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Solving Equations IGCSE – Chapter 2

Solving Equations IGCSE – Chapter 2. Note 1 : Solving Equations. The value of x that makes the equation true is called the solution. To solve : Isolate x by itself Keep the equation balanced (do the same to both sides) Line up the equal signs at each step Start each step on a new line.

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Solving Equations IGCSE – Chapter 2

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  1. Solving EquationsIGCSE – Chapter 2

  2. Note 1: Solving Equations The value of x that makes the equation true is called the solution. To solve: • Isolate x by itself • Keep the equation balanced (do the same to both sides) • Line up the equal signs at each step • Start each step on a new line

  3. Solve the following equations (addition & subtraction) 1.) x + 4 = 7 2.) x – 41 = 37 3.) x –11= -9 4.) x + 4 = -1 5.) x + 6 = -18 6.) x – 6 = -8 x = -9+11 x = 7 - 4 x = 37 + 41 x = 78 x = 2 x = 3 x = -1 -4 x = -8 + 6 x = -18 -6 x = -2 x = -5 x = -24 * Show three lines of working

  4. Solve the following equations (undoing multiplication) 1.) -3x = 21 2.) 7x = -4 3.) 6x = 9 4.) -13x = 26 5.) -8x = -48 6.) -x = -11 x = x = x = x = x = -7 x = x = x = x = -2 x = 6 x = 11 * Show three lines of working

  5. Solve the following equations(undoing division) 1.) = 2 2.) = -1 3.) = -50 4.) = 15 5.) 0.2x = 3 6.) = x = -50 x 2 x = 2 x 8 x = -1 x 3 x = 16 x = -100 x = -3 x = 15 x 5 x = 3 x 5 x = 3 x = 75 x = 15 * Show three lines of working

  6. Solve the following equations IGCSE Ex 11 Pg 53 1.) 2x + 3 = 11 2.) 2x + 5 = -3 3.) 8x – 1 = -9 4.) 10x -11 = 39 5.) 4x + 6 = -18 6.) -6x + 1 = 49 8x = -9+1 2x = 11 - 3 2x = -3 - 5 2x = 8 2x = -8 8x = -8 x = 4 x = -4 x = -1 10x = 39+11 4x = -18 -6 -6x = 49 - 1 4x = -24 10x = 50 -6x = 48 x = 5 x = -6 x = -8 * Show three lines of working

  7. Note 2: Solving equations with fractions When solving equations involving fractions, multiply both sides of the equation by a suitable number to eliminate the fractions. * Multiply both sides by 14 = 14 14 * It is possible to skip this step by cross-multiplying = 7(x+3) = 2(2x-3) * Expand and simplify 7x+21 = 4x - 6 x = -9 3x = -27

  8. Note 2: Solving equations with fractions = -1 5 - * Subtract 5 from both sides e.g. = -6 - 1 * cross-multiplying -6 = -6x * Divide both sides by -6 x = 1

  9. Solving word problems by linear equations • Set the unknown quantity to be a letter (e.g. x) and state its units • Express the given statement as an equation • Solve the equation for x and give the answer in words. • Check your solution works (in the problem) e.g. The sum of 3 consecutive whole numbers is 96. Find the numbers. a.) Let the smallest # be x, the other numbers are (x+1) & (x+2) b.) x + (x+1) + (x+2) = 96 c.) 3x + 3 = 96 In words: The numbers are 31, 32 and 33. 3x = 93 IGCSE Ex 14 Pg 56-57 odd Ex 15 Pg 58-60 odd x = 31 d.) Check: 31 + 32 + 33 = 96

  10. Starter A tennis racket costs $12 more than a hockey stick. If the price of the two is $31, find the cost of the tennis racket. x - 12 Let x be the cost of the tennis racket, and a hockey stick costs ______ x + (x-12) = 31 2x = 31 + 12 2x = 43 The tennis racket costs $21.50 x = 21.50 Check: $21.50 + 9.50 = $31.00

  11. Solving word problems by linear equations e.g. A man is 32 years older than his son. Ten years ago he was three times as old as his son was then. Find the present age of each. a.) Let the age of the father be x and his sons age be (x - 32) b.) x -10 = 3(x-42) c.) x - 10 = 3x - 126 -2x = -116 In words: x = 58 The fathers age is 58 and his son is 26 d.) Check: 26 + 32 = 58 48 = 16 x 3

  12. Solving word problems by linear equations e.g. The length of a straight line ABC is 10m. If AB:BC = 2:3, find the length of AB C A B a.) Let x be the length of 1/5th of the line (in metres) and let AB be 2x. b.) = x c.) x = 2 AB = 2 x 2 In words: The length of portion AB is 4 metres. AB = 4 m d.) Check: Gives a ratio of 2:3 and 4m + 6m = 10m

  13. Solving word problems by linear equations e.g. A bus is travelling with 48 passengers. When it arrives at a stop, x passengers get off and 3 get on. At the next stop half of the passengers get off and 7 get on. There are now 22 passengers. Find x. a.) Let x be the number of passengers that originally get off the bus b.) +7 = 22 c.) = 15 -x + 51 = 30 In words: 21 people originally get off the bus x = 21 IGCSE Ex 15 Pg 58-60 odd Ex 16 Pg 61-62 odd d.) Check: 48-21+3 = 30 + 7 = 22

  14. Solving word problems by linear equations e.g. A man runs to a telephone and back in 15 minutes. His speed on the way to the phone is 5 m/s and his speed on the way back is 4m/s. Find the distance to the phone. S = a.) Let d be the distance to the phone (in metres) t = b.) = 900 s + 900 15 min = ____ s c.) = 900 s In words: 9d = 18000 d = 2000 m The distance is 2 km. d.) Check: + = 15 minutes

  15. Note 4: Simultaneous Equations In order to solve for the value of two unknowns in a problem, you must have two different equations that relate to the unknowns Substitution Substitution is often used when one of the equations contains a single unit quantity of an unknown. IGCSE Ex 17 Pg 63 odd e.g. 2x – 3y = 34 (1) Label the equations 5x + y = 0 (2) y = -5x Rearrange for the single unit quantity (y here) 2x – 3(-5x) = 34 Substitute this expression for y into (1) & solve for x 17x = 34 x = 2 Solve for y by substituting x = 2 into (2) 5(2) + y = 0 y = -10

  16. Note 5: Simultaneous Equations Elimination This method is often preferred and can be used if substitution is not suitable e.g. (1) x + 2y = 10 Label the equations x 2 to eliminate x in this case 2x + 3y = 14 (2) 2x + 4y = 20 Multiply (1) by an appropriate # to eliminate either x or y 2x + 3y = 14 Subtract (2) from (1) – notice we have eliminated the x term y = 6 x + 2(6) = 10 Solve for x by substituting y = 6 into either (1) or (2) x + 12 = 10 x = -2 *try graphing this

  17. Using the Cover-up method 2x + 3y = 14 x + 2y = 10

  18. Note 5: Problems Solved using Simultaneous Equations The first OPEC oil shock occurred in 1973, gas prices doubled from 15 cents /L to 30 cents /L e.g. A motorist buys 24 L of petrol & 5 L of oil for $10.70, while another motorist buys 18 L of petrol and 10 L of oil for $12.40. Find the cost of 1 L of petrol and 1 L of oil at this garage. Let x be the cost of 1 L of petrol, let y be the cost of 1 L of oil (in cents) 24x + 5y = 1070 (1) 18x + 10y = 1240 (2) 48x + 10y = 2140 *multiply (1) by 2 18x + 10y = 1240 IGCSE Ex 19 Pg 65 odd Ex 20 Pg 66-67 odd 30x = 900 x = 30 *subst into (2) y = 70 18(30) + 10y = 1240

  19. Note 6: Factorising The reverse of expanding is factorising e.g.Factorise the following: x2 + 10x a.) x is the common factor to x2 and 10x x (x + 10) The factors are x and (x+10) 5y is the common factor to 5y2 and 25y 5y2 + 25y b.) 5y(y + 5) The factors are 5y and (y+5) c.) 3a2b3 + 12ab2 3ab2 is the common factor to 3a2b3 and 12ab2 3ab2(ab + 4) The factors are 3ab2 and (ab+4)

  20. Note 6: Factorising e.g.Factorise the following by dividing into pairs: a.) ah + ak + bh + bk Divide into pairs Factorise each pair a(h+k) +b(h+k) (h+k) is common to both terms (h+k) (a+b) The factors are (h+k) and (a+b) Divide into pairs ms+2mt2 – ns - 2nt2 b.) m(s+2t2) – n(s+2t2) Factorise each pair (s+2t2) (m-n) (s+2t2) is common to both terms The factors are (s+2t2) and (m-n)

  21. IGCSE Ex 21 Pg 68 odd Ex 22 Pg 69 odd Ex 23 Pg 69 odd Note 6: Factorising e.g.Factorise simple quadratics: a.) x2 + 6x + 8 Find 2 numbers which multiply to 8 and add to 6 (x+2) (x+4) 4 and 2 Put these number into brackets The factors are (x+2) and (x+4) b.) x2 - 5x - 24 Find 2 numbers which multiply to -24 and add to -5 (x-8) (x+3) -8 and 3 Put these number into brackets The factors are (x-8) and (x+3) x2 + 5x - 24 c.) Find 2 numbers which multiply to -24 and add to 5 (x+8) (x-3) 8 and -3 Put these number into brackets

  22. Note 7: Difference of two Squares Expand (x + y)(x – y) = x2 – y2 Remember this result The middle term cancels out a.) a2-16b2 (a)2-(4b)2 (a-4b) (a+4b) The factors are (a-4b) and (a+4b) 36a3b - 4ab3 b.) There is a common factor of 4ab 4ab(9a2 - b2) 4ab[(3a)2 – b2] 4ab [(3a – b)(3a+b)] Check your solution by expanding

  23. Note 7: Difference of two Squares Evaluate Factorise a.) 812 – 802 = (81-80) (81+80) = (1) (161) = 161 12342-12352 b.) = (1234-1235) (1234+1235) = (-1) (2469) = -2469 Check your solution

  24. Factorisation of quadratic expressions with a ≠ 1 Mulitply the coefficient of x2 and the constant. 3 x 8 Find 2 numbers that multiply to give this value and add to give the coefficient of x Write the quadratic with the x-term split into two x-terms using these numbers Factorise the pairs of terms IGCSE Ex 24 Pg 70 odd Ex 25 Pg 71 odd Factorise again, taking the bracket as the common factor Check by expanding

  25. Starter A wallet containing $40 has three times as many $1 notes as $5 notes. Find the number of each kind. Let x be the number of $1 notes and y be the number of $5 notes (1) 1x + 5y = 40 x + 5y = 40 (2) 3y = x x – 3y = 0 8y = 40 (1) – (2) y = 5 3(5) = x There are 15 $1 notes and 5 $5 notes x = 15

  26. Note 8: Quadratic Equations Quadratic equations always have an x2 term and generally have 2 different solutions If a x b = 0, this means that eithera or b (or both) must be zero. e.g. Solve the equation x2 + x – 12 = 0 (x-3)(x+4) = 0 factorise first! either x - 3 = 0 or x + 4 = 0 x = 3 x = -4 2 solutions

  27. Note 8: Quadratic Equations Quadratic equations always have an x2 term and generally have 2 different solutions 12 x -14 = -168 24, -7 e.g. Solve the equation 12x2 + 17x – 14 = 0 12x2 + 24x – 7x – 14 = 0 split the x term 12x (x+2) -7 (x+2) = 0 factorise pairs (12x-7) (x+2) = 0 either 12x - 7 = 0 or x + 2 = 0 x = -2 x =

  28. Note 8: Quadratic Equations SOLVE THE EQUATIONS e.g. x2+ 9x = 0 9x2 -16 = 0 e.g. x (x+9) = 0 factorise 9x2= 16 either x= 0 or x + 9 = 0 x2= x = -9 x = x = ± IGCSE Ex26 Pg71-72 odd Ex27 Pg72 odd

  29. Use the quadratic formula to solve 2x2 + 7x + 5 = 0 5 a = 2 b = 7 c = x = -7 ± √9 4 7 72 (2)(5) x = -7 ± 3 (2) 4 x = -5 Two Solutions or x = -1 2

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