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Cambridge IGCSE0607

Equations

International Mathematics 0607 Chapter Notes

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Equations
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In this chapter

  1. 1Linear equations and simple formulas
  2. 2Simultaneous equations and the calculator
  3. 3Fractional equations
  4. 4Solving quadratic equations
  5. 5Changing the subject of harder formulas
  6. 6Unfamiliar equations on the calculator

1. Linear equations and simple formulas

To construct an expression, write the words as algebra: "a number that is 2 more than n" is n + 2. Set an expression equal to the value you are given and you have an equation.

To solve a linear equation in one unknown, do the same to both sides until the letter is alone. Expand any bracket first. Collect the letters on one side and the numbers on the other. A term that crosses to the other side changes sign.

Changing the subject of a simple formula uses the same steps. The new subject appears once, with no power or root, and ends up alone on one side.

5 - 2x = 3(x + 7) ⇒ 5 - 2x = 3x + 21 ⇒ -16 = 5x

A = 3b - 2c ⇒ 3b = A + 2c ⇒ b = (A + 2c)/3

Examiner insight

“Some made errors in their first step, for example rearranging 8x − 3 = − 11 to 8x = − 11 – 3.” “There were occasional slips in the signs of their terms after they collated like terms.”

Worked example 14 marks

Ryan is paid textdollar12 each day to pick asparagus. Ryan is also paid textdollar1.50 for each kilogram of asparagus he picks.

(b) Ryan works for D days and picks a total of A kg of asparagus. Ryan is paid a total of textdollarP. Write down a formula for P in terms of D and A.

(c) Ryan works for 5 days and is paid a total of textdollar189. Work out how many kilograms of asparagus Ryan picks in these 5 days.

  1. 1

    (b) daily pay for D days is 12D, pay for A kg is 1.5A [1]

  2. 2

    P = 12D + 1.5A [1]

  3. 3

    (c) substitute: 189 = 12 × 5 + 1.5A, so 1.5A = 129 [1]

  4. 4

    A = 129 ÷ 1.5 = 86, so he picks 86 kg [1]

2. Simultaneous equations and the calculator

Two unknowns need two equations. To eliminate one letter, multiply one or both equations so that the letter has the same coefficient in each. Multiply every term. Then subtract the equations if those two terms have the same sign, and add if the signs differ.

Substitute the value you find into either equation to find the other letter. Then check that your pair of values fits both equations.

A graphic display calculator solves an equation, even an unfamiliar one such as 2^(x) = x². Enter the left-hand side as one function and the right-hand side as another. Draw both graphs. Read off the x-coordinate of each point where they meet.

4x + 3y = 18 and 2x - 3y = 0 ⇒ add: 6x = 18

Examiner insight

“Some candidates made algebraic errors in the manipulation of the equations. Often, this was to choose to add the equations when they should be subtracted to eliminate one of the variables.” “Candidates should check their answers in both equations – this step was rarely seen.”

Worked example 14 marks

Solve the simultaneous equations.

6x + 7y = 32

4x - 3y = 6

  1. 1

    make the x terms equal: (× 2) gives 12x + 14y = 64 and (× 3) gives 12x - 9y = 18 [1]

  2. 2

    both are +12x, so subtract: 14y - (-9y) = 64 - 18, so 23y = 46 [1]

  3. 3

    y = 2 [1]

  4. 4

    substitute: 6x + 14 = 32, so x = 3 [1]

3. Fractional equations

A fractional equation has numbers or linear expressions such as 2x + 1 in its denominators. Multiply both sides by the common denominator, so that no fraction is left. Put brackets round each numerator as you multiply.

Then expand the brackets, collect the terms and solve. The equation you reach may be linear or quadratic.

x/(2x + 1) = 4 ⇒ x = 4(2x + 1)

x/(x + 2) = 3/(x - 6) ⇒ x(x - 6) = 3(x + 2)

Examiner insight

“Putting the left-hand side over a common denominator and then eliminating fractions was more common than the more succinct method of multiplying both sides by (2x – 5) (x + 1) as a first step.” “The majority of candidates were able to score the first method mark for finding the common denominator but were then careless with the signs in the expansion of their brackets in the numerator.”

Worked example 13 marks

Solve.

(t + 1)/7 = (4 - t)/2

  1. 1

    multiply both sides by 14: 2(t + 1) = 7(4 -t) [1]

  2. 2

    2t + 2 = 28 - 7t, so 9t = 26 [1]

  3. 3

    t = 26/9, which is 28/9 [1]

4. Solving quadratic equations

Write the equation as ax² + bx + c = 0. To solve by factorisation, split the middle term into two terms that multiply to give ac and add to give b. Factorise in pairs, then set each bracket equal to zero. A question that says "by factorisation" earns only partial credit for the formula.

The quadratic formula is given. When you must show all your working, write the formula with the values of a, b and c substituted. You may be expected to give solutions in surd form.

On a calculator paper a graphic display calculator also solves a quadratic. Draw y = ax² + bx + c and read off the x-values where the graph crosses the x-axis.

x = (-b ± √(b² - 4ac))/2a

(x + 1)² = 9 ⇒ x + 1 = ± 3 ⇒ x = 2 or x = -4

Examiner insight

“Many candidates tried to split the middle term into two terms that added to give –13 and multiplied to give –30.” “A few ignored the instruction to use factorisation and used the formula.”

Worked example 13 marks

Solve by factorisation.

6x² - 13x - 5 = 0

  1. 1

    ac = -30 and b = -13: use -15 and +2, so 6x² - 15x + 2x - 5 = 0 [1]

  2. 2

    3x(2x - 5) + 1(2x - 5) = 0, so (3x + 1)(2x - 5) = 0 [1]

  3. 3

    x = -1/3 or x = 5/2 [1]

5. Changing the subject of harder formulas

If the new subject is squared, get the squared term alone first, then take the square root of the whole of the other side. If it is under a root, get the root alone and square both sides.

If the new subject appears twice, clear any fraction and expand. Collect every term that contains the new subject on one side and everything else on the other. Then factorise it out and divide by the bracket. The final answer must not have the new subject on both sides.

y = (3x + 1)/(x - 2) ⇒ xy - 2y = 3x + 1 ⇒ xy - 3x = 1 + 2y

x(y - 3) = 1 + 2y ⇒ x = (1 + 2y)/(y - 3)

Examiner insight

“Consequently, a significant number of candidates gave answers which had x as a function of x.” “However, the factorisation of the left-hand side proved to be very demanding.”

Worked example 13 marks

X = c² - 9d²

(b) Rearrange the formula to write d in terms of X and c.

  1. 1

    get the term in d alone: 9d² = c² - X [1]

  2. 2

    divide by 9: d² = (c² - X)/9 [1]

  3. 3

    square root: d = √((c² - X)/9) [1]

6. Unfamiliar equations on the calculator

Some equations, such as 2^(x) - 1 = 1/x³, are solved with a graphic display calculator. Enter the left-hand side as one function and the right-hand side as another. Draw both graphs, and change the view until you can see every point where they meet. Read off the x-coordinate of each intersection.

Show a sketch of the two graphs as your working. Give every solution, to the accuracy asked.

Examiner insight

“Many gave answers without any working shown. Candidates should show or refer to a sketch if that is their method of solving a problem.”

Worked example 13 marks

Solve 2^(x) - 1 = 1/x³.

Give each answer correct to 2 decimal places.

  1. 1

    enter Y₁ = 2^(x) - 1 and Y₂ = 1/x³, then sketch both graphs [1]

  2. 2

    the graphs meet twice: once where x is negative and once where x is positive [1]

  3. 3

    x = -1.21 and x = 1.00 [1]

Recap

  • Construct an equation from the words, then solve it: same operation on both sides.
  • Elimination: multiply every term, then subtract (same signs) or add (different signs).
  • Check both values in both equations.
  • Calculator: graph each side and read the x-coordinate of each intersection.
  • Extended: multiply both sides of a fractional equation by the common denominator.
  • Extended: solve a quadratic by factorisation, by the formula or by calculator.
  • Extended: if the new subject appears twice, collect it on one side and factorise.

End-of-chapter exercise

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

  1. Do not use a calculator. Solve 7 - 3x = 2(x + 11).3 marks
  2. A rectangle has length (2x + 3) cm and width (x - 1) cm. Its perimeter is 40 cm. Find x.3 marks
  3. Do not use a calculator. Solve the simultaneous equations. 2x + 3y = 13 2x - y = 13 marks
  4. Do not use a calculator. Two pens and one notebook cost textdollar3.50. One pen and three notebooks cost textdollar6.00. Find the cost of one pen and the cost of one notebook.4 marks
  5. Use a graphic display calculator to solve x³ = 3x + 1. Give your answers correct to 2 decimal places.3 marks
  6. Do not use a calculator. P = 4q - 3r Make q the subject. Then find q when P = 7.4 and r = 1.6.3 marks
  7. Do not use a calculator. Solve x² - 6x + 2 = 0. Give your answers in the form p ± √q.3 marks
  8. Do not use a calculator. Solve 3/(x + 1) + 2/(x - 2) = 2.4 marks
  9. A prize of textdollar600 is shared equally between n people. If 5 more people had shared the prize, each person would have received textdollar6 less. (a) Show that n² + 5n - 500 = 0. (b) Find the amount each of the n people receives.6 marks
  10. Three positive lengths f, u and v, in centimetres, are connected by the formula 1/f = 1/u + 1/v. (a) Make v the subject of the formula. (b) In one case f = 6 and v is 5 more than u. Find u.7 marks

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