Algebra 3

Question 1

How do you solve a simultaneous equation (by elimination)?

4 steps

Answer 1

1) multiply 1 or both equation so they have the same number of xs or ys (regardless of sign)
2) + or - resulting equations to cancel out the unknowns that are the now the same size
3) solve the resulting equation
4) substitute this into either equation (depending on which looks easiest) and solve resulting equation

eg. 1) 3x + 4y = 29 2) 5x - 2y = 5

2) x2 –> 10x - 4y = 10
1) ——-> 3x + 4y =29 +
————————————
13x = 39
( / 13)
x = 3

sub x into 1)

3x + 4y = 29
3(3) + 4y = 29
9 + 4y = 29
                 (-9)
4y = 20
           ( / 4)
y = 5

Question 2

How do you solve a linear inequality?

Answer 2

• solve as a normal equation
• if we x or / by a negative we change the direction of inequality sign
• sometimes shown on a number line

eg. solve 3x + 2 ≤ 8 and show on a number line

3x + 2 ≤ 8
               (-2)
3x ≤ 8
         ( / 3)
x ≤ 2

Question 3

How do you expand single brackets (1 step)

Answer 3

• multiple term on outside of brackets by each term on the inside of the brackets
  eg. 2(a + 5)

2 x a = 2a
2 x 5 = 10

2a + 10

Question 4

How do you expand double brackets? (2 steps)

Answer 4

1) multiple each term in the left bracket by each term in the right bracket
2) collect the terms

eg. (q + 4) (p + 3)

q x p = pq
q x 3 = 3q
4 x p = 4p
4 x 3 = 12

pq + 3q + 4p + 12

Question 5

How do you expand triple brackets? (3 steps)

Answer 5

1) multiple the first two brackets as double brackets
2) simplify the resulting expression
3) multiply these terms by the 3rd bracket and simplify

eg. (2x + 1) (x - 3) (3x + 2)

2x x x = 2x²
2x x 3 = -6x
1 x x = x
1 x -3 = -3

(2x² - 6x + x - 3) (3x + 2)
(2x² -5x -3) (3x + 2)

2x² x 3x = 6x³
2x² x 2 = 4x²
-5x x 3x = -15x
-5x x 2 = -10x
-3 x 3x = -9x
-3 x 2 = -6

6x³ + 4x² - 15x - 10x - 9x - 6
6x³ + -11x² - 19x - 6

Question 6

How do you factorise quadratics? (single x²)

Answer 6

• the reverse of expanding a double bracket

1) there will be an x at the front of each bracket
2) we need to find 2 numbers that multiply to give the number at the end and add to give the number of xs (middle number)

eg. x² + 5x + 6

(x + ) (x + )

3 x 2 = 6
3 + 2 = 5

so

(x + 3) (x + 2)

Question 7

How do you find the difference of 2 squares?

Answer 7

• 2 terms are involved -> the 2 squares
• the second term is a negative

1) square root each term
- one bracket will be a + and one will be a -
eg. x² - 25

5 x 5 = 25

so

(x + 5) (x - 5)

Question 8

How do you factorise quadratics in the form ax² + bx + c?

Answer 8

1) separate the ax

Question 9

How do you complete the square?

Answer 9

• we need to rearrange the expression into the form (x + a)² + b

1) half the number of xs to get a
2) subtract a² and combine it with any other number in the original expression

eg. x² + 2x + 6

(x + )² - + 6

2x / 2 = 1 so

(x + 1)² - + 6

1² = 1 so

(x +1)² - 1 + 6

Question 10

How do solve quadratic equations by 4 term factorising (by grouping)?

Answer 10

• the terms are factorised in pairs
• sometimes changing the sign outside the brackets or the term inside the brackets is required

1) factorise each half of the equation
eg. ab + ac + b² +bc

a(b + c)

Question 11

How do you solve a quadratic equation by factorising?

Answer 11

1) factorise equation
eg. x² + 5x = 0

x(x + 5) = 0

2) separate term outside of bracket from terms inside bracket

so x = 0 or x + 5 = 0

3) simplify expression of terms that were inside of bracket

x + 5 = 0
(-5)
x = -5

so x = 0 or x = -5

Question 12

How do you solve equations by using the quadratic formula?

Answer 12

• quadratic formula is x = -b ± √b² - 4ac / 2a
• a is the number of x²
• b is the number of xs
• c is the number

eg. x² - 5x + 3 = 0

a = 1 b = -5 c = 3

so

x = -(-5) ± √(-5)² - 4(1)(3) / 2(1)

(solve on calculator)

x = 5 + 3.6055 / 2 or x = 5 - 3.6055 / 2

x = 4.18 or x = 0.70