CHAPTER 1

NUMBERS AND EQUATIONS

(1) Natural numbers are numbers that are only positive whole numbers. They are also called counting numbers eg 1, 2, 3…

(2) Integers are positive and negative whole numbers eg …. -2, -1, 0, 1, 2, …

(3) Rational numbers are a set of negative and positive whole numbers and fractions. They are also called Quotients eg … -2, -1, -½ , 0, ½ , 1,3/2 , 2, 3, …..

(4) Irrational numbers are numbers which cannot be expressed as a fraction eg √5 = 2.23606……. other examples of irrational numbers are: √2,

(5) Real numbers are a set of rational and irrational numbers. The Real numbers make the number line to be complete.

(6) Complex numbers are a set of numbers that have both the real and imaginary parts eg a + ib.

SIMULTANEOUS LINEAR EQUATIONS IN TWO VARIABLES

1. Substitution Method

Example I: Solve the simultaneous equations below:

3x – 2y =5            (1)

2x + y = 7             (2)

Solution:

From (2), make y the subject of the formula

y = 7 – 2x

Substitute for y in (1)

3x – 2(7 – 2x) = 5

3x – 14 + 4x = 5

7x = 19

x = 19/7

Substitute for x in y = 7 – 2x

y = 7 – 2(19/7)

y = 7 – (38/7) = 11/7

Hence x = 19/7, y = 11/7

(2) Elimination Method

Example II: Solve the pair of Simultaneous equation

3x – 2y = 5                     (1)

2x + y = 7                       (2)

Solution:

(1) x 2, 6x – 4y = 10      (3)

(2) x 3, 6x + 3y = 21      (4)

(3) – (4)      -7y = -11

                   y = -11/ -7 = 11/7

Solving for x, we get x = 19/7

Hence x = 19/7, y = 11/7

(3) Inconsistent Equations

Example III: Solve the pair of equations

2x – 3y = 4                    (1)

4x – 6y = 5                    (2)

(1) x 2, 4x – 6y = 8       (3)

From the equations above you can conclude that no values of x and y satisfy both equations. Hence, no solution. Graphically, their lines will be parallel, which do not intersect.

Questions:

1. Solve the following pair of simultaneous equations:

(a) 3x + 4y = -2   (b) x + y = 5   (c) x + y = 7

     x + y = 0              x – y = 1       y = 1 – x

(d) x + 2y = 5   (e) 2x + 3y = 4    (f) 2x – 5y = 0

     x – 3y = -1        3x – 2y = -7       3x = 4y + 7

2. If the equations ax + by = -4 and bx + ay = 1 are satisfied by x = 1, y = 2, find the values of a and b

Note: Try solving these simultaneous equations. If you have some difficulties, contact Andrew on +2348036670623 or by email andrewukauwa797@gmail.com

QUADRATIC EQUATIONS

A quadratic equation in one variable is an equation in which their highest power of the variable is 2 eg 3x² – 5x = 7. The standard form of a quadratic equation is ax² + bx +c = 0 where a ≠ 0. The equation will always have solutions or roots but these may be equal or complex.

1. Solution by factorization

Example I: Solve 2x² – 5x = 0

Solution:

Factorize the LHS

x(2x – 5) = 0

Hence, x = 0 or 2x – 5 = 0

Either x = 0 or x = 5/2

Example 2: Solve x² – 5x = 6

Solution:

Standard form will be x² – 5x – 6 = 0

Factorize LHS eg (look for 2 number such that if you multiply them, it will give you (-6) and if you add them, it will give you (-5). Hence they are -6 and +1, therefore

(x – 6)(x + 1) = 0

Hence, x = 6 or x = -1

Questions:

1. Solve the following equations:

(a) x² – x – 2 = 0  (b) x² = 3x  (c) 1 – 3x² = 2x

(d) p² = 2(p + 4)  (e) 2x² + 3x = 2  (f) x – 12/x = 1

2. Find the value of the ratio a : b if 2a² + ab = 3b²

3. Solve the equation a² – 8a + 12 = 0. Hence find the four values of x which satisfy the equation (x² – x)² – 8(x² – x) + 12 = 0.

Note: Try solving these quadratic equations. If you have some difficulties, contact Andrew on +2348036670623 or by email andrewukauwa797@gmail.com

2. Solution by completing the square

(x – 3)² = x² – 6x +9

Hence x² – 6x = (x – 3)² – 9 = (x – 3)² – (-3)²

The key is half the coefficient of x, hence x² – bx + c

(x – half the coefficient of x) – (half the coefficient of x)²

= {x – (b/2)}² – (b/2)²

Example I: Complete the square for x² +3x

Solution:

x² +3x = {x + (3/2)}² – (3/2)²

= {x + (3/2)}² – (9/4)

Example 2: Complete the square for 2x² - x

Solution:

We can only complete the square if the coefficient of x² is 1. So we rewrite the expression as:

2{x² – (x/2)} = 2 [{x² – (1/4)}² – ( ¼)²]

2 [{x² – (1/4)}² – ( 1/16)

2 {x² – (1/4)}²  – ( 1/8)

Exercise 1.4

3. Solution by Formula Method

If ax² + bx + c = 0

Then x =  –b (+ -) √(b² – 4ac)

                           2a

Example 1: Solve the equation 2x² – 3x – 1

Solution:

From the above equation, a = 2, b = -3, c = -1

Hence, x =  –b (+ -) √(b² – 4ac) = – (–3) (+ -) √{(-3)² – 4(2)(-1)}

                                  2a                                2(2)

x = 3 (+ -) 4.123 = -0.28 or 1.78

                4

Example 2: Solve the equation 1 – 3x² = 5x

Solution:

Rewrite in Standard Form

3x² + 5x – 1 = 0

From the above equation, a = 3, b = 5, c = -1

Hence, x =  –b (+ -) √(b² – 4ac = – (5) (+ -) √{(5)² – 4(3)(-1)}

                                2a                                2(3)

x = -5 (+ -) 6.083 = -1.85 or 0.18

                6

SIMULTANEOUS LINEAR AND QUADRATIC EQUATIONS

Example I: Solve the equations

x + y = 5                      (1)

x² + y² = 13                 (2)

From (1), x = 5 – y      (3)

Substitute for x in equations (2)

(5 – y)² + y² = 13

25 – 10y + y² + y² = 13

2y² - 10y + 12 = 0

Dividing through by 2

y² – 5y + 6 = 0

Using factorization method

(y – 3)(y – 2) = 0

y = 3 or 2

From (3), substitute for y

x = 5 – 3 or 5 – 2

x = 2 or 3

Hence, x = 2, y = 3 or x = 3, y = 2

The coordinates of the points where line (x + y = 5) cuts the circle x² + y² = 13 are (2, 3) and (3, 2)

Example 2: Solve the equations

x – 2y = 1                    (1)

x² + y² = 29                  (2)

Solution:

From (1), x = 1 + 2y    (3)

Substitute for x in equation (2)

(1 + 2y)² + y² = 29

1 + 2y + 2y + 4y² + y² = 29

5y² + 4y + 1 = 29

5y² + 4y + 1 = 29

5y² + 4y + 28 = 0

Using the formula method to find y

y =  – (4) (+ -) √{(4)² – 4(5)(-28)}  = -4 (+ -) √{16 + 560}

                           2(5)                                       10

x = -4 (+ -) 24 = 20 or  -28

             10          10       10

y = 2 or -14

                5

Substituting for y in (3)

When y = 2, x = 1 + 4 = 5

When y = -14 = x = 1 – (28/5) = - 23

                   5                                  5

SPECIAL CASES OF EQUATIONS INVOLVING ONE LINEAR AND ONE QUADRATIC

Example I: Solve the equations:

3x – 2y = 4                              (1)

6x² – xy – 2y² + 36 = 0         (2)

Solution:

Factorizing equation (2)

(3x – 2y)(2x + y) + 36 = 0

But 3x – 2y = 4

Hence, 4(2x + y) + 36 = 0

Divide through by 4

2x + y + 9 = 0                          (3)

2x + y = -9                               (4)

4 x eqn. (4), 4x + 2y = -18       (5)

(1) + (5), 7x = -14

x = -14/7 = -2

Substituting for x in (1)

3(-2) – 2y = 4

-6 – 2y = 4

2y = -6 - 4 = -10

y = -10/2 = -5

Therefore, x = -2, y = -5

Example 2: Solve the equations:

x + 3y = 2                                (1)

x² – 2xy + y² = 36                    (2)

Solution:

Factorizing equation (2)

(x - y)(x - y) = 36

(x - y)² = 36

x - y = 6 or -6

x = y + 6 or y - 6                                  (3)

Substituting for x in equation (1)

Either (y + 6) + 3y = 2 or (y – 6) + 3y = 2

Either 4y = -4 or 4y = 8

Hence, y = -1 or 2

Substitute for y in equation (1)

When y = -1, x = 2 – (-3) = 5

When y = 2, x = 2 – (6) = -4

Therefore, x = 5, y = -1 and x = -4, y = 2