Solving equations

Solving Equations

Russell ShawGCSE Mathematics

February 2010

Solving Equations• Simple Equations

a + 6 = 10• Visually

a = 4• Or Mathematically

a + 6 = 10 a = 10 – 6

a = 4

Solving Equations• Balancing Method– Keep the balance level – What you do to one side, do to the other

• Mathematically:a + 6 = 10

a + 6 – 6 = 10 – 6 a = 4

Solving Equations• Opposite operation– If its “add” on one side carry it over and “subtract”– If its “subtract” on one side carry it over and “add”– If its “multiply” on one side carry it over and

“divide”– If its “divide” on one side carry it over and

“multiply”

Solving Equations• Try these:

y + 3 = 511 = y – 10

2a = 6 p/5 = 3

5t = 20 b/4 = 12


y + 3 = 5 y = 211 = y – 10 y = 21

2a = 6 a = 3 p/5 = 3 p = 15

5t = 20 t = 4 b/4 = 12 b = 48

Solving Equations• Combining operations:

2y + 3 = 21• Need to get “unknowns” on one side and known on

the other2y = 21 -3 = 182y = 18

• Now get the unknown by itself y = 18/2 = 9

• Check your answer 2 x 9 + 3 = 18 + 3 = 21 √


2y + 5 = 1511 = 2y – 9

2a - 4 = 6 p/5 + 4 = 14

5t + 5 = 20


2y + 5 = 15 y = 511 = 2y – 9 y = 10

2a - 4 = 6 a = 5 p/5 + 4 = 14 p = 50

5t + 5 = 20 t = 3

Solving Equations• Brackets:

3(2p + 5) = 45 • Expand the brackets first• Then solve as normal

3 x 2p + 3 x 5 = 45 6p + 15 = 45 6p = 30

p = 5Check!!

Solving Equations• Letters on both sides:

4(2x – 3) = 2x + 6• Expand the brackets first

8x – 12 = 2x + 6

• Get the “unknowns” on one side and known on the other 6x = 18

• Then solve as normal x = 3


2y + 3 = 5y – 6 5t + 7 = 3t + 10

2(a + 3) = 7 3(5r + 2) = 12r – 8 6(g + 7) = 3(4g + 2)


2y + 3 = 5y – 6 y = 3 5t + 7 = 3t + 10 t = 3/2

2(a + 3) = 7 a = 1/2 3(5r + 2) = 12r – 8 r = -14/3 6(g + 7) = 3(4g + 2) g = 6

Solving Equations• Word Formulae

pay = rate of pay x hours worked + bonus

If rate of pay = £7/hour and hours worked = 40 and a bonus of £20 is given, how much is the pay?

pay = £7 x 40 + £20 = £280 + £20 = £300

• A word formulae uses words to represent relationships between quantities

Solving Equations• Algebraic Formulae

pay = rate of pay x hours worked + bonus

We could represent pay as P, rate of pay as R, hours worked as H and bonus as B

So P = R x H + BP = RH + B

If R = £8/hour and H = 30 and B = £10, how much is the pay?pay = £8 x 30 + £10 = £240 + £10 = £250

An algebraic formulae uses letters to represent relationships between quantities

Solving Equations• Try this:– Write a word formula and an algebraic formula for

the area of a rectangle

– Use your formula to find the area of a rectangle with length = 10 and width = 3

Solving Equations• More complicated formulae

s = ½ a t2

Find s when t = 4 and a = 10 s = ½ x 10 x 42

s = ½ x 10 x 16 s = ½ x 160 s = 80

Solving Equations• Rearranging to change the subject of a

formula– You will need to be able to rearrange formulae.

H = (4t + 6)/s– Make t the subject of the formula

H = (4t + 6)/s H x s = (4t + 6)

4t + 6 = Hs 4t = Hs – 6

t = (Hs – 6)/4


formula• Once you can rearrange you can solve

problems– If Perimeter = 30cm and rectangle has length 8 cm

and width y cm – find y.• P = 2 x l + 2 x w = 2l + 2w• So P – 2l = 2w• So w = (P – 2l)/2• w = (30 – 2 x 8) / 2 = 7 cm


formula• Try this:

F = (9 x C) + 32 5What is the temperature in Farenheit when it is 30O C

Rearrange so that the formula says C = …

What is the temperature in Centigrade when it is 212O F


formula• Try this:

F = (9 x C) + 32 5What is the temperature in Farenheit when it is 30O C

F = (9 x 30 / 5) + 32 = 86O Farrange so that the formula says C = …

C = 5 x (F – 32) / 9What is the temperature in Centigrade when it is 212O F

C = 5 x (212-32) / 9 = (5 x 180) / 9 = 5 x 20 = 100O C

Solving Equations• Inequalities• Values maybe more than… or less than…• > means greater than• < means less than • ≥ means greater than or equal to• ≤ means less than or equal to

Solving Equations• Inequalities• Things maybe more than… or less than…• 6 is greater than 4 6 > 4• 5 is less than 10 5 < 10 • x is greater than or equal to 5 x ≥ 5• 2 is less than or equal to y 2 ≤ y

Solving Equations• Inequalities on a number line• We can represent inequalities on a number line. • If the number is “included” – using a ≥ or ≤ symbol – we make the

end point a solid circle• x ≥ -1

• If the number is “excluded” – using a > or < symbol – we make the end point an empty circle

• x < 2

Solving Equations• Inequalities on a number line• We can represent dual inequalities on a number

line. • -3 < x ≤ 4 (-3 is less than x, and x is less than or

equal to 4)

• -4 ≤ x ≤ 3

Solving Equations• Inequalities on a number line• Draw a number line from -10 to 10• Try these:

x ≥ 5x < 3x < -5x > 0-3 < x < 7-9 ≤ x < 3

Solving Equations• Solving Inequalities• You maybe asked to give the integer solutions to

an inequality

• -3 < x ≤ 4 -3 is NOT included (<), BUT 4 is included (≤) so the

solution is: -2 , -1, 0, 1, 2, 3, 4

Solving Equations• Solving Inequalities• Find integer solutions to:

-2 ≤ n ≤ 10 < n < 5

• Write an inequality for the integers listed:-2, -1 , 0 , 1, 2, 3, 4

1, 2, 3, 4, 5, 6

Solving Equations• Solving Inequalities• Find integer solutions to:

-2 ≤ n ≤ 1 -2, -1, 0, 10 < n < 5 1, 2, 3, 4

• Write an inequality for the integers listed:-2, -1 , 0 , 1, 2, 3, 4 -2 ≤ n < 5 or -1 < n ≤ 4 or ..

1, 2, 3, 4, 5, 6 0 < n < 7 or 1 ≤ n < 7 or ..

Solving Equations (H)• Solving Inequalities• We can solve inequalities – just like equations• Solve 3x + 3 > 18 3x > 18 – 3 = 15 x > 15/3 = 5 x > 5• These can have brackets included too –just like

normal equations!

Solving Equations (H)• Solving Inequalities• We can solve inequalities – just like equations• When solving you can:– Add or subtract same quantity to both sides– Multiply or divide both sides by the same positive

quantity• BUT you can’t:– Multiply or divide both sides by a negative quantity

Solving Equations (H)• Solving Inequalities• You can solve two sided inequalities by treating it

as two separate inequalities.• 7 ≤ 3x -2 < 10

So solve: 7 ≤ 3x -2 and 3x -2 < 10Writing the answer as 3 ≤ x < 4

Solving Equations (H)• Solving Inequalities• Try these:

3p – 5 ≤ 41 + 6b > 7

8 – 2m ≥ 1 – 4m7 < 4p + 3 ≤ 273 < 2q – 7 < 7

Solving Equations (H)• Solving Inequalities• Try these:

3p – 5 ≤ 4 p ≤ 31 + 6b > 7 b > 1

8 – 2m ≥ 1 – 4m m ≥ -7/27 < 4p + 3 ≤ 27 1 < p ≤ 63 < 2q – 7 < 7 5 < q < 7

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Solving equations