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Unit III · Pair of linear equations in two variables
7 min readPathway sheet2-mark or 3-mark question

Solving by elimination

Multiply the equations so one variable has equal coefficients, then add or subtract to remove it. Solve, then substitute back.

Same signsSubtract the equations
Opposite signsAdd the equations
Equal coefficientsMultiply, e.g. by 3 and by 2, to make them equal
Best whenNo variable has coefficient 1

Method

  1. 1
    Write both equations in the form ax + by = c.
  2. 2
    Multiply one or both equations so that x (or y) has the same number in front in both.
  3. 3★
    If the equal coefficients have the same sign, subtract. If they have opposite signs, add.
  4. 4
    Solve the equation in one variable.
  5. 5
    Substitute back into either original equation to find the other variable.
  6. 6
    Check in both equations.

★ key step

Example 1: multiply one equation

Problem

Solve by elimination: 5x + 2y = 16 and 3x − 4y = 20.

Working
  1. Make the y terms equal: multiply the first equation by 2: 10x + 4y = 32.
  2. The y terms are +4y and −4y, opposite signs, so add.
  3. (10x + 4y) + (3x − 4y) = 32 + 20, so 13x = 52 and x = 4.
  4. Substitute in 5x + 2y = 16: 20 + 2y = 16, so 2y = −4 and y = −2.
  5. Check: 3(4) − 4(−2) = 12 + 8 = 20.
Answer

x = 4, y = −2

Example 2: the swapped-coefficients trick

Problem

Solve: 4x + 3y = 25 and 3x + 4y = 24.

Working
  1. Add the equations: 7x + 7y = 49, so x + y = 7.
  2. Subtract the second from the first: x − y = 1.
  3. Add these two: 2x = 8, so x = 4. Then y = 7 − 4 = 3.
  4. Check: 4(4) + 3(3) = 25 and 3(4) + 4(3) = 24.
Answer

x = 4, y = 3

Example 3: a fraction word problem

Problem

If 1 is subtracted from the numerator of a fraction, it becomes 1/2. If 7 is added to its denominator, it becomes 1/3. Find the fraction.

Working
  1. Let the fraction be x/y.
  2. (x − 1)/y = 1/2 gives 2x − 2 = y, so 2x − y = 2.
  3. x/(y + 7) = 1/3 gives 3x = y + 7, so 3x − y = 7.
  4. The y terms are equal with the same sign, so subtract: (3x − y) − (2x − y) = 7 − 2, so x = 5.
  5. Then y = 2(5) − 2 = 8.
  6. Check: (5 − 1)/8 = 4/8 = 1/2 and 5/(8 + 7) = 5/15 = 1/3.
Answer

The fraction is 5/8.

Substitution or elimination?

SubstitutionBest when a coefficient is 1 or −1, as in x + 2y = 7.
EliminationBest when all coefficients are larger, as in 5x + 3y = 21. It avoids fractions.

Both methods give the same answer. If the question names a method, you must use that method.

In real life

Suppose 3 samosas and 2 teas cost ₹ 50, and 2 samosas and 3 teas cost ₹ 45 (example prices). Elimination solves this in two lines: a samosa costs ₹ 12 and a tea ₹ 7.

Exam angle

1 marks: “To eliminate y from 2x + 3y = 7 and 5x − 2y = 8, what should you multiply the equations by?” Multiply the first by 2 and the second by 3 (giving +6y and −6y), then add.

2 marks: “Solve by elimination: 3x + 2y = 11 and 2x + 3y = 4.” Adding: 5x + 5y = 15, so x + y = 3. Subtracting: x − y = 7. So x = 5, y = −2.

3 marks: “Solve by elimination: 9x − 4y = 8 and 13x + 7y = 101.” Multiply by 7 and by 4: 63x − 28y = 56 and 52x + 28y = 404. Add: 115x = 460, x = 4. Then 36 − 4y = 8, y = 7.

Remember it

Same signs subtract, opposite signs add. (Like 'SS' and 'OA'.)

Read alongside the books on your syllabus: NCERT, Mathematics: Textbook for Class X · NCERT, Exemplar Problems: Mathematics, Class X. Written from standard textbooks; faculty review in progress. Spotted an error? Report it.