Building Confidence in Equation Solving

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?: 

What is an equation? An equation 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 ? becomes your age = 5 + 2 or using a letter to represent the unknown quantity a = 5 + 2 2

What does solving an equation mean?: 

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 4 a – 1 = 6 a + 3 becomes 1 = a Notice it doesn’t matter on which side the unknown letter appears. 3

Why use an equation? : 

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. 4

Barriers to solving equations: 

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

Some methods: 

Some methods Give each method for solving simple equations a name: Shifting Scaling at Third Level Simplifying Sorting at Fourth Level Ungrouping Undoing 6

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. 1 Shifting too much by 1, get rid of + 1 term on LHS = = x x 7 ( i ) x + 1 = 4 x + 1 – 1 = 4 - 1 x = 3 by taking 1 from each side of the equation

1 Shifting: 

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

1 Shifting: 

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

2 Scaling: 

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. 10 t oo big, get rid of the multiplyer 5 on LHS by dividing each side by 5 = = 5 y y ( i ) 5 y = 10 = y = 2 check: yes, 5(2) = 10

2 Scaling: 

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. 2 Scaling 11 by multiplying each side by 4 t oo small, get rid of  4 on LHS (ii) = 2 4( ) = 4(2) x = 8 check : yes , = 2

2 Scaling: 

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

2 Scaling : 

2 Scaling If Angie buys n oranges, 25 n = 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 13 A problem (iv) Angie has £1.30 and wants to buy some oranges at 25p each. How many can she buy?

3 Simplifying: 

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

3 Simplifying: 

3 Simplifying 15 Notice that the answer should always be simplified if possible. by collecting together numbers t oo many numbers, make one number on LHS (ii) 3 + 4 + 3 = 6 r 15 = 6 r = r r = ,2 or 2·5 check : yes , 6( )= 15

3 Simplifying: 

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

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? 3 Simplifying 17 If Jill has £ p, p + 2 p = 25·2 3 p = 25·2 p = 8·4 yes, 2(8 · 4) + 8 · 4 = 16 · 8 + 8 · 4 = 25 · 2 So Jill has £8.40

4 Sorting: 

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. 18 Unknowns on both sides, get rid of s term as smaller than 4 s = = ( i ) s + 6 = 4 s s + 6 - s = 4 s - s 6 = 3 s 2 = s check : yes , 2 + 6 = 8 and 4(2) = 8 by taking s from each side

4 Sorting: 

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

4 Sorting: 

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

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? 4 Sorting 21 If Tom’s son is y years old, 4 y = 27 + y 3 y = 27 y = 9 yes , 4(9) = 36 and 9 + 27 = 36 So Tom’s son is now 9 years old

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. ( i ) 3 x + 1 = 2( x – 1) 3 x + 1 = 2( x ) – 2(1) 3 x + 1 = 2 x – 2 3 x + 1 - 2 x – 1 = 2 x – 2 – 2 x - 1 x = -3 check : yes , 3(-3) + 1 = -8 and 2(-3 - 1 ) = -8 5 Ungrouping 22 by multiplying out brackets ungroup the x term

5 Ungrouping: 

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

5 Ungrouping: 

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

5 Ungrouping: 

(iv) y + = 5 y – 3 2 y + 2( ) = 2(5 y ) – 2(3) 2 y + ( y + 1) = 10 y – 6 3 y + 1 = 10 y – 6 3 y + 1 - 3 y + 6 = 10 y – 6 – 3 y + 6 7 = 7 y y = 1 check: 1 + = 1 + 1 = 2 and 5(1) – 3 = 2 5 Ungrouping 25 by multiplying each side by 2 ungroup the y + 1 expression 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.

5 Ungrouping: 

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

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. 6 Undoing 27 by taking the square root of each side undo the square function ( i ) t 2 + 1 = 10 t 2 + 1 -1 = 10 -1 t 2 = 9  t 2 =  9 t = ±3 check: yes , 3 2 = 9 and (-3) 2 = 9 isolate the square function term by taking 1 from each side

6 Undoing: 

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

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 15m 2 . What are the dimensions of the largest cube which could be painted? 5 Undoing 29 If the dimension is d cm , there would be 24 faces each with area d 2 , so 24 d 2 = 15(100 x 100) d 2 = = 6250 d = 6250 = ±79 to nearest integer check: 24(79 cm x 79cm ) = 149784cm 2 or 14.9784m 2 So each cube could be 79cm x 79cm x 79cm