Linear Equations

What makes an equation linear

A linear equation in one variable is one where the variable only appears to the first power: no x2x^2, no x\sqrt{x}, no 1x\dfrac{1}{x}. After simplifying, such an equation can be written as

ax+b=0  ⇒  x=−ba(a≠0)ax + b = 0 \;\Rightarrow\; x = -\dfrac{b}{a} \quad (a \neq 0)

so whenever a≠0a \neq 0 it has exactly one solution. If the xx terms cancel completely (a=0a = 0), there is either no solution, as with x+1=x+2x + 1 = x + 2, or every xx works, as with x+1=x+1x + 1 = x + 1.

The four steps

  1. Clear fractions by multiplying every term on both sides by the LCM of the denominators.
  2. Expand brackets and simplify each side.
  3. Collect the variable terms on one side and the numbers on the other. A term changes sign when it crosses the equals sign.
  4. Divide by the coefficient of xx, then substitute back into the original equation to check.

Worked example: the variable on both sides

Solve 3x−5=x+73x - 5 = x + 7.

Solution: Subtract xx from both sides and add 5 to both sides:

3x−x=7+5  ⇒  2x=12  ⇒  x=63x - x = 7 + 5 \;\Rightarrow\; 2x = 12 \;\Rightarrow\; x = 6

Check: 3(6)−5=133(6) - 5 = 13 and 6+7=136 + 7 = 13.

Worked example: an equation with mixed fractions

Solve x−213=512x - 2\frac{1}{3} = 5\frac{1}{2}.

Solution: Isolate xx by adding 2132\frac{1}{3} to both sides:

x=512+213=(5+2)+(12+13)=7+3+26=756x = 5\tfrac{1}{2} + 2\tfrac{1}{3} = \left(5+2\right) + \left(\tfrac{1}{2}+\tfrac{1}{3}\right) = 7 + \tfrac{3+2}{6} = 7\tfrac{5}{6}

Worked example: fractions with the variable on top

Solve x−23+x+12=4\dfrac{x-2}{3} + \dfrac{x+1}{2} = 4.

Solution: The LCM of 3 and 2 is 6. Multiply every term, including the 4, by 6:

2(x−2)+3(x+1)=24  ⇒  2x−4+3x+3=24  ⇒  5x=25  ⇒  x=52(x-2) + 3(x+1) = 24 \;\Rightarrow\; 2x - 4 + 3x + 3 = 24 \;\Rightarrow\; 5x = 25 \;\Rightarrow\; x = 5

Check: 33+62=1+3=4\dfrac{3}{3} + \dfrac{6}{2} = 1 + 3 = 4.

Turning words into an equation

Common phrases, with x as the unknown number
PhraseExpression
A number increased by 5x+5x + 5
5 less than a numberx−5x - 5
A number subtracted from 55−x5 - x
Twice a number2x2x
One fourth of a numberx4\dfrac{x}{4}
Three consecutive integersx,  x+1,  x+2x, \; x+1, \; x+2
Age 12 years from nowx+12x + 12

Worked example: a word problem

One fourth of a number, added to one sixth of it, is 15. Find the number.

Solution: Let the number be nn:

n4+n6=15\dfrac{n}{4} + \dfrac{n}{6} = 15

Using a common denominator of 12: 3n12+2n12=5n12=15\dfrac{3n}{12} + \dfrac{2n}{12} = \dfrac{5n}{12} = 15, so:

n=15×125=36n = \dfrac{15 \times 12}{5} = 36

Worked example: an age problem

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

Solution: Let the son be xx years old, so the father is 3x3x. In 12 years they will be x+12x + 12 and 3x+123x + 12:

3x+12=2(x+12)  ⇒  3x+12=2x+24  ⇒  x=123x + 12 = 2(x + 12) \;\Rightarrow\; 3x + 12 = 2x + 24 \;\Rightarrow\; x = 12

The son is 12 and the father is 36. Check: in 12 years they will be 24 and 48, and 48=2×2448 = 2 \times 24.

Worked example: consecutive integers

Find three consecutive integers whose sum is 78.

Solution: Call them xx, x+1x+1 and x+2x+2, as in the table above. Their sum is 78:

x+(x+1)+(x+2)=78  ⇒  3x+3=78  ⇒  3x=75  ⇒  x=25x + (x+1) + (x+2) = 78 \;\Rightarrow\; 3x + 3 = 78 \;\Rightarrow\; 3x = 75 \;\Rightarrow\; x = 25

So the integers are 25, 26 and 27. Check: 25+26+27=7825 + 26 + 27 = 78. A shortcut: three consecutive integers add up to 3 times the middle one, so the middle one is 78÷3=2678 \div 3 = 26. The same method gives 110, 111 and 112 for a sum of 333.

Common mistakes

  • Multiplying only some terms by the LCM. In the fractions example, forgetting to multiply the 4 by 6 gives 5x−1=45x - 1 = 4 and the wrong answer x=1x = 1.
  • Losing a minus sign across a bracket. −(x−3)-(x - 3) is −x+3-x + 3, not −x−3-x - 3.
  • Writing "5 less than x" as 5−x5 - x. It's x−5x - 5. Read the phrase as "start with x, then take 5 away".
  • Skipping the check. Substituting the answer back into the original equation catches sign slips before they cost marks.

Where this comes up in exams

The 2025-26 CBSE Class IX syllabus opens its linear equations unit with a recall of linear equations in one variable, before moving on to equations in two variables.

Checked against CBSE Mathematics curriculum, Classes IX-X (2025-26). Syllabi can change from year to year, so confirm with the latest official notification for your exam.

Continue learning

Linear equations are the entry point before quadratic equations, the same isolate-the-variable habit carries over once a squared term is involved. With two unknowns, you need two equations solved together.