Linear Equations: Solving Methods and Word Problems

By Engr. Henry Akpan

Finding the Unknown

"I think of a number, multiply it by 3 and add 7, and the answer is 22. What is my number?" Puzzles like this, along with questions about ages, money, prices, perimeters, speeds and mixtures, all lead to linear equations: equations in which the unknown appears only to the first power, with no x², √x or 1/x. They are the foundation of algebra, and every later topic, from simultaneous equations to calculus, relies on solving them quickly and accurately. This tutorial covers linear equations as tested in WAEC, NECO, BECE and JAMB mathematics: the balance method, equations with brackets, unknowns on both sides, fractions, and turning word problems into equations, with worked examples on numbers, ages, money, geometry, speed, work and mixtures.

The Balance Method

An equation is like a balanced scale: both sides are equal. You may do anything to an equation, provided you do the same thing to both sides. To solve, undo the operations on the unknown in reverse order until it stands alone.

Worked example 1

Solve 3x + 7 = 22.

  1. Subtract 7 from both sides: 3x = 15.
  2. Divide both sides by 3: x = 5.

Check by substituting back: 3(5) + 7 = 22. Always check, because it takes only seconds and catches most mistakes.

Worked example 2

Solve x/4 − 3 = 2.

Add 3: x/4 = 5. Multiply by 4: x = 20.

Equations with Brackets

Expand the brackets first, then collect like terms.

Worked example 3

Solve 5(x − 2) = 3x + 4.

  1. Expand: 5x − 10 = 3x + 4.
  2. Subtract 3x from both sides: 2x − 10 = 4.
  3. Add 10: 2x = 14.
  4. Divide by 2: x = 7.

Check: 5(7 − 2) = 25, and 3(7) + 4 = 25.

Worked example 4: a minus sign before a bracket

Solve 4 − 2(3x − 1) = 3(x + 4) − 15.

Expand carefully: −2 × −1 = +2, so the left side is 4 − 6x + 2 = 6 − 6x. The right side is 3x + 12 − 15 = 3x − 3.

So 6 − 6x = 3x − 3, which gives 9 = 9x, and x = 1.

Unknowns on Both Sides

Collect all the terms containing the unknown on one side, and all the numbers on the other. It is often easiest to move the smaller x term, to avoid negative coefficients.

Worked example 5

Solve 7x − 5 = 2x + 20.

Subtract 2x: 5x − 5 = 20. Add 5: 5x = 25. So x = 5.

Equations with Fractions

Multiply every term by the lowest common multiple (LCM) of the denominators, to clear all the fractions at once.

Worked example 6

Solve (x + 1)/3 − (x − 2)/4 = 1.

The LCM of 3 and 4 is 12. Multiply every term by 12: 4(x + 1) − 3(x − 2) = 12.

Expand: 4x + 4 − 3x + 6 = 12, so x + 10 = 12, and x = 2.

Check: (3)/3 − (0)/4 = 1.

The brackets matter. Without them, the minus sign in front of the second fraction would apply only to x, not to the −2, and the answer would be wrong.

Worked example 7: decimals

Solve 0.3x + 1.2 = 0.5x − 0.4.

Multiply every term by 10: 3x + 12 = 5x − 4. So 16 = 2x, and x = 8.

From Words to Equations

Most marks in this topic come from word problems. Follow these steps:

  1. Read carefully, and decide what the unknown is.
  2. Let x stand for it, and write the other quantities in terms of x.
  3. Write an equation from the information that connects them.
  4. Solve it.
  5. Answer the question that was asked, in words with units, and check that the answer makes sense.

Useful translations: "sum" means +, "difference" means −, "product" means ×, "is" means =, "exceeds by" means "is larger by", and "twice" means 2 ×.

Worked Word Problems

Example 8: consecutive numbers

The sum of three consecutive whole numbers is 87. Find the numbers.

Let the smallest number be x. Then the numbers are x, x + 1 and x + 2.

x + (x + 1) + (x + 2) = 87, so 3x + 3 = 87, and x = 28.

The numbers are 28, 29 and 30.

Example 9: ages

A father is three times as old as his son. In 12 years' time, he will be twice as old as his son. Find their present ages.

Let the son's age be s, so the father's age is 3s. In 12 years, they will be s + 12 and 3s + 12.

3s + 12 = 2(s + 12), so 3s + 12 = 2s + 24, and s = 12.

The son is 12, and the father is 36. Check: in 12 years they will be 24 and 48, and 48 is twice 24.

Example 10: money

A trader has 15 notes, all ₦200 or ₦500 notes, worth ₦4,800 altogether. How many of each does she have?

Let the number of ₦200 notes be x, so there are 15 − x ₦500 notes.

200x + 500(15 − x) = 4,800, so 200x + 7,500 − 500x = 4,800, which gives −300x = −2,700, and x = 9.

She has nine ₦200 notes and six ₦500 notes. Check: 1,800 + 3,000 = 4,800.

Example 11: perimeter

The length of a rectangular garden is 5 m more than its width, and its perimeter is 50 m. Find its dimensions.

Let the width be w, so the length is w + 5. Perimeter: 2w + 2(w + 5) = 50, so 4w + 10 = 50, and w = 10.

The garden is 10 m wide and 15 m long.

Example 12: speed

A lorry leaves a depot at 60 km/h. One hour later, a car leaves the same depot along the same road at 90 km/h. How long does the car take to catch up, and how far from the depot does this happen?

Let the car travel for t hours. The lorry has then travelled for t + 1 hours. They have covered the same distance when 90t = 60(t + 1), so 30t = 60, and t = 2.

The car catches up after 2 hours, 90 × 2 = 180 km from the depot.

Example 13: mixtures

How many litres of a 20% salt solution must be added to 10 litres of a 50% solution to make a 30% solution?

Let x litres be added. The amount of salt is the same whether you count it before or after mixing: 0.20x + 0.50 × 10 = 0.30(10 + x).

So 0.2x + 5 = 3 + 0.3x, which gives 2 = 0.1x, and x = 20 litres.

Special Cases

  • If the unknown cancels out and leaves a true statement, such as 2(x + 3) = 2x + 6, which reduces to 6 = 6, the equation is an identity: every value of x works.
  • If it cancels out and leaves a false statement, such as 2x + 3 = 2x + 5, which reduces to 3 = 5, there is no solution.

Common Mistakes with Linear Equations

  • Doing something to only one side, or to only one term on a side.
  • Sign errors when expanding, such as −2(3x − 1) = −6x − 2 instead of −6x + 2.
  • Multiplying only some terms by the LCM when clearing fractions.
  • Dropping the brackets around numerators when combining fractions.
  • Answering with x instead of the actual question, such as finding the son's age when the father's age was asked.

Linear Equations: Practice Questions

  1. Solve 4x − 9 = 15.
  2. Solve 3(2x + 1) = 4(x + 5) − 3.
  3. Solve (2x − 1)/5 = (x + 2)/3.
  4. When 8 is subtracted from five times a number, the result is 27. Find the number.
  5. A pen costs ₦150 more than a pencil. Three pens and two pencils cost ₦1,200. Find the cost of a pencil.
  6. Ngozi is 4 years older than her brother. The sum of their ages is 30. How old is each?
  7. The angles of a triangle are x, 2x and 3x. Find x.

Answers: 1. 4x = 24, so x = 6. 2. 6x + 3 = 4x + 17, so x = 7. 3. 3(2x − 1) = 5(x + 2), so 6x − 3 = 5x + 10, and x = 13. 4. 5n − 8 = 27, so n = 7. 5. 3(p + 150) + 2p = 1,200, so 5p = 750, and a pencil costs ₦150 (and a pen ₦300). 6. b + (b + 4) = 30, so the brother is 13 and Ngozi is 17. 7. 6x = 180, so x = 30°.

Conclusion

A linear equation is solved by keeping it balanced, doing the same operation to both sides, until the unknown stands alone. Expand brackets carefully, collect the unknown on one side, and clear fractions by multiplying every term by the LCM of the denominators. For word problems, define the unknown, express the other quantities in terms of it, form an equation, solve it, and answer the actual question with units, checking that the answer makes sense. These steps solve problems about numbers, ages, money, shapes, speeds, work and mixtures.

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