WBBSE Solutions For Class 8 Maths Chapter 19 Formation Of An Equation And Its Solution

Formation Of An Equation And Its Solution

Today we will play an interresting game. Shibani has collected some marbles in a bag made of cloth. Dhruba, Mahua, Ashoke, Murad and I will play with marbles in the room on the roof of Sibani’s house.

The game is like this : first Shibani will distribute some marbles to Dhruba and Mahua in a certain way and then I will tell the number of marbles each of them has got.

To Mahua, Shibani gives 18 marbles more than twice the number of marbles she gives to Dhruba. Let’s find the number of marbles Sibani might give to Mahua.

Let Shibani gives × marbles to Dhruba.

∴ Shibani gives – (2 × x + 18) = 12x + 18I marbles to Mahua.

Number of marbles 1 2 3 4 5 6 7 ……………… x
given to Dhruba

Number of marbles 20 22 24 26 28 30 32 12x + 18 marbles to Mahua.
given to Mahua.

∴ Shibani can give 20, 22, marbles to Mahua.

I come to know that Shibani has given 108 marbles to Mahua.

2x + 18 = 108

or, 2x = 108 – 18

or, 2x = 90

or, x= 90/2

∴ x = 45

So, Shibani has given 45 marbles to Dhruba.

∴ x = 45 is the root of the equation : 2x + 18 = 108.

Read and Learn More WBBSE Solutions For Class 8 Maths

Now, Shibani gives Mahua 4 less marbles than half of marbles she gives to Murad. Let’s find out the number of marbles which Shibani can give to Mahua.

Let’s think that Shibani gives x marbles to Murad.

∴ Shibani gives \(\frac{x}{4}\)-4  marbles to Mahua.

Number of marbles given 8 10 12 20 28 ……… n to Ashoke

Number of marbles given 0 1 2 6 10 \(\frac{n}{2}\)-4 to Mahua

After counting, we see that Shibani has given 86 marbles to Mahua.

= \(\frac{x}{2}-4=86\)

∴ Shibani has given 180 to Murad.

∴ x = 180

We get that the root of the equation.

= \(\frac{x}{2}-4=86\) is 180

Again, Shibani gives to Mahua 3 less marbles than \(\frac{5}{2}\) part of marbles she gives to Ashoke. Let’s find the number of marbles which Shibani can give to Mahua.

Let’s think that Shibani gives x marbles to Ashoke.

∴ Shibani gives \(\frac{5 x}{2}-3\) marbles to Mahua.

Number of marbles 2 4 8 10 20 ………………. n given to Ashoke

Number of marbles 2 7 17 22 47 ……………… \(\frac{5 x}{2}-3\) given to Mahua

If Shibani gives 127 marbles to Mahua, let’s find the number of marbles Shibani has given to Ashoke.

Let’s find the root of the equation

= \(\frac{5 x}{2}-3=127\)

Formation Of An Equation Exercise 19.1

Question 1. Shibani has given some marbles to Murad and me. If the number of marbles of Murad is 2 less than \(\frac{x}{2}\) part of the marbles of mine, then let’s calculate the possible number of marbles of Murad.
Solution:

Given

Shibani has given some marbles to Murad and me. If the number of marbles of Murad is 2 less than \(\frac{x}{2}\) part of the marbles of mine

Let Shibani has given me x marbles.

∴ Murad has been given \(\frac{7 x}{3}-2\) marbles.

Number of marbles 3 6 9 15 18 21

Number of Murad’s marbles 5 12 19 33 40 47

∴ Murad can be given 5,12,19,33,40,47……………marbles.

B.T.P.,

= \(\frac{7 x}{3}-2=40\)

or, \(\frac{7 x}{3}=40+2\)

or, x = \(\frac{42 \times 3}{7}\)

or, x = 18

If Murad has 40 marbles, then Shibani has given me 18 marbles.

This type of n number of designs contain 4n number of sticks.

If we use 80 sticks then the number of squares is 20

Therefore, 4n = 80. The root of the equation is 20

Again, Rokeya made another design –

To make such n number of ‘H’ the number of sticks will be needed is (3 x n + 2).

Let’s form an equation and find the root of the equation to find the number of ‘IT if we form H,s by 35 sticks. Rokeya made another design

This type of n number of designs contain 4n number of sticks. Now I will make another design with matchsticks and find the number of sticks required to from n number of such designs.

Make equation and write in language.

x = -19

2x + 38 = 0

t = 21

5t-105 = 0

All of us will contribute some money for organising a feast. I have 2, 100 rupee notes. I want to change these two notes. I want to change these two notes in the form of 5 and 10 rupee notes.

Shibani’s father changed my money and gave me 32 notes in total.

Let’s form an equation and solve it to get how many notes of each type I have.

Let, I have x number of 5 rupee notes.

∴ I have (32 – x) number of 10 rupee notes.

Value of 1 number of 5 rupee notes is Rs. 5.

∴ Value of 1 number of 10 rupee notes is Rs. 10.

Value of x number of 5 rupee notes is Rs 5x.

∴ Value of (32 – x) number of 10 rupee notes is Rs. 10(32 – x).

So I have Rs. 5x and Rs. 10(32 – x) and my total money is Rs. 200.

’ We have, 5x + 10 (32-x) = 200

or, 5x + 320 – 10x = 200

or, 5x – 10x = 200-320

or, -5x = -120 -120

or, x = \(\frac{-120}{-5}\)

∴  x = 24

So I have 24 number of 5 rupee notes and 32-24 = 8 number of 10 rupee notes.

Question 2. If I change 3 hundred rupee notes in the form of 5 and 10 rupee notes then I get 48 notes in total. Let’s find out how many notes of each type I would have.
Solution:

Given

If I change 3 hundred rupee notes in the form of 5 and 10 rupee notes then I get 48 notes in total.

Let I have x notes of Rs. 5.

∴ I have 10 number of (48-x) notes.

Value of 5 notes of Rs. x is Rs. 5x

Value of 10 notes of Rs. (48-x) is Rs. 10 (48-x)

B.T.P.,

5x + 10 (48 – x) = 300

or, 5x + 480 – 10x = 300

or, -5x = 300 – 480

or, -5x = -180 .

or, x = \(\frac{-180}{-5}\)

or, x = 36

∴ I have 36 notes of Rs. 5 and (48-36) = 12 notes of 10.

Question 3. There are 35 students in Dhruba’s class. Their average age is 14 years. New 7 more students got admission and then the average age becomes is 13.9 years. Let’s find the equation and calculate the average age of the seven new students of Dhruba’s class.
Solution:

Given

There are 35 students in Dhruba’s class. Their average age is 14 years. New 7 more students got admission and then the average age becomes is 13.9 years.

Let the average age of the 7 new students is x years.

∴ Total age of the 7 new students is (7xx) years = 7x years.

Total age of the previous 35 students was 35×14 years = 490 years.

Now the total number of students is Dhruba’s class is (35+7)

= 42

∴ Total age of these 42 students is = (7x + 490) years.

Average age of these 42 students is = 13.9 years.

∴ Total age of these 42 students is = (13.9 x 42) years.

B.T.P., 7x +490 = 13.9×42

or, 7x = 13.9 × 42 – 490

or, x = \(\frac{13.9 \times 42-490}{7}\)

∴ x = 13.4

∴The average age of the 7 new students is h 3.4l years.

Question 4. Manash has written a fraction whose denominator is 1 more than twice of the numerator. If we add 4 to the numerator and the denominator, then the fraction will be \(\frac{7}{11}\) Let’s form an equation and find out the fraction written by Manash.

Solution: 

Manash has written a fraction whose denominator is 1 more than twice of the numerator. If we add 4 to the numerator and the denominator, then the fraction will be \(\frac{7}{11}\)

Let the numerator of the fraction = x

Hence the denominator = 2x + 1

∴ The fraction = \(\frac{x}{2 x+1}\)

Let’s add 4 to both the numerator and the denominator of the fraction and let’s see what we get,

= \(\frac{x+4}{2 x+1+4}=\frac{x+4}{2 x+1+4}=\frac{x+4}{2 x+5}\)

B.T.P.,

= \(\frac{x+4}{2 x+5}=\frac{7}{11}\)

or, 14x+35 = 11x+44

or, 14x-11x = 44-35

or, 3x = 9

∴ x = 3

The required fraction  = \(\frac{x}{2 x+1}=\frac{3}{2 x 3+1}=\frac{3}{7}\)

Question 5. Ashoke has written a fraction whose numerator is 2 less than the denominator. If we add 1 to the numerator and to the denominator then the fraction will be \(\frac{4}{5}\); let’s form an equation and hence find the fraction written by Ashoke.

Solution:

Given

Ashoke has written a fraction whose numerator is 2 less than the denominator. If we add 1 to the numerator and to the denominator then the fraction will be \(\frac{4}{5}\);

Let denominator is x.

∴ Number of fraction = x-2

∴ Fractions = \(\frac{x-2}{x}\)

B.T.P.,

= \(\frac{x-2+1}{x+1}=\frac{4}{5}\)

or,  \(\frac{x-1}{x+1}=\frac{4}{5}\)

or, 5x – 5 = 4x + 4

or, 5x – 4x = 4 + 5

or, x = 9

∴ Denominator = 9

∴ Numerator = 9-2 = 7

∴ Fractions = \(\frac{7}{9}\)

Question 6. Murad has written a two digit number where the sum of the two digits is 11; if we add 63 to the number then the digits change their positions. Let’s form an equation and hence try to find the two digit number written by Murad.

Given

Murad has written a two digit number where the sum of the two digits is 11; if we add 63 to the number then the digits change their positions.

Let the digit in the unit’s position of the two digit number is x.

∴ Digit in tenth place = (11 — x)

∴ Two digit number = 10x number in tenth place + number in unit place

= 10 × (11-x) + x

= 110 – 10x + x

Two digit number = 110- 9x

If we interchange the digits of the two digit number, then the number becomes

= 10 × x+11-x

= 10x +11-x

= 9x + 11

B.T.P.,

= 110-9x+63 = 9x+11

or, -9x-9x = 11-110-63

or, -18x = -162

or, x = \(\frac{-162}{-18}\) = 9

∴ x = 9

∴ Digit in unit place = 9

∴ Digit in tenth place = 11 -9 = 2

∴ Two digit number = 10×2 + 9

= 20 + 9

Two digit number = 29

Question 7. Half of a number is 6 more than \(\frac{1}{5}\) part of that number. Let’s form an equation and hence find the number.
Solution:

Given

Half of a number is 6 more than \(\frac{1}{5}\) part of that number.

Let the number be x.

∴ Half of the number = \(\frac{x}{2}\)

Half of the number is 6 more than of that \(\frac{1}{5}\) number.

Therefore, \(\frac{x}{2}-\frac{x}{5}\) = 6

or, \(\frac{5 x-2 x}{10}c\) = 6

or, \(\frac{3 x}{10}\) = 6

∴ x = 20

The required number is 20

Formation Of An Equation Exercise 19.1

Let’s form equation and work out.

Question 1. 2 more than twice the number that Sima has written is equal to 5 less than thrice the number she has written. Find the number she has written.

Solution:

Given

2 more than twice the number that Sima has written is equal to 5 less than thrice the number she has written.

Let Sima has written number x.

∴ Twice of the numbers = 2x

∴ Thrice of the number = 3x

B.T.P., required equation –

2x + 2 = 3x – 5

or, 2x – 3x = – 5 – 2

or, -x = -7

or, x = 7

∴ Number written by Sima = 7

Equation = 2x + 2 = 3x – 5, Number = 7

Question 2. Let’s write down three consecutive integers such that 5 less than the sum of the numbers is equal to 11 more than twice the second number. Let’s find the three consecutive integers.

Solution:

Let three consecutive numbers be x, x + 1 and x + 2.

Twice of the second number = 2 (x + 1).

B.T.P., required equation –

x + (x + 1) + (x + 2)-5 = 2 (x + 1)+11

or, x + x+1+x + 2- 5 = 2x + 2+11

or, 3x – 2x = 2 + 11 + 5 – 1-2

or, x = 15

∴ First number = 15

Second number = 15 + 1 = 16

Third number = 15 + 2 = 17

Required equation : x + x+1 + x + 2- 5 = 2 (x+1) +11

Three consecutive numbers are 15, 16 and 17.

Question 3. Let’s find a number such that one-fourth of the number is 1 less than one-third of the number.

Solution:

Let the number is = x

= \(\frac{1}{3}\) rd of the number = \(\frac{x}{3}\)

= \(\frac{1}{4}\) th of the number =\(\frac{x}{4}\)

B.T.P.,

= \(\frac{x}{3}-\frac{x}{4}=1\)

or, \(\frac{4 x-3 x}{12}=1\)

or, x = 12

Required equation : \(\frac{x}{3}-\frac{x}{4}=1\)

Required number = 12

Question 4. Let’s find a fraction whose denominator is 2 more than the numerator; and when 3 is added to the numerator and 3 is subtracted from the denominator, the fraction becomes equal to \(\frac{7}{3}\)

Solution.

Let the numerator of the fraction is x.

∴ The denominator of fraction = x + 2.

∴ Fraction = \(\frac{x}{x+2}\)

B.T.P., required equation-

= \(\frac{x+3}{x+2-3}=\frac{7}{3}\)

= \(\frac{x+3}{x-1}=\frac{7}{3}\)

or, 7x-7 = 3x+9

or, 7x-3x = 9+7

or, 4x = 16

or, x = \(\frac{16}{4}\) = 4

Required equation = \(\frac{x+3}{x+2-3}=\frac{7}{3}\) and fraction = \(\frac{4}{6}\)

Question 5. Sucheta wrote a fraction whose denominator is 3 more than the numerator. Again, she added 2 to the numerator and subtracted 1 from the denominator and also subtracted 1 from the numerator and added 2 to the denominator to get 2 fractions whose product is \(\frac{2}{5}\). Let’s write the fraction written by Sucheta.

Solution:

Given

Sucheta wrote a fraction whose denominator is 3 more than the numerator. Again, she added 2 to the numerator and subtracted 1 from the denominator and also subtracted 1 from the numerator and added 2 to the denominator to get 2 fractions whose product is \(\frac{2}{5}\).

Let the numerator of fraction is x.

∴ Denominator = x + 3

∴ Fraction = \(\)

B.T.P., required equation

= \(\frac{x+2}{x+3-1} \times \frac{x-1}{x+3+2}=\frac{2}{5}\)

= \(\frac{x+2}{x+2} \times \frac{x-1}{x+5}=\frac{2}{5}\)

= \(\frac{x-1}{x+5}=\frac{2}{5}\)

or, 5x – 5 = 2x + 10

or, 5x – 2x = 10 + 5

or, 3x = 15

or X= \(\frac{15}{3}\) = 5

∴ Numerator of the fraction = 5 and denominator =5 + 3 = 8

∴ Required fraction = \(\frac{5}{8}\)

Required equation :  \(\frac{x+2}{x+3-1} \times \frac{x-1}{x+3+2}=\frac{2}{5}\)

Fraction written by Sucheta = \(\frac{5}{8}\)

Question 6. Raju wrote a number having two digits where the digit in the tenth position is thrice the digit in the unit position and the new number obtained by reversing the positions of the digits is 36 less than the initial number. Let’s find the number written by Raju.

Solution:

Given

Raju wrote a number having two digits where the digit in the tenth position is thrice the digit in the unit position and the new number obtained by reversing the positions of the digits is 36 less than the initial number.

Let the digit in the unit’s place of the two digit number is x.

∴ Digit in the tenth’s place = 3x

∴ Two digit number = 10x digit in the tenth’s place + digit in the unit’s place.

= 10 × 3x + x

= 30x + x

Number got by interchanging the digits of the two digit number-

= 10 × x + 3x

= 10x + 3x

B.T.P., required equation

30x + x – 36 = 10x + 3x

or, 31 x – 13x = 36

or, 18x = 36

or, x = \(\frac{36}{18}\) = 2

∴ Digit in the units place of the two digit number = 2 and digit in the tenths place = 3×2 = 6

Required number = 10×6 + 2 = 62

Required equation : 30x + x – 36 = 10x +3x

Required number = 62

The number written by Raju = 62

Question 7. If the sum of two numbers is 89 and their difference is 15, let’s find the two numbers.

Solution:

Given

If the sum of two numbers is 89 and their difference is 15,

Let the greater number is x.

∴ The smaller number = 89 – x

B.T.P., required equation –

= x – (89 – x) = 15

or, x – 89 + x = 15

or, 2x = 15 + 89

or, 2x = 104

or, x  = \(\frac{104}{2}\)= 52

∴ Greater number = 52, smaller number = 89 – 52 = 37

Required equation : x – (89-x) = 15

Required number : 52 and 37.

Question 8. Divide 830 into two parts in such a way that 30% of one part will be 4 more than 40% of the other.

Solution:

Let one part is x.

∴ Second part = (830 – x)

30% of one part = 30% of x

= \(\frac{x \times 30}{100}=\frac{3 x}{10}\)

40% of second part = 40% of (830 – x)

= \(\frac{(830-x) \times 40}{100}=\frac{2(830-x)}{5}\)

B.T.P., required equation-

= \(\frac{x \times 30}{100}=\frac{(830-x) \times 40}{100}+4\)

= \(\frac{3 x}{10}=\frac{2(830-x)}{5}+4\)

= \( \frac{3 x}{10}=\frac{1680-2 x+20}{5}\)

= \(\frac{3 x}{2}=\frac{1660-2 x}{1}\)

= \(\frac{3 x}{2}=\frac{1660-2 x}{1}\)

or, 3x = 3360-4x

or, 3x+4x = 3360

or, x = \(\frac{3360}{7}\) = 480

∴ One part = 480

Second part = 830 – 480 = 350

Required equation: = \(\frac{x \times 30}{100}=\frac{(830-x) \times 40}{100}+4\)

Two parts are respectively 480 and 350.

Question 9. Divide 56 into two parts so that thrice of the first part becomes 48 more than one third of the second part.

Solution:

Let the first part is x.

∴ Second part = 56 – x

Thrice of the first part = 3x

= \(\frac{1}{3}\) rd of the second part = (56 – x)

∴ B.T.P., required equation –

3x = \(\frac{56-x}{3}\) + 48

or, 9x = 56 – x + 48 x 3

or, 9x + x = 56 + 144

or, 10x = 200

or x= \(\frac{200}{20}\) =20

First part is 20 and the second part is 56-20 = 36 ,

Required equation : 3x = \(\frac{56-x}{3}\)+ 48

Required numbers are 20 and 36.

Question 10. \(\frac{1}{5}\) part of a pole lies in mud, \(\frac{3}{5}\) part lies in water and the remaining 5 metres lies above the water. Let’s find the length of the pole and write it.

Solution:

Given

\(\frac{1}{5}\) part of a pole lies in mud, \(\frac{3}{5}\) part lies in water and the remaining 5 metres lies above the water.

Let the length of the pole is x meters.

∴ Part of pole in mud = \(\frac{1}{5}\) part x m of = \(\frac{x}{5}\)

∴ Part of pole in water = \(\frac{3}{5}\) part x m of = \(\frac{3x}{5}\)m

B.T.P., required equation-

= \(x-\left(\frac{x}{5}+\frac{3 x}{5}\right)=5\) = 5

= \(x-\frac{4 x}{5}\) = 5

= \(\frac{5 x-4 x}{5}\) = 5

or, x = 25

∴ length of the pole = 25 meters

B.T.P., required equation : \(x-\frac{4 x}{5}\) = 5

Length of the pole = 25 m

Question 11. At present, my father’s age is 7 times of my age. After 10 years my father’s age will be 3 times of my age. Let’s find and write the present age of my father and me.

Solution:

Given

At present, my father’s age is 7 times of my age. After 10 years my father’s age will be 3 times of my age.

Let my present age is x years.

∴ Present age of my father = 7x years.

10 years later my age – (x + 10) years.

10 years later father’s age = (7x + 10) years.

B.T.P., required equation-

7x + 10 = 3 (x + 10)

or, 7x + 10 = 3x + 30

or, 7x – 3x = 30 – 10

or, 4x = 20

or, x = \(\frac{20}{4}\) = 5

My present age = 5 years

Present age of my father = 7 × 5 years = 35 years

Required equation : 7x + 10 = 3 (x + 10)

My present age is 5 years and the present age of my father is 35 years.

Question 12. My uncle changed a 1000 rupee cheque from a bank in the form of 5 rupee and 10 rupee notes, if he received 137 notes in total, then find the number of 5 rupee notes that he got.

Solution:

Given

My uncle changed a 1000 rupee cheque from a bank in the form of 5 rupee and 10 rupee notes, if he received 137 notes in total

Let uncle got 10 number of Rs. x notes.

∴ (137 – x) number of Rs. 5 notes got.

B.T.P., required equation –

10x + 5 (137-x) = 1000

or, 10x + 685 – 5x = 1000

or, 5x = 1000 – 685

or, 5x = 315

or, x = \(\frac{315}{5}\) = 63

∴ Number of 5 rupee notes = (137-x) = 7 (137 – 63) = 74

Required equation : 10x + 5 (137 – x) = 100

Number of Rs. 5 notes = 74

Question 13. Salem uncle of our village used half of his savings to buy a house after his retirement from a government job. One day after falling into trouble, he sold his house and got 5% more than its cost price. If he would have taken 3450 rupees more than he would get 8% more than his cost price. Let’s find the amount of money Salem uncle used to buy his house, and his entire savings.

Solution:

Given

Salem uncle of our village used half of his savings to buy a house after his retirement from a government job. One day after falling into trouble, he sold his house and got 5% more than its cost price. If he would have taken 3450 rupees more than he would get 8% more than his cost price.

Let the total savings of Salem uncle be Rs. x

Expenditure on buying house = \(\frac{1}{2}\) part of Rs. x Rs. = Rs. \(\frac{x}{2}\)

5% of the cost price of house = 5% of Rs. \(\frac{x}{2}\) = Rs. \(\frac{x}{2} \times \frac{5}{100}\) x

8% of the cost price of house = 8% of Rs. \(\frac{x}{2}\) = Rs. \(\frac{x}{2} \times \frac{8}{100}\)

B.T.P., required equation –

= \(\frac{x}{2} \times \frac{5}{100}+3450=\frac{x}{2} \times \frac{8}{100}\)

or, \(\frac{x}{40}+3450=\frac{x}{25}\)

or, \(\frac{x}{40}-\frac{x}{25}=-3450\)

or, \(\frac{5 x-8 x}{200}=-3450\)

or, \(\frac{-3 x}{200}=-3450\)

or, \(x=\frac{3450 \times 200}{2}\)

or, x = 230000

Cost price of the house = = \(\frac{230000}{2}\) = Rs.115000

Amount of the savings of Salem Uncle = Rs. 230000

Required equation : – = \(\frac{x}{2} \times \frac{5}{100}+3450=\frac{x}{2} \times \frac{8}{100}\)

Salem uncle bought house of Rs. 115000 and his total savings was Rs. 230000.

Question 14. There was food provision for twenty days in a refugee camp of the village Gopalpur. After 7 days, 100 more refugees joined the refugee camp and they consumed the food grains within 11 days. Let’s write the number of refugees that were present initially.

Solution.

Given

There was food provision for twenty days in a refugee camp of the village Gopalpur. After 7 days, 100 more refugees joined the refugee camp and they consumed the food grains within 11 days.

Let there were x number of people in the refugee camp.

∴ For x people 20 days’ food was there. After 7 days for x people (20-7) or 13 days’ food was there.

7 days later number of people = (x + 100)

∴ (x + 100) people will eat x people’s food for 13 days.

B.T.P., required equation

= \(\frac{(20-7) x}{x+100}=11\)

or, \(\frac{13 x}{x+100}=11\)

or, 13x = 11x+1100

or, 13x-11x=1100

or, 2x = 1100

or, x = \(\frac{1100}{2}\) = 550

∴ Earlier there were 550 number of people in the refugee camp.

Requied equation \(\frac{(20-7) x}{x+100}=11\) and 550 number of people.

Question 15. Let’s find the roots of the equations :

1. \(\frac{5}{3 x+4}=\frac{4}{5(x-3)}\)

Solution:

or, 25x-75 = 12x+16

or, 25x-12x = 16+75

or, 13x = 91

or, x = \(\frac{91}{13}\) = 7

or, x = 7

2. 14(x-2)+3(x+5) = (x+8)+5

Solution:

or, 25x-75 = 12x+16

or, 25x-12 = 16+75

or, 13x = 91

or, x = \(\frac{42}{14}\) = 3

∴ x = 3

3. \(\frac{x}{2}+5=\frac{x}{3}+7\)

Solution:

= \(\frac{x}{2}+5=\frac{x}{3}+7\)

= \(\frac{x}{2}-\frac{x}{3}=7-5\)

or, \(\frac{3 x-2 x}{6}=2\)

or, x = 12

4. \(\frac{x+1}{8}+\frac{x-2}{5}=\frac{x+3}{10}+\frac{3 x-1}{20}\)

Solution:

or, \(\frac{5(x+1)+8(x-2)}{40}=\frac{2(x+3)+3 x-1}{-1^{20}}\)

or, \(\frac{5 x+5+8 x-16}{2}=\frac{2 x+6+3 x-1}{1}\)

or, \(\frac{13 x-11}{2}=\frac{5 x+5}{1}\)

or, 13x-11 = 10x+10

or, 13x-10x = 10+11

or, 3x = 21

or, x = \(\frac{21}{3}\) = 7

∴ x = 7

5. \(\frac{x+1}{4}+3=\frac{2 x+4}{5}+2\)

Solution:

or, \(\frac{x+1+12}{4}=\frac{2 x+4+10}{5}\)

or, \(\frac{x+13}{4}=\frac{2 x+14}{5}\)

or, 16x+2 = 3x+80

or, 16x-3x = 80-2

or, 13x = 78

or, x = \(\frac{78}{13}\) = 6

∴ x = 6

6. \(\frac{3}{5}(x-4)-\frac{1}{3}(2 x-9)=\frac{1}{4}(x-1)-2\)

Solution:

or, \(\frac{3(x-4)}{5}-\frac{(2 x-9)}{3}=\frac{(x-1)}{4}-2\)

or, \(\frac{9 x-36-10 x+45}{15}=\frac{x-1-8}{4}\)

or, \(\frac{9-x}{15}=\frac{x-9}{4}\)

or, 15x-135 = 36-4x

or, 15x+4x = 36+135

or, 19x = 171

or, x = \(\frac{171}{19}\) = 9

∴ x = 9

7. \(\frac{x+5}{3}+\frac{2 x-1}{7}=4\)

Solution:

or, \(\frac{7 x+35+6 x-3}{21}=4\)

or, 13x+32 = 84

or, 13x = 84-32

or, 13x = 52

or, x = \(\frac{52}{13}\) = 4

∴ x = 4

8. 25 + 3(4x-5) +8(x+2) = x+3

Solution:

or, 25 + 12x-15 + 8x+16 = x + 3

or, 12x + 8x-x = 3 + 15-25-16

or, 20x =18-41

or, 19x = -23

or, \(x=\frac{-23}{19}=-1 \frac{4}{19}\)

∴  x = \(-1 \frac{4}{19}\)

9. \(\frac{x-8}{3}+\frac{2 x+2}{12}+\frac{2 x-1}{18}=3\)

Solution:

or, \(\frac{12(x-8)+3(2 x+2)+2(2 x-1)}{36}=3\)

or, 12x – 96 + 6x + 6 + 4x – 2 = 108

or, 22x = 108+ 96+ 2-6

or, 22x = 206 – 6

or, 22x = 200

or, \(x=\frac{200}{22}=\frac{100}{11}\)

∴ \(x=9 \frac{1}{11}\)

10. \(\frac{t+12}{6}-t=6 \frac{1}{2}-\frac{1}{12}\)

Solution:

or, \(\frac{t+12-6 t}{6}=\frac{13}{2}-\frac{1}{12}\)

or, \(\frac{12-5 t}{6_1}=\frac{78-1}{72_2}\)

or, \(\frac{12-5 t}{1}=\frac{77}{2}\)

or, 24 -10t = 77

or, -10t = 77 – 24

or, -10t = 53

or, t = \(-\frac{53}{10}\)

or, t = \(-5\frac{3}{10}\)

11. \(\frac{x+1}{2}-\frac{5 x+9}{28}=\frac{x+6}{21}+5-\frac{x-12}{3}\)

Solution:

or, \(\frac{14 x+14-5 x-9}{28}=\frac{x+6+105-7 x+84}{21}\)

or, \(\frac{14 x+14-5 x-9}{-28}=\frac{x+6+105-7 x+84}{-243}\)

or, \(\frac{9 x+5}{4}=\frac{195-6 x}{3}\)

or, 27x + 15 = 780 – 24x

or, 27x + 24x = 780 – 15

or, 51x = 765

or, x = \(\frac{765}{51}\) = 15

∴ x = 15

12. \(\frac{9 x+5}{14}+\frac{8 x-7}{7}=\frac{18 x+11}{28}+\frac{5}{4}\)

Solution:

or, \(\frac{9 x+5+16 x-14}{14}=\frac{18 x+11+35}{28}\)

or, \(\frac{25 x-9}{1}=\frac{18 x+46}{2}\)

or, 50x-18 = 18x+46

or, 50x-18x = 46+18

or, 32x = 64

or, x = \(\frac{64}{32}\) = 2

or, x = 2

13. \(\frac{3 y+1}{16}+\frac{2 x-3}{7}=\frac{y+3}{8}+\frac{3 y-1}{14}\)

Solution:

or, \(\frac{3 y+1}{16}=\frac{y+3}{8}=\frac{y+3}{8}+\frac{3 y-1}{14}\)

or, \(\frac{3 y+1-2 y-6}{16}=\frac{3 y-1-4 y+6}{14}\)

or, \(\frac{y-5}{8}=\frac{5-y}{7}\)

or, 7y – 35 = 40 + 8y

or, 7y + 8y = 40 + 35

or, 15y = 75

or, y = \(\frac{75}{15}\) = 5

∴ y = 5

14. 5x -(4x-7)(3x-5) = 6-3(4x-9)(x-1)

Solution:

or, 5x-12 x²+ 20x + 21 x – 35 = 6-3 (4x²-4x – 9x + 9)

or, -12x² + 46x – 35 = 6 – 12x² + 12x + 27x – 27

or, -12x² + 12x² + 46x – 12x – 27x = 6 + 35-27

or, 46x – 39x = 41-27

or, 7x = 14

or, x =\(\frac{14}{7}\)= 2

∴ x = 2

15. 3(x-4)² + 5(x-3)²=(2x-5) (4x-1)-40

Solution:

or, 3 (x² – 8x + 16) +5 (x² – 6x + 9) = 8x² – 2x – 20x + 5-40

or, 3x² – 24x + 48 + 5x² – 30x + 45 = 8x² – 22x – 35

or, 8x² – 8x² – 24x – 30x + 22x =-35 – 48 – 45

or, -32x = – 128

or x = \(\frac{-128}{-32}\) =4

∴ x = 4

16. 3(y-5)²+5y = (2y-3)²-(y+1)² + 1

Solution:

or, 3 (y-5)² + 5y = (2y-3)²– (y+1 )²+1

or, 3(y²-10y+25) + 5y = 4y²– 12y + 9 – (y² + 2y+ 1) + 1

or, 3y² – 30y + 75 + 5y = 4y² – 12y+ 9- y²– 2y – 1 + 1

or, 3y²+y²-4y²-30y + 12y + 2y = 9 – 75

or, -11 y = -66

or, y = \(\frac{-66}{-11}\) = 6

Question 17. Let’s write mathematical stories and form equations.

Solution:

x = 5 → 2x-10 = 0

y = -11 → 5y+55 = 0

t = \(\frac{7}{8}\) → 8t-7 = 0

x = 24 → 3x-72 = 0

x = 7 → 7x-49 = 0