#### Solving Algebraic Equations

- An algebraic expression with an equal to sign is called an
*algebraic equation*. For example, x + 8 = 3 is an equation in x. - The unknown in an equation is shown by a letter. For example, m is the unknown in the

equation 2m + 3 = 17. - If a value is substituted for an unknown, the equation may be true or false. For example, 2y = 6 is true when y = 3, and is false when y = 5.
- To solve an equation using the balance method, always do the same to both sides of the equation.
- When solving an equation, you may cut out steps in the method when you can do them mentally.
- Check if your solution is correct by substituting the value to see if it makes the equation true.

**SIMULTANEOUS EQUATIONS**

- In an equation, such as y = 3x – 4, the unknowns x and y are called variables.
- Simultaneous linear equations can be solved algebraically; either by substitution or by elimination.
- Simultaneous linear equations can also be solved graphically by drawing the graphs of the two equations on the same axes; their point of intersection gives the solution.

**QUADRATIC EQUATIONS**

- A quadratic equation is equation in the form ax² + bx + c = 0, where a, b, and c are constants

and a ≠ 0. - The highest power in a quadratic equation is 2.
- The values of x that satisfy the quadratic equation are called the roots of the equation.
- The roots of a quadratic equation can be found by:
- factorization:
- the quadratic equation is simply factorized and set equal to zero. For example:
- Given the quadratic equation ax² + bx + c = 0, if (dx + e) and (fx + g) are the factors,
- Then (dx + e)(fx + g) = 0.
- Then either (dx + e) = 0, i.e., x = –
^{e}/_{d} - Or (fx + g) = 0, i.e., –
^{g}/_{f}

- the quadratic equation is simply factorized and set equal to zero. For example:

- factorization:

__Examples__

__Examples__

- Say whether the following are true or false:
- x + 2 = 5, when x = 3.
- When x = 3; x + 2 = 3 + 2 = 5
- Thus, x + 2 = 5 is true when x = 3

- 3x + 1 = 10 when x = 2
- When x = 2; 3x + 1 = (3 × 2) +1
- = 6 + 1
- = 7
- 7 ≠ 10 (‘≠’ means not equal to)
- Thus, 3x + 1 = 10 is false when x = 2.

- x + 2 = 5, when x = 3.
- Is the equation
^{x}/_{4}= 5 true when x = 24?- When x = 24;
^{x}/_{4}=^{24}/_{4}= 6 - 6 ≠ 5
- Thus,
^{x}/_{4}= 5 is not true when x = 24.

- When x = 24;
- Solve 18 – x = 7.
- The problem is to find a number which when taken from 18 gives 7.
- The number is 11.
- ∴ x = 11 is the solution.

- Solve
^{1}/_{3}y = 7.- The unknown y is on the Left Hand Side (LHS).
- It is divided by 3 to give
^{1}/_{3}y on the LHS. - Multiply both sides by 3
^{1}/_{3}y × 3 = 7 × 3- ∴ y = 21.

- Solve 5x – 6 = 29.
- The unknown x is contained in the LHS.
- Add 6 to both sides to leave 5x alone on the LHS: 5x – 6 + 6 = 29 + 6
- ∴ 5x = 35
- The unknown ‘x’ is on the LHS, and it is multiplied by 5.
- Divide both sides by 5 to obtain x on the LHS:
^{5x}/_{5}=^{35}/_{5} - ∴ x = 7
- To Check: when x = 7; LHS = 5 × 7 – 6
- = 35 – 6
- = 29
- ∴ 5x – 6 = 29 is true when x = 7

- Solve 2x + 7 = 12.
- Subtract 7 from both sides: 2x + 7 – 7 = 12 – 7
- ∴ 2x = 5
- Divide both sides by 2:
^{2x}/_{2}=^{5}/_{2} - ∴ x = 2
^{1}/_{2} - To check: When x = 2
^{1}/_{2} - LHS = 2 × 2
^{1}/_{2}+7 - = 5 + 7
- = 12
- ∴ 2x + 7 = 12 when x = 5

- Solve the equations (Substitution Method)

3a + b = 10

2a + 4b = 0- 3a + b = 10 – (i)

2a + 4b = 0 – (ii) - from (i), b = 10 – 3a – (iii)
- Substitute (10 – 3a) for b in (ii): 2a + 4(10 – 3a) = 0
- Clear the brackets: 2a + 40 – 12a = 0
- Collect like terms: 2a – 12a = -40
- -10a = -40
- ∴ a = 4
- Substitue a = 4 in (iii): b = 10 – (3 × 4)
- b = 10 – 12
- ∴ b = -2
- Therefore, a = 4, and b = -2

- 3a + b = 10 – (i)
- Solve the equations (Elimination Method)

3f = 4 – 4e

2e = 5f + 15- 3f = 4 – 4e – (i)

2e = 5f + 15 – (ii) - Arrange the equations so that the unknowns are in alphabetical order on the LHS and the numbers are on the RHS:
- 4e + 3f = 4 – (iii)

2e – 5f = 15 – (iv) - Multiply (iv) by 2: 4e – 10f = 30 – (v)
- Subtract (v) from (iii) to eliminate terms in e: 13f = -26f

∴ f = -2 - Substitute -2 for f in (ii): 2e = 5(-2) + 15
- 2e = -10 + 15
- 2e = 5
- ∴ e = 2
^{1}/_{2}

- 3f = 4 – 4e – (i)
- Solve the equation x² + 6x – 7 = 0
- First step: Factorize x² + 6x – 7. Find two numbers such that their product is -7 (+1 × -7) and their sum is +6.
**Factors of -7****Sum of Factors**a. -1 and +7 +6 b. +1 and -7 -6 - Of these, only a gives the required result. Therefore x² + 6x – 7 is the same as x² – x + 7x – 7
- = x(x – 1) + 7(x – 1)
- = (x – 1) (x + 7)
- ∴ (x – 1) (x + 7) = 0
- Second Step: Equate each bracket to zero and solve for x.
- ⇒ x – 1 = 0
- ∴ x = 1
- ⇒ x + 7 = 0
- ∴ x = -7
- ∴ x = 1 and – 7

- First step: Factorize x² + 6x – 7. Find two numbers such that their product is -7 (+1 × -7) and their sum is +6.
- Solve 3y² + 13y + 12 = 0.
- Factors which when multiplied would give 36 (3 × 12) and when summed up would give 13 are +9 and + 4
- ∴ 3y² + 13y + 12 is the same as 3y² + 9y + 4y + 12
- ∴ 3y² + 9y + 4y + 12 = 0
- ⇒ 3y(y + 3) + 4(y + 3) = 0
- (3y + 4) (y + 3) = 0
- ⇒ 3y + 4 = 0
- ∴ y = –
^{4}/_{3} - ⇒ y + 3 = 0
- ∴ y = -3
- ∴ y = -3 and –
^{4}/_{3}

- Solve 4 – 3z – z²
- Factors which when multiplied would give -4 (4 × -1) and when summed would give -3 are -4 and +1.
- ∴ 4 – 3z – z² is the same as 4 – 4z +z – z²
- ∴ 4 – 4z +z – z² = 0
- ⇒ 4(1 – z) + z(1 – z) = 0
- ∴ (4 + z) (1 – z) = 0
- ⇒ 4 + z = 0
- ∴ z = -4
- ⇒ 1 – z = 0
- ∴ z = 1
- ∴ z = 1 and -4