Classwork Series and Exercises {Mathematics – JSS2}: Solving Equations

JSS 2 Mathematics Second Term Week 5

Topic: SOLVING EQUATIONS

Solving Equations (1)

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.

3x/3 = 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

a. 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

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

2x = 24

Divide both sides by 2.

2x/2 = 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:

a. 3x – 8 = 10

b. 4x – 1 = 1

c. 27 =10x – 3

Solving Equations (2)

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.

        – 9x/-9 = -23/-9

          x = 23/9 = 2 5/9

Check: when x = 23/9,

LHS = 25 – 9 X 23/9 = 25 – 23 = 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

a. 13 – 6 = 1

b. 4b + 24 = 0

c. 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)

Subtarct 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 + 5 = 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

Exercise

a. 5(x – 4) – 4(x + 1) = 0

b. 3(2x + 3) – 7(x + 2) = 0

c. 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.

4m/5 – 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 X (4m/5) – 15 X (2m/3) = 15 X 4

                3 X 4m – 5 X 2m = 15 X 4

                           12 – 10m = 60

                                    2m = 60

Divide both sides by 2.

m =30

check: when m = 30,

LHS = 4 X 30/5 – 3 X 30/3 = 120/5 – 60/3

         = 24 – 20 = 4 = LHS

Example

Solve the equation 3x – 2/6 – 2x + 7/9 = 0.

The LCM of 6 and 9 is `8.

18(3x – 2)/6 – 18(2x + 7)/9 = 18 X 0

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

         = 12 – 2/6 – 8 + 7/9

         = 10/6 – 15/9 = 5/3 -5/3 = 0 = RHS

Exercise

a. x/3 = 5

b. x/5 = ½

c. 4/3 = 2z/15

d. 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:

1. Choose a letter for the unknown

2. Write down the information of the question in algebra form.

3. Make and equation.

4. Solve the equation

5. Give the answer in written form

6. 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 4n. 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

1. John thinks of a number. He doubles it. His result is 58. What numbe did John think of?

2. Six boys each have the same number of sweets. The total number of sweets is 78. How many sweets did each boy have?

3. A number is multiplied by 6 and then 4 is added. The result is 34. Find the first number.

Word problems with brackets

Example

1. 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

2. 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 chicken 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.

Exercise

1. I think of a number. I double it. I divide the result by 5. My answer is 6. What number did I think of?

2. I subtract 17 from a certain number and then divide the result by 5. My final answer is 3. What was the original number?

3. I add 9 to a certain number and then divide the sum by 16. Find the number if my final answer is 1.