Linear equations

The linear equation is the simplest equation, yet most word problems reduce to it: problems on motion, work, ages, mixtures and prices all end up in the form ax = b. More complicated equations (rational, with absolute values, with parameters) also often split into several linear ones. So you should know the solving procedure automatically and understand why each step is taken. This lesson covers the definition, the solving steps, fractional coefficients, parameters, equations that reduce to linear ones, and word problems.

§1. Definition and number of solutions

An equation of the form ax = b is called linear (x is the unknown, a and b are numbers). The name comes from the graph of y = ax − b being a straight line: solving the equation means finding where this line crosses the Ox axis.
A root of an equation is a number that turns it into a true equality. To solve an equation is to find all roots or prove there are none.

In graph language: a ≠ 0 — the line crosses Ox once; a = 0 — the line is parallel to Ox or coincides with it.

§2. Solving steps and equivalent transformations

Two equations are equivalent if they have the same roots. Solving means turning the equation by equivalent transformations into x = number. Allowed transformations:

• add the same number or expression to both sides (moving a term to the other side with the opposite sign is exactly this);
• multiply or divide both sides by a non-zero number.

Procedure:

1. Expand the brackets.
2. Move terms with the unknown to one side and numbers to the other (signs change).
3. Collect like terms.
4. Divide by the coefficient and check the answer.

Why may we not multiply by zero? Multiplying x = 2 by 0 gives 0 = 0, which is true for every x — extra “roots” appear.

Worked example

Solve 3(x − 2) = 2x + 5.

1. Expand: 3x − 6 = 2x + 5.
2. Move terms: 3x − 2x = 5 + 6.
3. x = 11.
4. Check: left side 3·9 = 27, right side 22 + 5 = 27 — correct.

§3. Fractional coefficients

If the equation contains fractions, multiply both sides by the LCM of the denominators — the fractions disappear and the work continues with integers.

Worked example

Solve (x − 1)/3 − (x + 2)/4 = 1.

1. Multiply by LCM(3, 4) = 12: 4(x − 1) − 3(x + 2) = 12.
2. Expand: 4x − 4 − 3x − 6 = 12.
3. x − 10 = 12.
4. x = 22.

Careful: the minus in front of a fraction applies to the whole numerator — −3(x + 2) = −3x − 6, not −3x + 6.

§4. Equations with a parameter

If a coefficient contains a letter (a parameter), check the three cases of §1: first bring the equation to the form A(a)·x = B(a), then treat separately the values for which A(a) = 0.

Worked example

For which a does (a − 2)x = a² − 4 have infinitely many solutions?

1. Infinitely many solutions require a − 2 = 0 and a² − 4 = 0.
2. At a = 2 both hold: 0·x = 0.
3. At a = −2: −4x = 0 → the unique solution x = 0.
4. Answer: a = 2. (For every other a ≠ 2 the unique solution is x = a + 2.)

§5. Equations that reduce to linear ones

If the unknown appears in a denominator, the equation is rational. Method: first write the condition that the denominator is non-zero, then multiply by the common denominator to get a linear equation, and check the root against the condition.

Worked example

Solve (x² − 9)/(x − 3) = 6.

1. Condition: x ≠ 3.
2. Factorise the numerator: (x − 3)(x + 3)/(x − 3) = x + 3.
3. x + 3 = 6 → x = 3.
4. x = 3 violates the condition. Answer: no roots.

§6. Word problems

Algorithm: name the unknown → write the conditions as an equation → solve → check that the answer makes sense (an age cannot be negative, a number of people is a whole number).

Worked example

A father is 40 and his son is 10. In how many years will the father be twice as old as the son?

1. In t years the father is 40 + t, the son 10 + t.
2. Equation: 40 + t = 2(10 + t).
3. 40 + t = 20 + 2t → t = 20.

Worked example

One pipe fills a pool in 6 hours, another in 3 hours. How long do they take together?

1. Rates: 1/6 and 1/3 (pools per hour).
2. Together: 1/6 + 1/3 = 1/2 pool per hour.
3. Time: 1 : 1/2 = 2 hours.

Worked example

How many grams of a 30% solution must be added to 200 g of a 10% solution to obtain a 25% solution?

1. Add x g; substance: 0.1·200 + 0.3x = 20 + 0.3x.
2. Substance in the new solution: 0.25(200 + x).
3. 20 + 0.3x = 50 + 0.25x → 0.05x = 30.
4. x = 600 g.

Check yourself

1. Solve 5x − 7 = 3x + 9.

2. Solve x/2 + x/3 = 10.

3. How many solutions does 0·x = 7 have?

4. Two cars 300 km apart drive towards each other at 70 and 80 km/h. When do they meet?

5. A boat covers 48 km downstream in 3 h and 48 km upstream in 4 h. Speed of the current?

6. For which a does ax + 3 = 2x + a have no roots?

7. Solve (2x + 1)/(x − 3) = 3.

8. A rectangle has perimeter 46 and its length is 5 more than its width. Find the width.

Common mistakes

• Not changing the sign when moving a term to the other side.
• Applying the minus in front of a fraction only to the first term of the numerator.
• Ignoring the zero-coefficient case in an equation with a parameter.
• Forgetting the “denominator ≠ 0” condition in a rational equation and keeping an extraneous root.
• Not converting units (minutes/hours, m/km) in a word problem.
• Dividing both sides by an expression containing the unknown and losing a root.