil and eraser prices

Let p and e be the respective unit prices in shillings. The equations are 4p + 3e = 165 and 2p + 5e = 135. Double the second: 4p + 10e = 270. Subtract the first: 7e = 105, giving e = 15. Then 4p + 45 = 165, so p = 30.

A pencil costs KSh 30 and an eraser costs KSh 15. Check: 4(30) + 3(15) = 165 and 2(30) + 5(15) = 135. Both original purchases are satisfied.

Practice solution 2: large and small envelopes

Let L be the number of large envelopes and S the number of small envelopes. L + S = 28 and 9L + 6S = 216. Substitute S = 28 − L into the money equation: 9L + 6(28 − L) = 216. This gives 3L + 168 = 216, so L = 16 and S = 12.

The answer is 16 large and 12 small envelopes. Check: 16 + 12 = 28 and 9(16) + 6(12) = 144 + 72 = 216. Both numbers are non-negative whole counts.

Practice solution 3: rectangle dimensions

Let L and W be the length and width in centimetres. The equations are 2L + 2W = 96 and L = W + 8. Simplify the first to L + W = 48, then substitute: W + 8 + W = 48. Therefore W = 20 and L = 28.

The dimensions are 28 cm by 20 cm. Check: 2(28) + 2(20) = 96 cm, and 28 − 20 = 8 cm. The answer includes both the measurements and their units.

Use your mistakes to choose the next question

If your equations were wrong, practise translating the statements before doing more elimination. If the equations were correct but the arithmetic failed, redo the calculation with one transformation on each line. If only the final answer was wrong, check whether you swapped the variables or forgot the units.

For broader practice, see Grade 9 topical questions and answers or KCSE Mathematics topic tests revision, as appropriate to your course. Both existing resource pages require a member login. Select questions that match the methods your teacher has covered; the worked questions on this page remain available here.

Frequently asked questions

How do I choose the two unknowns in a word problem?

Choose the two quantities the question asks you to find. Define each letter precisely, including whether it represents a count, a unit price or a measurement. State the units for prices and measurements, and keep those definitions throughout the solution.

Why do I need two independent conditions?

One linear equation in two unknowns generally allows many pairs. A second independent condition can identify one pair. An equation obtained only by multiplying the first equation repeats the same information and cannot, by itself, determine a unique solution.

When should I use substitution?

Use substitution when one unknown is already isolated or can be isolated easily. Replace that unknown in the other equation with the full equivalent expression. Put the expression in brackets where needed, then solve for the remaining unknown.

When should I use elimination?

Use elimination when you can conveniently make the coefficients of one unknown equal or opposite. Subtract equations with equal coefficients or add equations with opposite coefficients to cancel that unknown. Apply any multiplying factor to every term on both sides.

Can a correct algebraic answer be wrong for the story?

Yes. An algebraic solution may require a negative number of objects, a fractional whole-object count or a length that cannot fit the stated situation. Check both equations and the restrictions imposed by the context before accepting the answer.

Does a negative answer always mean I made a mistake?

No. Negative values can describe quantities such as temperatures or signed positions. They cannot describe a count of physical objects. Recheck the equations and arithmetic, then decide whether the value is allowed by the meaning of the unknown.

What if the second equation is just twice the first?

Then the equations describe the same condition and do not determine a unique pair by themselves. If the left side doubles but the stated right side is different from twice the original total, the equations are inconsistent and have no algebraic solution.

How should I check a simultaneous-equations word problem?

Substitute both values into both original conditions. Check the totals, comparisons and units, then ask whether the quantities are possible in the story. Finish with a sentence naming each quantity, so the reader knows which value belongs to which unknown.

Simultaneous Equations Word Problems: Set Up, Solve and Check