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

K. WILLIAM

S RM 206

K. WILLIAM

S RM 206

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

K. WILLIAM

S RM 206

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 = 74a – 1 = 6a + 3 becomes 1 = a

Notice it doesn’t matter on which side the unknown letter appears.

K. WILLIAM

S RM 206

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. WILLIAM

S RM 206

BARRIERS TO SOLVING EQUATIONS

Sometimes learners forget why they are learning these methods, or never understand how they work together.

K. WILLIAM

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SOME METHODSGive each method for solving simple equations a name:

ShiftingScaling

at Third Level Simplifying Sorting

at Fourth Level UngroupingUndoing

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S RM 206

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(i) x + 1 = 4 x + 1 – 1 = 4 - 1 x = 3 by taking 1 from

each side of the equation

K. WILLIAM

S RM 206

1 SHIFTING

too 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!

(ii) x – 3 = 2 x – 3 + 3 = 2 + 3 x = 5check: yes, 5 - 3 = 2

K. WILLIAM

S RM 206

(iii) 2 = r - 5 2 + 5 = r – 5 + 5 7 = rcheck: yes, 2 = 7 - 5

1 SHIFTING

Remember that the initial focus is on the unknown term, whichever side of the equation it appears.

too little, get rid of -5 term on RHS

by adding 5 to each side

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S RM 206

2 SCALINGScaling 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

multiplyer 5 on LHS

by dividing each side by 5

==5y y

55y

510

(i) 5y = 10 = y = 2check: yes, 5(2) = 10

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S RM 206

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

by multiplying each side by 4

too small, get rid of 4 on LHS

4x

48

4x

(ii) = 2 4( ) = 4(2) x = 8check: yes, = 2

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S RM 206

2 SCALING

31

too big, get rid of multiplyer 3 on LHS

by dividing each side by 3, the coefficient of y

(iii) 3y = 7 = y = or 2 check: yes, 3( ) = 7

33y

37

37

37

Notice that with practice, intermediate lines of working can be omitted.

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S RM 206

2 SCALING

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

51

25130

A problem (iv) Angie has £1.30 and wants to buy some oranges

at 25p each. How many can she buy?

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Simplify an equation by collecting together any like terms creating one letter term, one number term or removing fractions, e.g.

3 SIMPLIFYING

by collecting together like terms

too 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) 5q – 2q = 15 3q = 15

q = 5check: yes, 5(5) – 2(5) = 15

K. WILLIAM

S RM 206

615

3 SIMPLIFYING

Notice that the answer should always be simplified if possible.

25

by collecting together numbers

too many numbers, make one number on LHS

(ii) 3 + 4 + 3 = 6r 15 = 6r = r r = ,2 or 2·5check: yes, 6( )= 15

25

21

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3 SIMPLIFYING

3c

53

by scaling, multiplying every term by 3

get rid of the fraction

3c

56

51

53

53

(iii) 2c - = 1 3(2c) – 3( ) = 3(1) 6c – c = 3 5c = 3 c = or 0·6check: yes, 2( ) - /3 = - = 1

K. WILLIAM

S RM 206

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

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. WILLIAM

S RM 206

4 SORTINGWhen 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 s term 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

by taking s from each side

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S RM 206

(ii) 3x + 6 = 8x - 4 3x + 6 – 3x = 8x – 4 – 3x

6 = 5x – 4 6 + 4 = 5x – 4 + 4 10 = 5x

x = 2check: yes, 3(2) + 6 = 12 and 8(2) – 4 = 12

4 SORTING

Unknowns on both sides, get rid of 3x term as smaller than 8x

by taking 3x from each side

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4 SORTING

(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

Unknowns on both sides, get rid of -5p as smaller than -p

by adding 5p to each side

Notice that dealing with the letter terms first can help avoid negative coefficients for the unknown

K. WILLIAM

S RM 206

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

If Tom’s son is y years old,4y = 27 + y3y = 27 y = 9

yes, 4(9) = 36 and 9 + 27 = 36

So Tom’s son is now 9 years old

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S RM 206

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) 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

5 UNGROUPING

by multiplying out brackets

ungroup the x term

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5 UNGROUPING

by multiplying each side by 3

ungroup the 2r term

(ii) = 5 3( ) = 3(5) 2r – 1 = 15

2r – 1 + 1 = 15 + 1 2r = 16 r = 8

check: yes, = = 5

31 -2r

31 -2r

31 -2(8)

315

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(iii) 2(4a – 1) = 6

or =

4a – 1 = 3 4a = 4 a = 1

5 UNGROUPING

21) - 2(4a

26

check: yes, 2(4(1) – 1) = 2(3) = 6

by multiplying out brackets

by dividing each side by 2

ungroup the 4a term

2(4a) – 2(1) = 6 8a – 2 = 6 8a = 8 a = 1

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S RM 206

21 1

21y

(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 and 5(1) – 3 = 2

5 UNGROUPING

by multiplying each side by 2

ungroup the y +1 expression

21y

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.

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S RM 206

(v) Find the larger of two consecutive odd numbers such that the sum of the larger and double the smaller is 47.

5 UNGROUPING

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 =

47So the larger odd number is 173216589473

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S RM 206

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

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 = ±3check: yes, 32 = 9 and (-3)2 = 9

isolate the square function term

by taking 1 from each side

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6 UNDOING

by squaring each side

undo the square root function

(ii) 3s = 4(3s )2 = 4 2

9s = 16 s = ,1

check: yes, 3 = 3( ) = 4

916

97

916

34

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S RM 206

(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?

5 UNDOING

If the dimension is d cm, there would be 24 faces each with area d

2, so 24d

2 = 15(100 x 100) d

2 = = 6250 d = 6250 = ±79 to nearest integercheck: 24(79cmx79cm) = 149784cm 2 or 14.9784m 2

So each cube could be 79cmx79cmx79cm

2415x100x100