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Linear Equations

Algebra · Linear Equations

Syllabus tag: KCSE | Mathematics | Form 1 | Topic 17 Linear Equations

Lesson objectives

By the end of this topic, you should be able to:

  • Solve linear equations in one unknown.
  • Solve simultaneous linear equations by substitution.
  • Solve simultaneous linear equations by elimination.
  • Form and solve equations from word problems.

Linear Equations

An equation states that two expressions are equal. Solving it finds the value or values that make the statement true.

a) One unknown

One unknown One unknown 3(x − 2) = 12 expand first 3x − 6 = 12 brackets opened 3x = 18 add 6 to both sides x = 6 divide both sides by 3

Expand any brackets first.

Collect the unknown on one side and the numbers on the other.

Whatever you do to one side, do to the other.

Undo additions and subtractions before multiplications and divisions. That is the reverse of the usual order, because you are undoing.

b) Equations with fractions

Multiply every term by the LCM of the denominators to clear them.

For x/2 + x/3 = 5, the LCM is 6. Multiplying through gives 3x + 2x = 30, so 5x = 30 and x = 6.

Multiply every term, including the one with no fraction. Missing it is the usual error.

c) Two unknowns

One equation with two unknowns has infinitely many solutions. Two equations usually pin it down to one pair.

There are two standard methods.

d) Substitution

By substitution By substitution x + 2y = 11 the first equation rearrange x = 11 − 2y make x the subject 3x − y = 5 the second equation substitute 3(11 − 2y) − y = 5 replace x solve y = 4 then x = 3

Make one letter the subject in one equation.

Substitute that expression into the other equation.

Solve the single-letter equation that results, then find the second letter.

Choose the rearrangement that avoids fractions. A letter with a coefficient of 1 is the easiest to isolate.

e) Elimination

By elimination By elimination 2x + 3y = 16 first equation 2x − y = 4 second equation subtract 4y = 12 the x terms cancel solve y = 3 then x = 3.5

Multiply one or both equations so that one letter has the same coefficient in both.

Subtract if the signs match; add if they are opposite.

Solve for the remaining letter, then substitute back for the other.

f) Which method

Substitution suits equations where a letter already stands alone or has coefficient 1.

Elimination suits equations where coefficients already match, or match after one small multiplication.

Both give identical answers.

g) Word problems

Name each unknown and state clearly what it represents.

Write one equation for each separate fact given.

You need as many equations as unknowns.

Solve, then check in both original equations, and read the answer back in the words of the question.

h) Where this is used

Finding two unit prices from two bulk purchases. Splitting a total between two categories. Break-even calculations. Any situation with two unknowns and two facts about them.

Words to know

  • Equation -- a statement that two expressions are equal.
  • Linear equation -- one in which the unknown appears only to the power 1.
  • Solution (root) -- the value of the unknown that satisfies the equation.
  • Simultaneous equations -- two or more equations solved together for common values.
  • Elimination -- removing one unknown by adding or subtracting equations.

:::checkpoint Check yourself

  1. Solve 5(x − 3) = 20.
  2. Solve x/4 + x/2 = 9.
  3. Solve by elimination: 3x + y = 11 and x + y = 5.
  4. Why must you check in both equations rather than one? :::

Bridge to practice

The exercises begin with solving in one unknown, move through brackets, fractions, substitution and elimination, and finish with word problems. For every simultaneous pair, decide which method suits before starting — the choice usually saves more time than it costs.

Check yourselfPractise Linear Equations10 questions →Next in MathematicsCommercial Arithmetic (I)