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O-Level Mathematics · Secondary 3 and 4

Simultaneous Equations

Simultaneous equations are two equations that must both be true at the same time. In O-Level Mathematics you solve pairs of linear equations in two unknowns by elimination, by substitution and by reading the intersection of two straight-line graphs. You also turn word problems about costs, ages and numbers into a pair of equations. The same skills are used later when lines meet in coordinate geometry and in Additional Mathematics, where one of the equations may be a curve.

What the syllabus expects

  • Solve a pair of linear equations in two unknowns by elimination
  • Solve a pair of linear equations in two unknowns by substitution
  • Interpret the solution as the point of intersection of two straight-line graphs
  • Form a pair of simultaneous equations from information given in words and solve them
  • Check a solution by substituting into both original equations

What a solution means

A single equation such as x + y = 10 has endless solutions: x = 1 and y = 9, x = 4 and y = 6, and so on. A second equation, such as x − y = 4, narrows these down to the one pair that satisfies both, here x = 7 and y = 3. On a graph, each linear equation is a straight line, and the solution is the point where the two lines cross, (7, 3). This picture explains the unusual cases. If the two lines are parallel, they never meet and there is no solution, for example x + y = 4 and x + y = 6. If the two equations describe the same line, for example x + y = 4 and 2x + 2y = 8, every point on the line works. In exam questions you will almost always meet the normal case of exactly one solution, and your answer should state both values.

Elimination

Elimination works by adding or subtracting the equations so that one unknown disappears. First make the coefficients of one unknown equal in size by multiplying one or both equations. For 3x + 2y = 16 and 5x − 3y = 14, multiply the first by 3 and the second by 2 so that the y terms become +6y and −6y. Because the signs are different, add the equations, which removes y. If the signs had been the same, you would subtract instead. A helpful rule is: same signs subtract, different signs add. Number your equations (1) and (2) and label new ones (3) and (4), so that anyone reading can see which equation you multiplied. After finding one unknown, substitute it into whichever original equation is simplest to find the other.

Substitution

Substitution is quicker when one equation already gives one unknown in terms of the other, such as y = x − 4. Replace y in the other equation with (x − 4), keeping the brackets. With 2x + 3y = 13 this gives 2x + 3(x − 4) = 13, so 5x − 12 = 13, 5x = 25 and x = 5. Then y = 5 − 4 = 1. If neither equation is arranged this way, you can rearrange one of them, but choose an unknown with a coefficient of 1 to avoid fractions. For instance, from x + 2y = 7 you can write x = 7 − 2y and substitute that expression. Substitution is also the method you will need when one equation is linear and the other is a quadratic, so it is worth becoming fluent with it now.

Forming equations from words

In a word problem, start by defining both unknowns with units, for example 'let the cost of one adult ticket be $a and one child ticket be $c'. Each sentence that gives a total becomes one equation, so three adult and two child tickets costing $46 gives 3a + 2c = 46. Keep the units consistent: if some prices are in cents and others in dollars, convert first. Age problems need care with time. If a mother is now m years old and her son is s years old, then in 5 years' time their ages are m + 5 and s + 5, so 'in 5 years the mother will be twice as old as her son' becomes m + 5 = 2(s + 5). Finish by answering the question that was asked in a sentence, rather than leaving the answer as a = 12 and c = 5.

Worked example

Solve the simultaneous equations 3x + 2y = 16 and 5x − 3y = 14.

  1. Label the equations: 3x + 2y = 16 is (1) and 5x − 3y = 14 is (2).
  2. Multiply (1) by 3 to get 9x + 6y = 48, which is (3).
  3. Multiply (2) by 2 to get 10x − 6y = 28, which is (4).
  4. The y terms have different signs, so add (3) and (4): 19x = 76, which gives x = 4.
  5. Substitute x = 4 into (1): 12 + 2y = 16, so 2y = 4 and y = 2.
  6. Check in (2): 5(4) − 3(2) = 20 − 6 = 14, which is correct.

Answer: x = 4 and y = 2

Common mistakes

  • Multiplying only one side of an equation. When you multiply 3x + 2y = 16 by 3, every term changes, including the right-hand side, giving 9x + 6y = 48, not 9x + 6y = 16.
  • Adding when you should subtract. If both equations contain +6y, adding gives 12y instead of removing y. Check the signs of the matching terms before you combine the equations.
  • Sign errors when subtracting a whole equation. Subtracting 2x − 3y = 5 means subtracting 2x, adding 3y and subtracting 5. Writing the subtraction term by term, or multiplying by −1 and adding, avoids this.
  • Dropping the brackets in substitution. Replacing y with x − 4 in 2x + 3y = 13 must give 2x + 3(x − 4), not 2x + 3x − 4.
  • Stopping after one unknown. Both values are needed, and the answer to a word problem must be stated in context, such as 'an adult ticket costs $12'.
  • Checking in the wrong equation. Substituting into the equation you used to find the second unknown always seems to work, so check in the other original equation instead.

Exam tips

  • Before multiplying, decide which unknown needs the smaller multipliers. Eliminating the unknown whose coefficients are already factors of each other saves arithmetic.
  • If the question asks you to solve graphically, read the intersection point carefully from the grid and give values to the accuracy the scale allows, usually to the nearest tenth of a unit.
  • When a word problem gives two different totals, you almost always need two unknowns and two equations. Write a 'let' statement with units first, so your equations are clear to you and to anyone marking them.

Practice questions

1. Solve x + y = 10 and x − y = 4.

Show answer

x = 7 and y = 3. Adding the equations gives 2x = 14, so x = 7. Then 7 + y = 10 gives y = 3.

2. Solve 2x + 3y = 13 and y = x − 4.

Show answer

x = 5 and y = 1. Substitute: 2x + 3(x − 4) = 13, so 5x − 12 = 13, 5x = 25 and x = 5. Then y = 5 − 4 = 1. Check: 2(5) + 3(1) = 13.

3. Three adult tickets and two child tickets cost $46. Two adult tickets and five child tickets cost $49. Find the cost of one adult ticket and of one child ticket.

Show answer

An adult ticket costs $12 and a child ticket costs $5. Let the prices be $a and $c, so 3a + 2c = 46 and 2a + 5c = 49. Multiply by 2 and by 3 to get 6a + 4c = 92 and 6a + 15c = 147. Subtracting gives 11c = 55, so c = 5. Then 3a + 10 = 46 gives a = 12. Check: 2(12) + 5(5) = 49.

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