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Building Confidence in Equation Solving

Building Confidence in Equation Solving. A guide to solving simple equations to support Primary and Lower Secondary School learners and teachers from First Level Scotland. What is an equation?.

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Building Confidence in Equation Solving

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  1. Building Confidence in Equation Solving K. WILLIAMS RM 206 A guide to solving simple equations to support Primary and Lower Secondary School learners and teachersfrom First Level Scotland

  2. What is an equation? Anequation can be formed whenever a statement about an unknown quantity is written using mathematical symbols. I am 5 years old. You are 2 years older than me. How old are you? becomesyour age = 5 + 2 or using a letter to represent the unknown quantity a = 5 + 2 K. WILLIAMS RM 206

  3. What does solving an equation mean? Solving a simple equation requires changing the original statement to its simplest form, so that the unknown letter stands by itself and is matched by a single number, e.g. a = 5 + 2 becomes a= 7 4a – 1 = 6a + 3 becomes1 = a Notice it doesn’t matter on which side the unknown letter appears. K. WILLIAMS RM 206

  4. Why use an equation? Because there is a bank of methods which can be used to solve quite complicated mathematical statements which can be learned from an early age. Any type of equation can be solved by trial and error, but as the equation becomes more complicated this method becomes less viable. K. WILLIAMS RM 206

  5. Barriers to solving equations Sometimes learners forget why they are learning these methods, or never understand how they work together. K. WILLIAMS RM 206

  6. Some methods Give each method for solving simple equations a name: Shifting Scaling at ThirdLevelSimplifying Sorting at Fourth Level Ungrouping Undoing K. WILLIAMS RM 206

  7. 1 Shifting Shifting an equation can be thought of as adjusting the whole equation, either up or down, to achieve the removal of a number, so decluttering the equation, e.g. too much by 1, get rid of +1 term on LHS (i)x + 1 = 4 x + 1 – 1 = 4 - 1 x = 3 K. WILLIAMS RM 206 = = x x by taking 1 from each side of the equation

  8. 1 Shifting too little, get rid of -3 term on LHS • (ii) x– 3 = 2 • x –3 +3 = 2 + 3 • x = 5 • check: yes, 5-3 = 2 by adding 3 to each side K. WILLIAMS RM 206 Always check solutions, at least mentally, by substitution in the original equation!

  9. 1 Shifting too little, get rid of -5 term on RHS (iii) 2 = r - 5 2 +5 = r –5+5 7 = r check: yes, 2 = 7 - 5 by adding 5 to each side K. WILLIAMS RM 206 Remember that the initial focus is on the unknown term, whichever side of the equation it appears.

  10. 2 Scaling Scaling an equation can be thought of as magnifying or reducing the whole equation, so as to remove a coefficient of the unknown letter, e.g. too big, get rid of the multiplyer5 on LHS (i) 5y = 10 = y = 2 check: yes, 5(2) = 10 K. WILLIAMS RM 206 = 5y y = by dividing each side by 5

  11. 2 Scaling too small, get rid of  4 on LHS (ii) = 2 4( ) = 4(2) x = 8 check: yes, = 2 Notice that writing successive steps of a solution, one underneath the other so that the equals signs line up, helps to identify LHS & RHS of the equation. by multiplying each side by 4 K. WILLIAMS RM 206

  12. 2 Scaling too big, get rid of multiplyer3 on LHS (iii) 3y = 7 = y = or 2 check: yes, 3() = 7 by dividing each side by 3, the coefficient of y K. WILLIAMS RM 206 Notice that with practice, intermediate lines of working can be omitted.

  13. 2 Scaling A problem (iv) Angie has £1.30 and wants to buy some oranges at 25p each. How many can she buy? If Angie buys n oranges, 25n = 130 n= n = 5 check: yes, 25(5) = 125 so she has enough and can’t buy a bit of an orange Angie can buy 5 oranges K. WILLIAMS RM 206

  14. 3 Simplifying Simplify an equation by collecting together any like terms creating one letter term, one number term or removing fractions, e.g. too many letter terms, make one q term on LHS (i) 5q – 2q = 15 3q = 15 q = 5 check: yes, 5(5) – 2(5) = 15 K. WILLIAMS RM 206 by collecting together like terms Notice any letter can be used when there is no good reason to use a particular letter

  15. 3 Simplifying too many numbers, make one numberon LHS (ii) 3+4+3 = 6r 15 = 6r = r r = ,2 or 2·5 check: yes, 6( )= 15 by collecting together numbers K. WILLIAMS RM 206 Notice that the answer should always be simplified if possible.

  16. 3 Simplifying get rid of the fraction (iii) 2c - = 1 3(2c) – 3( ) = 3(1) 6c – c = 3 5c = 3 c = or 0·6 check: yes, 2( )- /3 = - = 1 by scaling, multiplying every term by 3 K. WILLIAMS RM 206

  17. 3 Simplifying A problem (v) Jack and Jill have £25.20 between them. Jack has twice as much money as Jill. How much money does Jill have? If Jill has £p, p + 2p = 25·2 3p = 25·2 p = 8·4 yes, 2(8·4)+8·4 = 16·8+8·4 = 25·2 So Jill has £8.40 K. WILLIAMS RM 206 17

  18. 4 Sorting When unknown terms appear on both sides of an equation, the unknown terms should be sorted onto one side of the equation, e.g. Unknowns on both sides, get rid of sterm as smaller than 4s = (i)s + 6 = 4s s +6-s = 4s-s 6 = 3s 2 = s check: yes, 2+6 = 8 and 4(2) = 8 K. WILLIAMS RM 206 by taking s from each side =

  19. 4 Sorting Unknowns on both sides, get rid of 3xterm as smaller than 8x (ii) 3x + 6 = 8x - 4 3x +6–3x = 8x –4–3x 6 = 5x – 4 6+4 = 5x –4+4 10 = 5x x = 2 check: yes, 3(2)+6=12 and 8(2)–4=12 by taking 3x from each side K. WILLIAMS RM 206

  20. 4 Sorting Unknowns on both sides, get rid of -5p as smaller than -p (iii) 7 – p = 3 – 5p 7–p +5p = 3–5p +5p 7 + 4p= 3 7+4p –7 = 3-7 4p= -4 p = -1 check: yes, 7–(-1)=8 and 3–5(-1)=8 by adding 5p to each side K. WILLIAMS RM 206 Notice that dealing with the letter terms first can help avoid negative coefficients for the unknown

  21. 4 Sorting A problem (iv) Tom was 27 years old when his son was born. Now he is four times as old as his son. How old is his son now? If Tom’s son is y years old, 4y = 27 + y 3y = 27 y = 9 yes, 4(9) = 36 and 9 + 27 = 36 So Tom’s son is now 9 years old K. WILLIAMS RM 206

  22. 5 Ungrouping When the unknown is bound up with other terms, e.g. in brackets or in a fraction, it is usually better to ungroup before any of the previous methods are used, e.g. ungroup the x term (i) 3x +1 = 2(x –1) 3x +1 = 2(x)–2(1) 3x +1 = 2x – 2 3x +1-2x –1 = 2x –2–2x -1 x = -3 check: yes, 3(-3)+1=-8 and 2(-3-1)=-8 by multiplying out brackets K. WILLIAMS RM 206

  23. 5 Ungrouping ungroup the 2r term (ii) = 5 3( ) = 3(5) 2r – 1 = 15 2r –1+1 = 15+1 2r = 16 r = 8 check: yes, = = 5 by multiplying each side by 3 K. WILLIAMS RM 206

  24. ungroup the 4a term 5 Ungrouping (iii) 2(4a – 1) = 6 or 2(4a) – 2(1) = 6 8a – 2 = 6 8a = 8 a = 1 by multiplying out brackets = 4a – 1 = 3 4a = 4 a = 1 K. WILLIAMS RM 206 by dividing each side by 2 check: yes, 2(4(1)–1) = 2(3) = 6

  25. ungroup the y +1expression 5 Ungrouping (iv) y + = 5y – 3 2y + 2() = 2(5y) – 2(3) 2y +(y +1) =10y –6 3y +1 =10y –6 3y +1-3y +6 =10y –6 –3y +6 7 = 7y y = 1 check: 1+ =1+1=2 and5(1)–3=2 by multiplying each side by 2 K. WILLIAMS RM 206 Notice that as equations become more complicated it’s a matter of judgment on which method to use first. It’s often a good idea to get rid of any fractions first. 25

  26. 5 Ungrouping (v) Find the larger of two consecutive odd numbers such that the sum of the larger and double the smaller is 47. If the larger odd number is n, n + 2(n – 2) = 47 n + 2n - 4 = 47 3n - 4 = 47 3n = 51 n = 17 check: 17+2(17–2)=17+30=47 So the larger odd number is 17 K. WILLIAMS RM 206 3 3 21 65 89 47

  27. 6 Undoing When the unknown is part of a function, e.g. a power or trig function, apply the inverse function. The function should be isolated from any other terms before the inverse is applied, e.g. isolate the square function term by taking 1 from each side (i) t2 +1=10 t2+1-1 =10-1 t 2=9 • t 2=9 • t = ±3 check: yes, 32=9and(-3)2 = 9 K. WILLIAMS RM 206 undo the square function by taking the square root of each side

  28. 6 Undoing undo the square root function (ii) 3s = 4 (3s)2 = 4 2 9s = 16 s = ,1 check: yes, 3= 3( ) = 4 by squaring each side K. WILLIAMS RM 206

  29. 5 Undoing (iii) A set for a school play involves 4 large cubes, to be painted red. Only one tin of paint is available which would cover 15m2. What are the dimensions of the largest cube which could be painted? If the dimension is dcm, there would be 24 faces each with area d2, so 24d2= 15(100x100) d2= = 6250 d =6250 = ±79 to nearest integer check: 24(79cmx79cm) = 149784cm2 or 14.9784m2 So each cube could be 79cmx79cmx79cm K. WILLIAMS RM 206

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