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Cambridge IGCSE0606

Simultaneous equations

Mathematics Additional 0606 Chapter Notes

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Simultaneous equations
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In this chapter

  1. 1Substitution: make one unknown the subject
  2. 2Substituting into a squared expression
  3. 3Products, and pairing the values
  4. 4Equations with fractions
  5. 5Elimination

1. Substitution: make one unknown the subject

To solve simultaneous equations in two unknowns, find the pair of values that satisfies both equations. Use elimination or substitution.

When one equation is linear, make one unknown the subject of it. Then substitute that expression into the other equation.

Choose the unknown with coefficient 1, so that no fractions appear. Substitute so the result has only one unknown.

Examiner insight

“Choosing to find y in terms of x from the second equation and substituting in the first meant that the substitution had to be made in just one place.”

Worked example 14 marks

Solve the simultaneous equations 2x + y = 7 and xy = 6.

  1. 1

    make y the subject: y = 7 - 2x [1]

  2. 2

    substitute: x(7 - 2x) = 6, so 2x² - 7x + 6 = 0 [1]

  3. 3

    factorise: (2x - 3)(x - 2) = 0, so x = 3/2 or x = 2 [1]

  4. 4

    y = 7 - 2x gives x = 3/2, y = 4 and x = 2, y = 3 [1]

2. Substituting into a squared expression

A non-linear equation often has x², y² and xy. After substituting, expand the whole bracket when it is squared. The cross term must be included.

Collect everything on one side to give a three-term quadratic equal to zero. Factorise it, or use the quadratic formula.

Do not cancel an unknown from both sides: factorise instead.

(a - b)² = a² - 2ab + b²

(a + b)² = a² + 2ab + b²

Examiner insight

“Some candidates would have done better if they had taken more care with the substitution as slips, such as omitting the square, were seen on occasion.”

Worked example 15 marks

Solve the simultaneous equations y - x + 3 = 0 and x² - 3xy + y² + 19 = 0.

  1. 1

    make y the subject: y = x - 3 [1]

  2. 2

    substitute: x² - 3x(x - 3) + (x - 3)² + 19 = 0 [1]

  3. 3

    expand: x² - 3x² + 9x + x² - 6x + 9 + 19 = 0, so x² - 3x - 28 = 0 [1]

  4. 4

    factorise: (x - 7)(x + 4) = 0, so x = 7 or x = -4 [1]

  5. 5

    y = x - 3 gives x = 7, y = 4 and x = -4, y = -7 [1]

3. Products, and pairing the values

Some pairs have no linear equation. When one equation is a product that contains the other, such as xy² and xy, write the longer product in terms of the shorter one and replace it by its value.

Every solution is a pair of values. Find the second unknown from the simpler equation, and write each x with its own y.

Keep every pair, including negative and fractional values.

Examiner insight

“Other candidates had a quadratic equation in terms of y and then solved this as if it were x, going on to incorrectly work out y from these ‘x’ values.”

Worked example 13 marks

Solve the simultaneous equations xy² = 4 and xy = 3.

  1. 1

    xy² = (xy) × y, so 3y = 4 [1]

  2. 2

    y = 4/3 [1]

  3. 3

    x = 3 ÷ 4/3 = 9/4 [1]

4. Equations with fractions

When an equation has fractions, substitute first. Then multiply every term by the common denominator to clear the fractions.

Expand, collect on one side and solve. Leave the answers exact unless the question asks for a decimal.

Examiner insight

“Sometimes these candidates omitted to multiply all terms when clearing the fraction or, again, they made sign errors.”

Worked example 15 marks

Solve the simultaneous equations x/y + 2y/x = 4 and y = x - 2. Give your answers in surd form.

  1. 1

    substitute: x/(x - 2) + (2(x - 2))/x = 4 [1]

  2. 2

    multiply every term by x(x - 2): x² + 2(x - 2)² = 4x(x - 2) [1]

  3. 3

    expand: 3x² - 8x + 8 = 4x² - 8x [1]

  4. 4

    x² = 8, so x = ± 2√2 [1]

  5. 5

    y = x - 2 gives x = 2√2, y = 2√2 - 2 and x = -2√2, y = -2√2 - 2 [1]

5. Elimination

Eliminate an unknown by adding or subtracting the equations. First multiply every term of an equation, the right-hand side included, so that one pair of coefficients matches.

When the unknowns appear only as x² and y², eliminate one square. Then take both square roots. Each value of x can go with each value of y, so check every pair.

Examiner insight

“Another common error was to forget to multiply every term in one of the equations in preparation for elimination of one variable.” “A common error was to forget the negative square roots in the final step, or to find all four roots and then reject the negative ones as not being possible.”

Worked example 14 marks

Solve the simultaneous equations x² + y² = 13 and x² - 2y² = -14.

  1. 1

    subtract: 3y² = 27 [1]

  2. 2

    y² = 9, so y = ± 3 [1]

  3. 3

    x² = 13 - 9 = 4, so x = ± 2 [1]

  4. 4

    four pairs (x, y): (2, 3), (2, -3), (-2, 3), (-2, -3) [1]

Recap

  • Solve by elimination or substitution. The answer is a pair of values.
  • With a linear equation, make one unknown the subject, then substitute into the other equation.
  • Expand a squared bracket in full: (a - b)² = a² - 2ab + b².
  • Collect to a three-term quadratic equal to zero, then factorise.
  • Clear fractions by multiplying every term. Leave answers exact.
  • To eliminate, multiply every term of an equation, then add or subtract.
  • Keep every pair, including negative and fractional values, and pair each x with its y.

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. A pair of equations has one linear equation and one non-linear equation. State what you do to the linear equation before you substitute.1 mark
  2. Do not use a calculator. Solve 3x + 2y = 7 and 5x - 3y = -1.3 marks
  3. Solve y = 3x - 1 and xy = 10.3 marks
  4. Do not use a calculator. Solve x - 2y = 3 and x² + y² = 5.4 marks
  5. Do not use a calculator. Solve 2x + 3y = 12 and xy = 6.4 marks
  6. Do not use a calculator. Solve x²y = 50 and xy = -10.3 marks
  7. Do not use a calculator. Solve x² + 3y² = 28 and 2x² - y² = 28.4 marks
  8. Do not use a calculator. Solve x/y + y/x = 5/2 and y = x - 1.5 marks
  9. Two numbers are x and y. When y is subtracted from twice x, the result is 3. The square of the difference between x and y, added to three times x, is 9. Find all the possible pairs of values of x and y.5 marks
  10. The product of two numbers x and y is 6. The square of x added to four times the square of y is 40. Find all the possible pairs of values of x and y.6 marks

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