Back to: MATHEMATICS JSS 2

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*In today’s Mathematics class, We will be discussing Solving Equations. We hope you enjoy the class!*

**Solving Equations**

2x – 9 = 15 is an **equation** in x. x is the **unknown** in the equation. 2x – 9 is on the left-hand side (LHS) of the equals sign and 15 is on the right-hand side (RHS) of the equal sign.

To **solve an equation **means to find the value of the unknown that makes the equation true.

##### The balance method (revision)

Think of the two sides of an equation as forming a balance. Keep the balance by doing the same operation to both sides of the equation.

**Example**

Solve 3x = 12

3x = 12

Divide both the LHS and RHS by 3, the coefficient of the unknown. This keeps the balance of the equation.

$\frac{3x}{3}=\frac{12}{3}$x = 4

x = 4 is the solution of the equation 3x = 12

check: when x = 4, LHS = 3 X 4 = 12 = RHS

**Example**

Solve 2x – 9 = 15.

2x – 9 = 15

The LHS contains the unknown. Add 9 to 2x – 9. This leaves 2x. 9 must also be added to the RHS to keep the balance of the equation.

2x – 9 = 15

Add 9 to both sides (+9 is the additive inverse of -9)

Simplify 2x = 24

The equation is now simpler. Divide the LHS by 2 to leave x. The RHS must also be divided by 2 to keep the balance of the equation.

2x = 24

Divide both sides by 2.

$\frac{2x}{2}=\frac{24}{2}$x = 12

x = 12 is the solution of the equation 2x – 9 = 15.

Check: when x = 12, LHS = 2 x 12 – 9 = 24 – 9 = 15 RHS.

**Exercise**

Use the balance method to solve the following:

- 3x – 8 = 10
- 4x – 1 = 1
- 27 =10x – 3

**Using directed numbers**

It is possible to use operations with directed numbers when solving equations.

**Example**

Solve 25 – 9x = 2

25 – 9x = 2

Subtract 25 from both sides.

25 – 25 – 9 = 2 – 25

– 9x = – 23

Divide both sides by -9.

$\frac{\u20139x}{\u20139}=\frac{\u201323}{\u20139}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}x=\frac{23}{9}=2\frac{5}{9}$ $check:whenx=\frac{23}{9}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}LHD=25\u20139\times \frac{23}{9}=25\u201323=2=RHS$

**Unknowns on both sides**

If an equation has unknown terms on both sides of the equal sign, collect the unknown terms on one side and the number terms on the side.

Example

Solve 5x – 4 = 2x + 11

5x – 4 = 2x + 11 (1)

Subtract 2x from both sides of (1).

5z – 2x – 4 = 2x – 2x + 11

3x – 4 = 11 (2)

Add 4 to both sides of (2).

3x – 4 + 4 = 11 + 4

3x = 15

Divide both sides of (3) by 3. (3)

x = 5

Check: x = 5,

LHS = 5 x 5 -4 25 – 4 =21

RHS = 2 x 5 + 11 = 10 + 11 = 21 = LHS

Note that equations (1), (2), and (3) are still equivalent.

Exercise

- 13 – 6 = 1
- 4b + 24 = 0
- 12 + 5a = 23

**Equations with brackets**

Always remove brackets before collecting terms.

Solve 3(3x – 1) = 4(x + 3)

3(3x – 1) = 4(x + 3) (1)

Remove brackets.

9x – 3 = 4x + 12 (2)

Subtract 4x from both sides and add 3 to both sides.

9x – 4x -3 + 3 = 4x – 4x + 12 + 3

5x = 15 (3)

Divides both sides by 5.

x = 3

Check: when x = 3,

LHS = 3(3 x 3 -1) = 3(9 – 1) = 3 X 8 = 24

RHS = 4(3 + 3) = 4 X 6 = 24 = LHS

**Example**

Solve 5(x + 11) + 2(2x – 5) = 0.

5(x + 11) + 2(2x – 5) = 0. (1)

5(x + 11) + 2(2x – 5) = 0.

Remove brackets.

5x + 55 + 4x – 10 = 0 (2)

Collect like terms.

9x + 45 = 0 (3)

Subtract 45 from both sides.

9x = -45 (4)

Divide both sides by 9.

x = -5

Check: when x = -5

LHS = 5(-5 + 11) + 2(2 X (-5) -5)

= 5 X 6 + 2(-10 -5)

= 30 + 2 X (-15) = 30 – 30 = 0 = RHS

** **

**EVALUATION**

- 5(x – 4) – 4(x + 1) = 0
- 3(2x + 3) – 7(x + 2) = 0
- 2(x + 5) = 18

**Equations with fractions**

Always clear fractions before collecting terms. To clear fractions multiply both sides of the equation by the LCM of the denominators of the fractions.

**Example**

Solve the equation 4m/5 – 2m/3 = 4.

$\frac{4m}{5}\u2013\frac{2m}{3}=4$The LCM of 5 and 3 is 15.

Multiply both sides of the equations by 15, i.e. multiply every term by 15.

$15\times \left(\frac{4m}{5}\right)\u201315\times \left(\frac{2m}{3}\right)=15\times 4\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}3\times 4m\u20135\times 2m=15\times 4$15 X (4m/5) – 15 X (2m/3) = 15 X 4

3 X 4m – 5 X 2m = 15 X 4

12m – 10m = 60

2m = 60

Divide both sides by 2.

m =30

check: when m = 30,

$LHS=4\times \frac{30}{5}\u20133\times \frac{30}{3}=\frac{120}{5}\u2013\frac{60}{3}$= 24 – 20 = 4 = LHS

**Example**

Solve the equation

$3x\u2013\frac{2}{6}\u20132x+\frac{7}{9}=0$**Solution**

The LCM of 6 and 9 is `8.

$\frac{18(3x\u20132)}{6}\u2013\frac{18(2x+7)}{9}=18\times 10$3(3x – 2) – 2(2x + 7) = 0

Clear brackets.

9x – 6 – 4x – 14 = 0

Collect like terms.

5x – 20 = 0

Add 20 to both sides.

5x = 20

Divide both sides by 5

x = 4

Check: when x = 4

LHS = 3 X 4 – 2/6 – 2 X 4 + 7/9

$LHS=3\times 4\u2013\frac{2}{6}\u20132\times 4+\frac{7}{9}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=12\u2013\frac{2}{6}\u20138+\frac{7}{9}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{10}{6}\u2013\frac{15}{9}=\frac{5}{3}\u2013\frac{5}{3}=0=RHS$

**Exercise**

- x/3 = 5
- x/5 = ½
- 4/3 = 2z/15
- x – 2/3 = 4

**Word Problems**

We can use equations to solve **word problems, **i.e. problems using everyday language instead of just numbers or algebra. There is always an **unknown **in a word problem. For example, if a question says *what is the length of the room?. *Then length is the unknown and the task is to find its numerical value.

**From words to algebra**

When solving a word problem:

- Choose a letter for the unknown
- Write down the information about the question in algebra form.
- Make an equation.
- Solve the equation
- Give the answer in written form
- Check the result against the information given in the question.

**Example**

I think of a number. I multiply it by 5. I add 15. The result is 100. What is the number I thought of?

Let the number be n

I multiply n by 5: 5n

I add 15: 5n + 15

The result is 100; 5n + 15 = 100 (1)

Subtract 15 from both sides of (1).

5n + 15 – 15 = 100 – 15

5n = 85 (2)

Divides both sides of (2) by 5.

5n/5 = 85/5

n = 17

The number is 17.

Check: 17 X 5 = 85; 85 + 15 = 100

**Example**

When 6 is added to four times a number, the result is 50. Find the number.

Step 1: What are we trying to find?

A number.

Step 2: Assign a variable for the number.

Let’s call it *n*.

Step 3: Write down what the variable represents.

Let *n* = a number

Step 4: Write an equation.

We are told 6 is added to 4 times a number. Since n represents the number, four times the number would be 4*n*. If 6 is added to that, we get 6 + 4n. We know that answer is 50, so now we have an equation 6 + 4n = 50

Step 5: Solve the equation.

6 + 4n = 50

4n = 44

n = 11

Step 6: Answer the question in the problem

The problem asks us to find a number. We decided that *n* would be the number, so we have *n* = 11. The number we are looking for is 11.

Step 7: Check the answer.

The answer makes sense and checks in our equation from Step 4.

6 + 4(11) = 6 + 44 = 50

**Exercise**

- John thinks of a number. He doubles it. His result is 58. What number did John think of?
- Six boys each have the same number of sweets. The total number of sweets is 78. How many sweets did each boy have?
- A number is multiplied by 6 and then 4 is added. The result is 34. Find the first number.

#### Word problems with brackets

**Example**

- If fish cost £2 and chips cost £1 and you went into the shop and asked for 2 fish and chips would you be expecting to pay £5 or £6?

Well, it all depends on what you actually wanted.

Was it?

2 x (fish and chips) = 2 x (£2 + £1) = 2 x £3 = £6

or

(2 x fish) + chips = (2 x £2) + £1 = £4 + £1 = £5

- I bought 3 boxes of eggs in the market. Each box contained 12 eggs. When I got home I found that 5 were broken and had to be thrown away. How many eggs did I have left?

(3 x 12) – 4 = 32

36 – 4 = 32

If I had not done the calculation in brackets first, I could have got 24 as an answer

3 x 12 – 4 = 24

3 x 8 = 24

and that would have been the wrong answer.

**Exercise**

**1)** The farmer has four chicken runs. In each run, there are 67 brown and fourteen black hens. How many chickens are there altogether?

**Hint:**(67+14) x 4 = 224

**2)** 124 cakes were bought, but there wasn’t enough so they decided to buy 4 times more. Then there were too many so they took 10 away. How many did they have in the end?

**Hint: **(124×4) – 10 = 486

##### Word Problems with fractions

I add 55 to a certain number and then divide the sum by 3. The result is four times the first number. Find the number.

Let the number n.

I add 55 to n: this gives n + 55

I divide the sum by 3: this gives n + 55/3

The result is 4n.

So, n + 55/3 = 4n (1)

Multiply both sides by 3.

3(n + 55)/3 = 3 X 4n (2)

n + 55 = 12n (3)

Collect terms.

55 = 12n – n

55 = 11n (4)

2x – 9 = 15 is an **equation** in x. x is the **unknown** in the equation. 2x – 9 is on the left-hand side (LHS) of the equals sign and 15 is on the right-hand side (RHS) of the equal sign.

So, n = 5, the number is 5.

**Weekend Assessment**

- I think of a number. I double it. I divide the result by 5. My answer is 6. What number did I think of?
- I subtract 17 from a certain number and then divide the result by 5. My final answer is 3. What was the original number?
- I add 9 to a certain number and then divide the sum by 16. Find the number if my final answer is 1.

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