8 Math Linear Equations

Finish Line & Beyond

Linear Equations in One Variable Class 8 1. An algebraic equation is an equality involving variables. It says that the value of the expression on one side of the equality sign is equal to the value of the expression on the other side. 2. In such equations, the expressions which form the equation contain only one variable. Further, the equations are linear, i.e., the highest power of the variable appearing in the equation is 1. 3. A linear equation may have for its solution any rational number. 4. An equation may have linear expressions on both sides. 5. Just as numbers, variables can, also, be transposed from one side of the equation to the other. 6. Occasionally, the expressions forming equations have to be simplified before we can solve them by usual methods. Some equations may not even be linear to begin with, but they can be brought to a linear form by multiplying both sides of the equation by a suitable expression. 7. The utility of linear equations is in their diverse applications; different problems on numbers, ages, perimeters, combination of currency notes, and so on can be solved using linear equations.

Exercise 1 Solve the following equations: 1. x − 2 = 7

⇒ x = 7+ 2= 9 y + 3 = 10 ⇒ y = 10 − 3 = 7

2.

3. 6 = z + 2

⇒ z = 6− 2 = 4 3 17 + x= 7 7 17 3 17 − 3 14 ⇒ x= − = = = 2 7 7 7 7

4.

5. 6 x = 12

⇒ x = 12 ÷ 6 = 2 t = 10 5 ⇒ t = 10 × 5 = 50

6.

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Finish Line & Beyond 2x = 18 3 ⇒ 2 x = 18 × 3 = 54 ⇒ x = 54 ÷ 2 = 27

7.

y 1.5 ⇒ y = 1.6 × 1.5 = 2.4

8.

1.6 =

9. 7 x − 9 = 16

⇒ 7 x = 16 + 9 = 25 25 ⇒ x= 7 14 y − 8 = 13 ⇒ 14 y = 13 + 8 = 21 21 3 ⇒ y= = 14 2

10.

17 + 6 p = 9 ⇒ 6 p = 9 − 17 = − 6 6 p= − = 1 6

11.

x 7 + 1= 3 15 x 7 7 − 15 8 ⇒ = − 1= = − 3 15 15 15 8 8 ⇒ x= − × 3= − 15 5

12.

Exercise 2 1. If you subtract

1 1 1 from a number and multiply the result by , you get . 2 2 8

What is the number? Answer: Let us assume the number to be x Then, as per question we have following equation:

1 1 1   x−  × = 2 2 8 

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Finish Line & Beyond 1 = 2 1 ⇒ x− = 2 1 ⇒ x= + 4 ⇒ x−

1 1 ÷ 8 2 1 1 × 2= 8 4 1 2+ 1 3 = = 2 4 4

2. The perimeter of a rectangular swimming pool is 154 metres. Its length is 2m more than twice its breadth. What are the length and breadth of the pool. Answer: Let us assume the breadth of swimming pool = Then as per question its length = 2 x + 2 metres. As we know Perimeter of Rectangle =

⇒ ⇒ ⇒ ⇒

x metres

2(length + breadth)

154 = 2( x + 2 x + 2) 154 = 2(3x + 2) = 6 x + 4 6 x = 154 − 4 = 150 x = 150 ÷ 6 = 25

So, Breadth = 25 m Length = 52 m Perimeter= 2(52+25)=154 m 3. The base of an isosceles triangle is is

4

4 cms. The Perimeter of the triangle 3

2 cms. Find the length of other two sides of the triangle. 15

Answer: Let us assume the two equal sides to be x cms Then Perimeter can be written as following equation:

2 4 = 2x + 15 3 62 4 62 − 20 42 ⇒ 2x = − = = 15 3 15 15 42 1 21 7 ⇒ x= × = = cms 15 2 15 5 4

4. Sum of two numbers is 95. If one exceeds the other by 15 find the numbers. Answer: Let us assume one number= Hence, another muber = x + 15

x

As per question:

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Finish Line & Beyond x + x + 15 = 95 ⇒ 2 x + 15 = 95 ⇒ 2 x = 95 − 15 = 80 ⇒ x = 80 ÷ 2 = 40 ⇒ x + 15 = 55 5. Two numbers are in the ration 5:3. If they differ by 18, find these numbers. Answer: Suppose one number = x Then another number = x + 18 As per question we get following equation:

x + 18 5 = x 3 ⇒ 3 x + 54 = 5 x ⇒ 5 x − 3 x = 54 ⇒ x = 27 ⇒ x + 18 = 45 45 : 27 = 5 : 3

6. Three consecutive integers add up to 51. What are these integers? Answers: Let the first integer = x, second = x+1 and third = x+3 So, x + x + 1 + x + 2 = 51

⇒ 3 x + 3 = 51 ⇒ 3 x = 48 ⇒ x = 16

1st Integer = 16 2nd Integer = 17 3rd Integer = 18 7. The sum of three consecutive multiples of 8 is 888. Find the multiple. Answer: Let us assume first multiple = 8 x Hence Second multiple = 8( x +

1) Third multiple = 8( x + 2)

Now we get following equation:

8 x + 8( x + 1) + 8( x + 2) = 888 ⇒ 8 x + 8 x + 8 + 8 x + 16 = 888 ⇒ 24 x + 24 = 888 ⇒ 24 x = 888 − 24 = 864 ⇒ x = 864 ÷ 24 = 36

Hence,

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Finish Line & Beyond 1st multiple = 8 × 36 = 288 2nd multiple = 8 × 37 = 296 3rd multiple = 8 × 38 = 304 288+296+304=888 8. Three consecutive integers are as such when they are taken in increasing order and multiplied by 2, 3, and 4 respectively, they add up to 74. Find these numbers. Answer: Suppose first integer= x Then second integer = x + 1 Third integer = x + 2 Now we get the following equation:

2 x + 3( x + 1) + 4( x + 2) = 74 ⇒ 2 x + 3 x + 3 + 4 x + 8 = 74 ⇒ 9 x + 11 = 74 ⇒ 9 x = 74 − 11 = 63 ⇒ x = 63 ÷ 9 = 7

Hence, the set of numbers is : 7, 8, 9

(14+24+36=74)

9. The number of boys and girls in a class is in 7:5 ratio. The number of boys is 8 more than that of girls. Fin their numbers. Answer: We can make following equation:

7 x − 5x = 8 ⇒ 2x = 8 ⇒ x = 8÷ 2 = 4 So, Boys = 7 × 4 = 28 Girls = 5 × 4 = 20

10. The ages of Rahul and Haroon are in the ratio of 5:7. Four years from now sum of their ages will be 56 years. Find their present age. Answer: Suppose Rahul’s current age = 5x Rahul’s age after 4 years = 5x+4 Haroon’s current age = 7x Haroon’s age after 4 years = 7x+4 As per question,

5 x + 4 + 7 x + 4 = 56 ⇒ 12 x + 8 = 56 ⇒ 12 x = 56 − 8 = 48 ⇒ x = 48 ÷ 12 = 4 So, Rahul’s Age = 5 × 4 = 20 Haroon’s Age = 7 × 4 = 28 11. Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung. The sum of their ages is 135. Find their ages.

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Finish Line & Beyond Answer: As per question we get following equation: x + x + 26 + x − 29 = 135 , where x is father’s age.

⇒ 3 x − 3 = 135 ⇒ 3 x = 132 ⇒ x = 132 ÷ 3 = 44

Father’s Age = 44 years Grandfather’s age = 70 years Baichung’s Age = 15 years 12. Fifteen years from now Ravi’s age will be 4 times his current age. What is his current age. Answer: We get following equation:

x + 15 = 4 x ⇒ 3 x = 15 ⇒ x = 15 ÷ 3 = 5

Ravi’s Current Age = 5 years. 13. A rational number is as such that when you multiply it by to it you get

−

5 2 and add 2 3

7 . What is the number? 12

Answer: We get following equation:

5 2 7 + = − 2 3 12 5x 7 2 − 7− 8 15 ⇒ = − − = = − 2 12 3 12 12 5 2 1 ⇒ x= − × = − 12 5 6 x×

14. Lakshmi is a cashier in a bank. She has notes of denominations of Rs. 100, 50 and 10 respectively. The ratio of number of these notes is 2:3:5 respectively. The total cash with Lakshmi is 4,00,000. How many notes of each denomination does she have? Answer: We can get following equation:

200 x + 150 x + 50 x = 400000 ⇒ 400 x = 400000 ⇒ x = 1000

So, 100 rupees notes = 2000 50 rupees notes = 3000 10 rupees notes = 5000

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Finish Line & Beyond 15. I have total Rs 300 in coins of denominations of Rs.1, Rs.2, and Rs. 5.The number of Rs. 2 coins is 3 times the number of Rs. 5 coins. The total number of coins is 160. How many coins of each denomination are with me. Answer: Suppose number of Rs. 5 coins = x Then number Rs. 2 coins = 3 x Suppose number of Rs. 1 coins = y We can get following equations: 5 x + 6 x + y = 300 ……………………………….(1)

x + 3 x + y = 160 ………………………………..(2) Subtracting equation 2 from 1 we get:

7 x = 140 ⇒ x = 140 ÷ 7 = 20

hence, Number of Rs. 5 coins = 20 Number of Rs. 2 coins = 60 Number of Rs. 1 coins = 80 Total amount = 5 × 20 + 2 × 60 + 1 × 80 = 100 + 120 + 80 = 300 16. The organizers in an essay competition decide that winner will get a prize of Rs. 100 and a participation who doesn’t win gets a prize of Rs. 25. The total prize money distributed is Rs. 3,000. Find the number of winners if the total number of participants is 63. Answer: Suppose number of winners = x and number losing participants = Then, x + y = 63 ………………………………….(1)

y

And, 100 x + 25 y = 3000 ……………………….(2) Multiplying equation (1) by 25 and subtracting the result from equation (2) we get

100 x + 25 y − (25 x + 25 y ) = 3000 − 1575 ⇒ 75 x = 1425 ⇒ x = 1425 ÷ 75 = 19

Number of winners = 19 Number of losing participants =

63 − 19 = 44

Alternate Method:

x + y = 63 ⇒ x = 63 − y …………………..(3) 100 x + 25 y = 3000 ⇒ 100 x = 3000 − 25 y 3000 − 25 y ⇒ x= ……………(4) 100

From equation (3) and (4) we get following equation:

3000 − 25 y 100 ⇒ 6300 − 100 y = 3000 − 25 y 63 − y =

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Finish Line & Beyond ⇒ 6300 = 3000 − 25 y + 100 y = 3000 + 75 y ⇒ 75 y = 6300 − 3000 = 3300 ⇒ y = 3300 ÷ 75 = 44 Hence, x = 63 − 44 = 19 Exrecise 3 Solve Following Equations:

3 x = 2 x + 18 ⇒ 3 x − 2 x = 18 ⇒ x = 18

1.

2.

5t − 3 = 3t − 5 5t = 3t − 5 + 3 = 3t − 2 3t − 5t = 2 − 2t = 2 t = −1

3.

5 x + 9 = 5 + 3x 5 x = 3x − 4 3x − 5 x = 4 − 2x = 4 x= −2

⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒

4z + 3 = 6 + 2z ⇒ 4z − 2z = 6 − 3 = 3 3 ⇒ z= 2

4.

5. 2 x − 1 = 14 − x

⇒ 2 x + x = 14 + 1 = 15 ⇒ x= 5 8 x + 4 = 3( x − 1) + 7 ⇒ 8 x = 3x − 3 + 7 − 4 ⇒ 8 x = 3x

6.

Only 0 can satisfy above condition, so the value of x=0 7.

x=

4 ( x + 10) 5

⇒ 5 x = 4 x + 10 ⇒ 5 x − 4 x = x = 10

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Finish Line & Beyond

8.

⇒ ⇒ ⇒ ⇒

2x 7x + 1= + 3 3 15 2x 7x − = 2 3 15 10 x − 7 x = 2 15 3 x = 30 x = 10

5 26 = − y 3 3 26 5 21 ⇒ 3y = − = = 7 3 3 3 7 ⇒ y= 3

9.

2y +

10.

3m = 5m −

8 5

8 5 4 ⇒ m= 5 ⇒ 2m =

Exercise 4 1. Amina thinks of a number and subtracts

5 from it. She multiplies the res2

ult by 8. The final result is 3 times her original number. Find the number. Answer: Let the number = x, then we can make following equation:

5   x −  × 8 = 3x 2  ⇒ 8 x − 20 = 3 x ⇒ 8 x − 3 x = 5 x = 20 ⇒ x = 20 ÷ 5 = 4 2. A positive number is 5 times another number. If 21 is added to both the numbers then one of the new numbers becomes twice of another new numbers. Find the original numbers. Answer: Let one of the numbers = x, then another number = 5x We can make following equation:

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Finish Line & Beyond 2( x + 21) = 5 x + 21 ⇒ 2 x + 42 = 5 x + 21 ⇒ 5 x − 2 x = 42 − 21 = 21 ⇒ 3 x = 21 ⇒ x= 7 ⇒ 5 x = 35

Required numbers are 7 and 35. 3. Sum of the digits of a two digit number is 9. When we interchange the digits the new number is 27 greater than the earlier number. Find the number. Answer: Suppose the digit at unit’s place = x Digit at ten’s place = y Then, x + y = 9 ………………………….(1)

10 y + x Second Number = 10 x + y And Difference, 10 x + y − (10 y + x) = 27 ⇒ 9 x − 9 y = 27 ⇒ 9( x − y ) = 27 ⇒ x − y = 3 …………………………………(2) First number =

Adding equation (1) & (2) we get

x + y + x − y = 9 + 3 = 12 ⇒ 2 x = 12 ⇒ x= 6 ∴ y = 9− 6 = 3

The numbers are 36 and 63 4. One of the digits of a two digit number is three times the other digit. If you interchange the digits and add the resulting number to original number you get 88 as final result. Find the numbers. Answer: Let the digits be x and y Then x= 3y The numbers are: 10x+y Or, 30y+y and, 10y+x Or, 10y+3y Hence, 31y+13y = 88 Or, 44y=88 Or, y= 2 So, x=6 Hence, numbers are 26 and 62.

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Finish Line & Beyond 5. Sahoo’s mother’s present age is six times Sahoo’s present age. Five year from now Sahoo’s age will be one-third of his mother’s age. Find their current age. Answer: Let Sahoo’s Age= x Mother’s Age=5x Sahoo’s Age 5 years from now= x+5 Mother’s Age 5 years from now= 5x+5 As per question 3(x+5)=5x+5 Or, 3x+15=5x+5 Or, 3x+10 = 5x Or, 2x = 10 Or, x = 5 So, Sahoo’s Current Age = 5 years Mother’s Current Age = 25 years

6. There is a narrow rectangular plot. The length and breadth of the plot are in the ratio of 11:4. At the rate of Rs. 100 per metre it will cost village panchayat Rs.75000 to fence the plot. What are the dimensions of the plot. Answer: Let us assume, Length = 11x Breadth = 4 x

Cost 75000 = = 750 metre Rate 100 2(11x + 4 x) = 750 ⇒ 11x + 4 x = 750 ÷ 2 = 375 ⇒ 15 x = 375 ⇒ x = 375 ÷ 15 = 25 ⇒ 11x = 275m and 4 x = 100m

Perimeter =

7. Hasan buys two kinds of cloth materials for school uniform. Shirt material cost him Rs. 50 per metre and trousers material cost him Rs. 90 per metre. For every 2 metres of the trousers material he buys 3 metres of shirt material. He sells them at 12% and 10% profit respectively. His total sale is Rs. 36,660. How much trousers material did he buy? Answer: Let us assume the length of trousers material = 2 x m Length of shirt material = 3 x m Profit on 1 metre of trousers cloth = 90 × 12% = 10.80 Hence Sales Price per metre for trousers cloth = 90 + 10.80 = 100.80 Profit on 1 metre of shirt cloth = 50 × 10% = 5 Hence Sales Price per metre for shirt cloth = 55

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Finish Line & Beyond Now we can get following equation:

2 x × 100.80 + 3 x × 55 = 36660 ⇒ 201.60 x + 165 x = 36660 ⇒ 366.60 x = 36660 ⇒ x = 36660 ÷ 366.60 = 100 ⇒ 2 x = 200 metre is the length of trousers material.

8. Half of a herd of deer are grazing in the field and three fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the total number of deer in the herd. Answer: Let total number deer= x

x 2 x 3 3x = Number of deer playing = × 2 4 8 Number of deer grazing =

Number of deer playing = 9

x 3x + + 9 2 8 4 x + 3 x + 72 ⇒ x= 8 ⇒ 8 x = 7 x + 72 ⇒ x = 72 ⇒ x=

9. A grandfather is 10 times older than his granddaughter. He is also 54 years older than her. Find their age. Answer: Let Granddaughter’s age = x Then Grandfather’s age = 10x So, 10 x = x + 54

⇒ 9 x = 54 ⇒ x= 6

Age of Granddaughter = 6 years Age of Grandfather = 60 years 10. A man’s age is three times his son’s age. Ten years ago his age was five times his son’s age. Find their current age. Answer: Let Son’s Age = x Then, Father’s Age = 3x Ten years ago Son’s Age = x-10 Ten years ago father’s Age = 3x-10

5( x − 10) = 3 x − 10 ⇒ 5 x − 50 = 3 x − 10

So,

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Finish Line & Beyond ⇒ 5 x = 3 x + 40 ⇒ 2 x = 40 ⇒ x = 20

Son’s Age = 20 years Father’s Age = 60 years

Exercise 5 Solve Following Equations: 1.

⇒ ⇒ ⇒ ⇒

x 1 x 1 − = + 2 5 3 4 x x 1 1 − = + 2 3 4 5 3x − 2 x 5 + 4 = 6 20 x 9 = 6 20 9 27 x= × 6= 20 10

x 3x 5 x − + = 21 2 4 6 6 x − 9 x + 10 x ⇒ = 21 6 ⇒ 7 x = 21 × 6 21 × 6 ⇒ x= = 18 7

2.

3.

⇒ ⇒ ⇒ ⇒ ⇒ 4.

8 x 17 5 x = − 3 6 2 8 x 5 x 17 x− + = − 7 3 2 6 6 x − 16 x + 15 x 17 − 42 = 6 6 5x 25 = − 6 6 5 x = − 25 x= −5 x+ 7−

3t − 2 2t + 3 2 − = − t 4 3 3

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Finish Line & Beyond

⇒ ⇒

3t − 2 2t + 3 2 − + t= 4 3 3 9t − 6 − 8t − 12 + 12t 2 = 12 3 13t − 18 2 = 12 3 2 13t − 18 = × 12 = 8 3 13t = 8 + 18 = 26 t = 13

5.

m−

⇒ ⇒ ⇒ ⇒

⇒ ⇒ ⇒ ⇒ ⇒ 6.

⇒ ⇒ ⇒ ⇒ 7.

⇒ ⇒ ⇒ ⇒ 8.

⇒ ⇒ ⇒ ⇒ ⇒

m− 1 m− 2 = 1− 2 3 m− 1 m− 2 m− + = 1 2 3 6m − 3m + 3 + 2m − 4 = 1 6 5m − 1 = 6 5m = 7 7 m= 5

3(t − 3) = 5(2t + 1) 3t − 9 = 10t + 5 3t − 10t = 5 + 9 = 14 − 7t = 14 t = −2 15( y − 4) − 2( y − 9) + 5( y + 6) = 0 15 y − 60 − 2 y + 18 + 5 y + 30 = 0 18 y − 12 = 0 18 y = 12 12 2 y= = 18 3 3(5 z − 7) − 2(9 z − 11) = 4(8 z − 13) − 17 15 z − 21 − 18 z + 22 = 32 z − 52 − 17 − 3z + 1 = 32 z − 69 32 z − (− 3 z ) = 1 + 69 = 70 35 z = 70 z= 2

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Finish Line & Beyond 9.

⇒ ⇒ ⇒ ⇒

0.25(4 f − 3) = 0.05(10 f − 9) f − 0.75 = 0.5 f − 0.45 0.5 f − f = 0.35 − 0.5 f = 0.35 f = 0.35 ÷ − 0.5 = − 0.7

Exercise 6 Solve Following Equations: 1.

⇒ ⇒ ⇒

⇒

8x − 3 = 2 3x 8x − 3 = 6x 8x − 6x = 3 2x = 3 3 x= 2

9x = 15 7 − 6x ⇒ 9 x = 105 − 90 x ⇒ 9 x + 90 x = 99 x = 105 105 35 ⇒ x= = 99 33

2.

z 4 = z + 15 9 ⇒ 9 z = 4 z + 60 ⇒ 5 z = 60 ⇒ z = 12

3.

4.

⇒ ⇒ ⇒ ⇒

3y + 4 = − 2 − 6y 15 y + 20 = 15 y = − 4 + 3 y = − 24 y= −8

2 5 − 4 + 12 y 12 y − 20

7y + 4 4 = − y+ 2 3 ⇒ 21 y + 12 = − 4 y − 8

5.

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Finish Line & Beyond ⇒ 21 y = − 4 y − 8 − 12 = − 4 y − 20 ⇒ − 4 y − 21 y = 20 ⇒ − 25 y = 20 20 4 ⇒ y= − = − 25 5 6. Hari and Harry’s age are in the ratio of 5:7. Four years later the ratio of their ages will be 3:4. Find their current age. Answer: Let Hari’s Age = 5x and Harry’s Age = 7x Four years from now Hari’s Age= 5x+4 Four years from now Harry’s Age = 7x+4 Then we get following equation:

5x + 4 3 = 7x + 4 4 ⇒ 20 x + 16 = 21x + 12 ⇒ 21x − 20 x = 16 − 12 ⇒ x= 4

So, Hari’s Age 5x =20 years Harry’s Age 7x = 28 years 7. If in a rational number denominator is greater than numerator by 8. If you increase the numerator by 17 and decrease the denominator by 1, you get

3 as result. Find the number. 2

Answer: Let the numerator be x, then denominator = x+8 So, the equation we get is as follows:

x + 17 3 = x+ 7 2 ⇒ 2 x + 34 = 3x + 21 ⇒ 3 x − 2 x = x = 34 − 21 = 13

Numerator = 13 and Denominator = 13+8=21 The rational number =

13 21

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Finish Line & Beyond

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