AlgebraMethods and Solving Equations

- Linear number patterns are sequences of numbers where the difference between terms is always the same (constant) - Rule generating a linear pattern is: Difference × n ± a constant

e.g. Write a rule (using n) to describe the following number patterns.

n Number of Squares (s)

Number of Dots

(d)

1 1 5

2 4 9

3 7 13

4 10 17

5 13 21

+ 3+ 3+ 3+ 3

+ 4

+ 4

+ 4

+ 4

Rule: s =

Rule: d =3×n 4×n

3×1= 33 = 1- 2

- 2

4×1= 44 = 5+ 1

+ 11. Find the difference between terms and if the same multiply by n

2. Substitute to find constant3. Check if rule works

3×4 – 2

4×4 + 1

LINEAR PATTERNS

SIMPLE QUADRATIC PATTERNS- Quadratic number patterns are sequences of numbers where the difference between terms is not the same and the rule contains a squared term - You need to halve the difference of the differences to find the squared term

e.g. Write a rule for the following pattern

n Term (T)

1 4

2 7

3 12

4 19

5 28

1. Find the difference between terms

+ 5

+ 3

+ 7+ 9

2. If difference is not the same, find the difference of the differences!

+ 2+ 2+ 2

Rule: T =

3. Halve the 2nd difference to find the n2 rule

1n2

4. Substitute to find constant

1×12 = 11 = 4+ 3

42 + 3 = 19

5. Check if rule works

+ 3

- i.e. If the difference of the differences is a 2, the rule contains 1n2

HARDER QUADRATIC PATTERNS- The squared term of the rule is found by halving the difference of the differences- i.e. If the difference of the differences is a 6, the rule contains 3n2- If the simple trial and error does not work, try this technique:

e.g. Write a rule for the following pattern

n Term (T)

1 5

2 13

3 25

4 41

5 61

1. Find the difference between terms

+ 12

+ 8

+ 16+ 20

2. If difference is not the same, find the difference of the differences!

+ 4+ 4+ 4

Rule: T =

3. Halve the 2nd difference to find the n2 rule

2n2

4. Subtract the n2 rule from the term

2×1= 22 = 3+ 1

2×42 + 2×4 + 1 = 41

5. Find the linear part of the rule

+ 1

Term (T) – 2n2

6. Check if the rule works

5 - 2×12

13 - 2×22

3

57

9

11

+ 2

+ 2

+ 2

+ 2

+ 2n

1. Multiplication

- Does x2 × x3 = x × x × x × x × x ? YES- Therefore x2 × x3 = x5

- How do you get 2 3 = 5 ?+

- When multiplying index (power) expressions with the same letter, ADD the powers.

e.g. Simplify

a) p10 × p2 b) a3 × a2 × a= p(10 + 2) = a(3 + 2 + 1)

= p12 1

No number = 1 i.e. p = 1p1

= a6

- Remember to multiply any numbers in front of the variables first

e.g. Simplify

a) 2x3 × 3x4 b) 2a2 × 3a × 5a4= 2 × 3= 2 × 3 × 5= 6

1

= 30

x(3 + 4)

a(2 + 1 + 4)x7

a7

POWER RULES

2. Division

YES- Does 6 = 1 ? 6

- Therefore x = 1 x

- Does x5 = x × x × x × x × x ? x3 x × x × x

YES = x × x × 1 × 1 × 1

- Therefore x5 = x2

x3

- When dividing index (power) expressions with the same letter, SUBTRACT the powers.

e.g. Simplify

a) p5 ÷ p = p(5 - 1) = x(7 - 4)

= p4

= x3

- Remember to divide any numbers in front of the variables first

e.g. Simplify

a) 12x5 ÷ 6x4 = 12 ÷ 6= 2

- How do you get 5 3 = 2 ?-

1 b) x7

x4

If the power remaining is 1, it can be left out of the answer

x(5 - 4) b) 5a7

15a2

÷ 5÷ 5

= 1 5

a(7 - 2)

= 1 5

xa5 or a5

5

3. Powers of powers

- Does (x2)3 = x2 × x2 × x2 ? YES

- Therefore (x2)3 = x6

- How do you get 2 3 = 6 ?×

- Does x2 × x2 × x2 = x6 ? YES

- When taking a power of an index expression to a power, MULTIPLY the powers

e.g. Simplify

a) (c4)6 b) (a3)3= c(4 × 6) = a(3 × 3)

= c24 = a9

- If there is a number in front, it must be raised to the power, not multiplied

e.g. Simplify

a) (3d2)3 b) (2a3)4× d(2 × 3)

= 27 a12= 33 × a(3 × 4)= 24

d6 = 16- If there is more than one term in the brackets, raise all to the power

e.g. Simplify

a) (x3y z4)3 = x(3 × 3)

= x9

1 y(1 × 3) z(4 × 3)

y3 z12

b) (4b2c5)2 b(2 × 2)

= 16

c(5 × 2)

b4 c10

= 42

4. Powers of zero- Any base to the power of zero has a value of 1

e.g. x0 = 1

- LIKE terms are those with exactly the same letter, or combination of letters and powers

LIKE terms: UNLIKE terms:2x, 3x, 31x4ab, 7ab

2x, 35x, 6x2

2ab, 2ace.g. Circle the LIKE terms in the following groups:

a) 3a 5b 6a 2c b) 2xy 4x 12xy 3z 4yx

While letters should be in order, terms are still LIKE if they are not.

LIKE TERMS

- We ALWAYS aim to simplify expressions from expanded to compact form- Only LIKE terms can be added or subtracted- When adding/subtracting just deal with the numbers in front of the letters

e.g. Simplify these expanded expressions into compact form:

a) a + a + a b) 5x + 6x + 2x

c) 3p + 7q + 2p + 5q

= (1 + 1 + 1)a= 3a

= (5 + 6 + 2)x= 13x

= (3 + 2)p= 5p + 12q

d) 4a + 3a2 + 7a + a2

1 1 1

(+ 7 + 5)q 1 = (4 + 7)a (+ 3 + 1)a2

= 11a + 4a2

- For expressions involving both addition and subtraction take note of signs

a) 4x + 2y – 3x b) 3a – 4b – 6a + 9b

c) 3x2 - 9x + 6x2 + 8x - 5

= (4 – 3)x= x + 2y = (3 - 6)a

= -3a + 5b

= 9x2 - x - 5

d) 4ab2 +2a2b – 5ab2 + 3ab

(- 9 + 8)x

= -ab2 + 2a2b + 3ab

+ 2y(- 4 + 9)b

= (3 + 6)x2 - 5

If the number left in front of a letter is 1, it can be left out

e.g. Simplify the following expressions:

ADDING/SUBTRACTING TERMS

- Does 6 × (3 + 5) = 6 × 3 + 6 × 5 ? 6 × 8 = 18 + 30 48 = 48

YES

- The removal of the brackets is known as the distributive law and can also be applied to algebraic expressions

- When expanding, simply multiply each term inside the bracket by the term directly in front

e.g. Expand

a) 6(x + y) b) -4(x – y)

c) -4(x – 6) d) 7(3x – 2)

e) x(2x + 3y) f) -3x(2x – 5)

= 6 × x

+ 6 × y = 6x

= -4 × x

- -4 × y = -4x

= -4 × x - -4 × 6 = -4x

= 7 × 3x - 7 × 2 = 21x

= x × 2x

+ x × 3y

= 2x2

1 1

= -3x × 2x - -3x × 5

= -6x2

Don’t forget to watch for sign changes!

+ 6y + 4y

+ 24 - 14

+ 3xy + 15x

EXPANDING

- If there is more than one set of brackets, expand them all then collect any like terms.

e.g. Expand and simplify

a) 2(4x + y) + 8(3x – 2y)

b) -3(2a – 3b) – 4(5a + b)

= 2 × 4x

+ 2 × y

+ 8 × 3x

- 8 × 2y = 8x + 2y + 24x - 16y

= 32x - 14y

= -3 × 2a - -3 × 3b - 4 × 5a + -4 × 1b = -6a + 9b - 20a - 4b

= -26a + 5b

- Involves replacing variables with numbers and calculating the answer- Remember the BEDMAS rules

e.g. If m = 5, calculate m2 – 4m - 3

= 52 – 4×5 - 3= 25 – 4×5 - 3= 25 – 20 - 3= 2

- Formulas can also have more than one variable

e.g. If x = 4 and y = 6, calculate 3x – 2y

= 3×4 - 2×6= 12 - 12= 0

e.g. If a = 2, and b = 5, calculate 2b – a 4

Because the top needs to be calculated first, brackets are implied

= (2 × 5 – 2) 4= (10 – 2) 4= 8 4

= 2

SUBSTITUTION

- Simply halve the power (as √x is the same as x½ )

e.g. Simplify

a) = 10x4

b)= 8x3

100x8

64x6= 100 x

8 = 64 x6

SQUARE ROOTS

ALGEBRAIC FRACTIONS1. Multiplying Fractions- Multiply top and bottom terms separately then simplify.

a) b × b2

2 5

e.g. Simplify:

= b3

10

b) y2 × 4 3 y

= 4y2

3y= 4y 3

= b × b2

2 × 5= y2 × 4

3 × y

2. Dividing Fractions- Multiply the first fraction by the reciprocal of the second, then simplify

Note: b is the reciprocal of 2 2 b

a) 2a ÷ a2

5 3

e.g. Simplify:

= 6a 5a2

= 6 5a

or 6a-1

5

= 2a 5

× 3a2

= 2a × 35 × a2

4. Adding/Subtracting Fractions a) With the same denominator:

- Add/subtract the numerators and leave the denominator unchanged. Simplify if possible.

a) 3x + 3x 10 10

e.g. Simplify:

= 3x + 3x 10= 6x 10

÷ 2÷ 2

= 3x 5

b) 6a - b 5 5

= 6a - b 5

b) With different denominators: - Multiply denominators to find a common term. - Cross multiply to find equivalent numerators. - Add/subtract fractions then simplify.

e.g. Simplify:a) a + 2a 2 3

= 3a + 4a 6= 7a 6

b) 2x – 5x 3 4

= 8x – 15x 12= -7x 12

= 2×3

3×a + 2×2a = 3×4

4×2x - 3×5x

EXPANDING TWO BRACKETS- To expand two brackets, we must multiply each term in one bracket by each in the second

e.g. Expand and simplify

a) (x + 5)(x + 2) b) (x - 3)(x + 4)

c) (x - 1)(x - 3) c) (2x + 1)(3x - 4)

= x2 + 2x+ 5x+ 10

To simplify, combine like terms

= x2 + 7x + 10

= x2 + 4x- 3x - 12

Remember integer laws when multiplying

= x2 + 1x – 12

= x2 - 3x - 1x + 3

= x2 – 4x + 3

= 6x2- 8x+ 3x- 4

= 6x2 – 5x – 4

PERFECT SQUARES- When both brackets are exactly the same

e.g. Expand and simplify

a) (x + 8)2

= x2 + 8x+ 8x + 64

To simplify, combine like terms

= x2 + 16x + 64

= (x + 8)(x + 8) b) (x - 4)2

= x2 - 4x - 4x + 16

= x2 - 8x + 16

= (x – 4)(x – 4)

c) (3x - 2)2

= 9x2 - 6x- 6x + 4

= 9x2 – 12 x + 4

= (3x – 2)(3x – 2)

Watch sign change when multiplying

Write out brackets twice BEFORE expanding

DIFFERENCE OF TWO SQUARES- When both brackets are the same except for signs (i.e. – and +)

e.g. Expand and simplify

a) (x – 3)(x + 3) b) (x – 6)(x + 6)

c) (2x – 5)(2x + 5)

= x2 + 3x- 3x - 9

Like terms cancel each other out

= x2 – 9

= x2 + 6x- 6x - 36

= x2 – 36

= 4x2 + 10x - 10x - 25

= 4x2 – 25

- Factorising is the reverse of expanding- To factorise: 1) Look for a common factor to put outside the brackets

2) Inside brackets place numbers/letters needed to make up original terms

e.g. Factorise

a) 2x + 2y b) 2a + 4b – 6c= 2( ) x + y = 2( ) a + 2b

e.g. Factorise

a) 6x - 15 b) 30a + 20= 3( ) 2x

- 5 = 10( ) 3a + 2

- Always look for the highest common factor

You should always check your answer by expanding it

e.g. Factorise

a) 6x + 3 b) 20b - 10= 3( ) 2x

+ 1 = 10( ) 2b - 1

- Sometimes a ‘1’ will need to be left in the brackets

- 3c

FACTORISING

e.g. Factorise

a) cd - ce b) xyz + 2xy – 3yz = c( ) d - e = y( ) xz + 2x

- Letters can also be common factors

e.g. Factorise

a) 5a2 – 7a5 b) 4b2 + 6b3= a2( ) 5 - 7a3 = 2b2( ) 2 + 3b

- Powers greater than 1 can also be common factors

- 3z

c) 4ad – 8a = 4 ( ) a d - 2

Factorising by grouping- When two groups within an expression have their own common factor

e.g. Factorise

2ac + 2bc + 3ad + 3bd = 2c( ) a + b + 3d( ) a + b = (2c + 3d)(a + b)

As both 2c and 3d are being multiplied by (a + b) we place them in separate brackets

FACTORISING QUADRATICSThe general equation for a quadratic is ax2 + bx + cWhen a = 1- You need to find two numbers that multiply to give c and add to give b

e.g. Factorise

a) x2 + 11x + 24

List pairs of numbers that multiply to give 24 (c)

1, 242, 123, 84, 6

Check which pair adds to give 11 (b)

Place numbers into brackets with x

= (x + 3)(x + 8)

b) x2 + 7x + 6

2, 31, 6

List pairs of numbers that multiply to give 6 (c)

Check which pair adds to give 7 (b)

Place numbers into brackets with x

= (x + 1)(x + 6)

To check answer, expand and see if you end up with the original question!

- Expressions can also contain negativese.g. Factorise

a) x2 + x – 12 1, 122, 63, 4

As the end number (c) is -12, one of the pair must negative.

Make the biggest number of the pair the same sign as b

---

Check which pair now adds to give b

= (x - 3)(x + 4) b) x2 – 6x – 16 1, 162, 84, 4

As the end number (c) is -16, one of the pair must negative.

Make the biggest number of the pair the same sign as b

Check which pair now adds to give b

---

= (x + 2)(x - 8)

c) x2 – 9x + 20 1, 202, 104, 5

As the end number (c) is +20, but b is – 9, both numbers must be negative

---

---

Check which pair now adds to give b

= (x - 4)(x - 5) d) x2 – 10x + 25 1, 255, 5

--

--

= (x - 5)(x - 5)= (x - 5)2

As the end number (c) is +25, but b is – 10, both numbers must be negative

SPECIAL CASES1. No end number (c)e.g. Factorise

a) x2 + 6x b) x2 – 10x

Add in a zero and factorise as per normal

+ 0

0, 6

= (x + 0)(x + 6)

= x(x + 6)

= x( )

2. No x term (b) (difference of two squares)e.g. Factorisea) x2 - 25

OR: factorise by taking out a common factor

Add in a zero x term and factorise

+ 0x-5, 5

= (x - 5)(x + 5)

OR: factorise by using A2 – B2 = (A – B)(A + B)

b) x2 – 100 - 10 + 10= (x )(x )

x - 10

c) 9x2 – 121

= (3x )(3x ) - 11 + 11

TWO STAGE FACTORISINGWhen a ≠ 11. Common factor- Always try to look for a common factor first.e.g. Factorisea) 2x2 + 12x + 16 = 2( )

1, 82, 4

= 2(x + 2)(x + 4)b) 3x2 – 6x – 9 = 3( )

1, 3- = 3(x + 1)(x – 3)

c) 3x2 + 24x = 3x( )x + 8 d) 4x2 – 36 = 4( )= 4(x )(x ) - 3 + 3

x2 + 6x + 8 x2 – 2x – 3

x2 – 9

2. No common factor (HARD)- Use the following technique

e.g. Factorise

a) 3x2 – 10x – 8

Multiply first and last terms

3x2 × - 8 = -24x2

Find two terms that multiply to -24x2 but add to -10x

1x, 24x2x, 12x3x, 8x4x, 6x

----

Replace 10x with the two new terms

= 3x2 + 2x – 12x – 8

Factorise 2 terms at a time.

= x( )3x + 2 -4( )3x + 2 = (x – 4)(3x + 2)

Write in two brackets

b) 2x2 + 7x + 3

2x2 × 3 = 6x2 1x, 6x2x, 3x

= 2x2 + x + 6x + 3 = x( )2x + 1 + 3( ) 2x + 1 = (x + 3)(2x + 1)

RATIONAL EXPRESSIONS- Look to factorise the numerator and/or denominator and then remove the common factor by dividing through.

e.g. Simplify

a) x2 + 7x + 12 x + 3

b) x – 5 . x2 – 2x – 15

= (x + 3)(x + 4)x + 3

Cancel out common factors e.g. x + 3 = 1 x + 3

= (x + 4)

= x – 5 .(x – 5)(x + 3)

As the numerator cancels out, you must leave a ‘1’ on top to signal this= 1 .

(x + 3)

- Terms containing the variable (x) should be placed on one side (often left)

e.g. Solve

a) 5x = 3x + 6 b) -6x = -2x + 12-3x -3x

2x = 6÷2÷2

x = 3

You should always check your answer by substituting into original equation

+2x +2x -4x = 12

÷-4÷-4 x = -3

Always line up equals signs and each line should contain the variable and one equals sign

- Numbers should be placed on the side opposite to the variables (often right)

e.g. Solve

a) 6x – 5 = 13 b) -3x + 10 = 31+5 +5

6x = 18÷6÷6

x = 3

-10 -10 -3x = 21

÷-3÷-3 x = -7

Always look at the sign in front of the term/number to decide operation

Don’t forget the integer rules!

SOLVING EQUATIONS- Remember that addition/subtraction undo each other as do multiplication/division

- Same rules apply for combined equations

e.g. Solve

a) 5x + 8 = 2x + 20 b) 4x - 12 = -2x + 24-2x -2x

3x + 8 = 20-8-8

3x = 12

+2x +2x 6x - 12 = 24

÷6÷6 x = 6

÷3÷3 x = 4

+12

+12 6x = 36

- Answers can also be negatives and/or fractions

e.g. Solve

a) 8x + 3 = -12x - 17 b) 5x + 2 = 3x + 1+12x +12x

20x + 3 = -17-3-3

20x = -20

-3x -3x 2x + 2 = 1

÷2÷2 x = -1 2

÷20÷20 x = -1

-2-2 2x = -1

Make sure you don’t forget to leave the

sign too!

Answer can be written as a decimal but easiest to leave as a fraction

- Expand any brackets first

e.g. Solve

a) 3(x + 1) = 6 b) 2(3x – 1) = x + 8

-3-3 3x = 3

÷3÷3 x = 1

3x+ 3= 6 6x - 2 = x + 8-x-x

5x - 2 = 8+2+2

5x = 10÷5÷5

x = 2- For fractions, cross multiply, then solve

e.g. Solve

a) x = 9 4 2

2x = 36÷2÷2

x = 18

b) 3x - 1 = x + 3 5 2

2(3x - 1)= 5(x + 3)

6x - 2 = 5x + 15-5x-5x

x - 2 = 15+2+2

x = 17

- For two or more fractions, find a common denominator, multiply it by each term, then solve

e.g. Solve 4x - 2x = 10 5 3

5 × 3 = 15×15×15×15

60x

5

- 30x

3

= 150

Simplify terms by dividing numerator by denominator

12x

- 10x

= 150

2x = 150 ÷2÷2

x = 75

e.g. Solve 5x - (x + 1) = 2x 6 4

×24×24×24

120x

6

- (24x + 24)

4

= 48x

20x

- 6x – 6

= 48x

14x – 6 = 48x -48x-48x

-34x – 6 = 0+ 6+ 6

-34x = 6÷ -34 ÷ -34

x = -6 34

- Inequations contain one of four inequality signs: < > ≤ ≥ - To solve follow the same rules as when solving equations- Except: Reverse the direction of the sign when dividing by a negative

e.g. Solve

a) 3x + 8 > 24 b) -2x - 5 ≤ 13-8-8

3x > 16÷3÷3

x > 16 3

+5+5-2x ≤ 18

÷-2÷-2 Sign reverses as dividing by a negative

x ≥ -9

As answer not a whole number, leave as a fraction

SOLVING INEQUATIONS

SOLVING QUADRATICSTo solve use the following steps:1. Move all of the terms to one side, leaving zero on the other2. Factorise the equation3. Set each factor to zero and solve.

e.g. Solve

a) (x + 7)(x – 2) = 0

x + 7 = 0 x – 2 = 0-7-7 +2+2

x = - 7 x = 2

b) (x – 4)(x – 9) = 0

x – 4 = 0 x – 9 = 0+4 +4 +9 +9

x = 4 x = 9

c) x2 + x – 2 = 0 (x – 1)(x + 2) = 0

1, 2-

x – 1 = 0 x + 2 = 0+1 +1 - 2 - 2

x = 1 x = - 2

d) x2 – 5x + 6 = 0

(x + 1)(x – 6) = 0

x + 1 = 0 x – 6 = 0-1 -1 +6 +6

x = -1 x = 6

1, 62, 3

--

e) x2 + 8x = 0 x( ) = 0

x = 0 x + 8 = 0- 8 - 8

x = -8

x + 8f) x2 – 11x = 0

x( ) = 0x = 0 x – 11 = 0

+ 11 + 11 x = 11

x – 11

g) x2 - 49 = 0 (x )(x ) = 0

x + 7 = 0- 7 - 7

x = -7

- 7 + 7x - 7 = 0

+7 +7 x = 7

h) 9x2 - 4 = 0 (3x )(3x ) = 0 - 2 + 23x - 2 = 0

+2 +2 3x = 2÷3 ÷3 x = 2/3

3x + 2 = 0 -2 -2 3x = -2÷3 ÷3 x = -2/3

i) x2 = 4x + 5

(x + 1)(x – 5) = 0

x + 1 = 0 x – 5 = 0-1 -1 +5 +5

x = -1 x = 5

1, 5--4x -5-4x -5x2 – 4x – 5 = 0

j) x(x + 3) = 180x2 + 3x = 180

-180 -180 x2 + 3x – 180 = 0

(x + 15)(x – 12) = 0x + 15 = 0 x – 12 = 0

-15 -15 +12 +12 x = -15 x = 12

e.g. Write an equation for the following information and solvea) A rectangular pool has a length 5m longer than its width. The perimeter of

the pool is 58m. Find its width

Draw a diagram

Let x = width-10-10

Therefore width is 12 m

x x

x + 5

x + 5

x + 5 + x + x + 5 + x = 58 4x + 10 = 58

4x = 48 ÷4÷4

x = 12

WRITING AND SOLVING

b) I think of a number and multiply it by 7. The result is the same as if I multiply this number by 4 and add 15. What is this number?

Let n = a number n = n + 15-4n-4n

3n = 15

Therefore the number is 5

7 4

÷3÷3n = 5

QUADRATIC EQUATIONS- Involves writing an equation from the information then solving

e.g. The product of two consecutive numbers is 20. What are they?If x = a number, then the next consecutive number is x + 1

x(x + 1) = 20x2 + x = 20

-20-20x2 + x – 20 = 0

1, 202, 104, 5

---

(x – 4)(x + 5) = 0x – 4 = 0 x + 5 = 0

+4 +4 -5 -5 x = 4 x = -5

The numbers are 4, 5 and -5, -4

e.g. A paddock of area 150 m2 has a length 5 m longer than its width. Find the dimensions of this paddock.

A = 150 m2 x

x + 5

x(x + 5) = 150x2 + 5x = 150

-150-150x2 + 5x – 150 = 0(x – 10)(x + 15) = 0

x – 10 = 0 x + 15 = 0+10 +10 -15 -15

x = 10 x = -15

The dimensions are 10 m by 15 m

- Involves rearranging the formula in order to isolate the new ‘subject’- Same rules as for solving are used

e.g.

a) Make x the subject of y = 6x - 2+2+2

y + 2 = 6x÷6÷6

y + 2 = x 6

All terms on the left must be divided by 6

b) Make R the subject of IR = V÷I÷I

Treat letters the same as numbers!

R = V I

c) Make x the subject of y = 2x2

÷2÷2y = x2

2Taking the square root undoes squaring

y2

=x

Remember: When rearranging or changing the subject you are NOT finding a numerical answer

REARRANGING FORMULA

SIMULTANEOUS EQUATIONS- Are pairs of equations with two unknowns

To solve we can use one of three methods:

1. ELIMINATION METHOD- Line up equations and either add or subtract so one variable disappearse.g. Solvea) 2x + y =

20 x – y = 4

To remove the ‘y’ variable we add as the signs are opposite.

+ (

)

3x = 24÷3÷3

x = 8 Now we substitute x-value into either equation to find ‘y’2 × 8 + y =

2016 + y = 20-16

-16y = 4

Check values in either equation8 – 4 = 4

b) 2x + y = 7

x + y = 4

To remove the ‘y’ variable we subtract as the signs are the same.

- ( )

x = 3

Now we substitute x-value into either equation to find ‘y’

2 × 3 + y = 7 6 + y = 7

-6-6y = 1

Check values in either equation3 + 1 = 4

- You may need to multiply an equation by a number to be able to eliminate a variablee.g. Solvea) 2x – y = 0 x + 2y =

5+ (

)

5x = 5÷5÷5

x = 1

Now we substitute x-value into either equation to find ‘y’

2 × 1 – y = 0 2 – y = 0

-2-2– y = -2

Check values in either equation 1 + 2 × 2

= 5

Multiply the 1st equation by ‘2’ then add to eliminate the y

× 2× 1

4x – 2y = 0 x + 2y = 5

÷-1

÷-1 y = 2

Note that it was possible to eliminate the ‘x’ variable by multiplying second equation by 2 and then subtracting

b) 4x – 2y = 28

3x + 3y = 12

- ( )

-18y = 36 ÷-

18÷-18y = -2

Now we substitute y-value into either equation to find ‘x’

4x – 2 × -2 = 28 4x + 4 =

28 -4-44x = 24

Check values in either equation 3 × 6 + 3 × -2 =

12

Multiply the 1st equation by ‘3’ and the 2nd by ‘4’ then subtractto eliminate the x

× 3× 4

12x – 6y = 84

12x + 12y = 48

÷4÷4x = 6

Note that it was possible to eliminate the ‘y’ variable by multiplying the 1st equation by 3 and the 2nd by 2 and then adding

2. SUBSTITUTION METHOD

- Make x or y the subject of one of the equations- Substitute this equation into the second

e.g. Solvea) y = 3x + 1

9x – 2 y = 4

b) x – y = 2 y = 2x +

3

Substitute what the subject equals in for that variable in the other equation

9x – 2(3x + 1) = 49x – 6x – 2 = 4

3x – 2 = 4+2+2

3x = 6÷3÷3

x = 2

Now we substitute x-value into first equation to find ‘y’

y = 3×2 + 1y = 7

To check values you can substitute both values into second equation

x – (2x + 3) = 2x – 2x – 3 = 2

-x – 3 = 2+3+3

-x = 5÷-1

÷-1x = -5

1As we are subtracting more than one term, place in brackets and put a one out in front.

Now we substitute x-value into second equation to find ‘y’

y = 2×-5 + 3y = -7

To check values you can substitute both values into second equation