NCERT Solutions of Chapter: Pair Of Linear Equations In Two Variables. NCERT Solutions along with worksheets and notes for Class 10.
Exercise 3.1
(1) Aftab tells his daughter” Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be” (Isn’t this interesting?) Represent this situation algebraically and graphically.
Ans-
Let age of Aftab now = x years
Age of Aftab’s daughter now = y years
Age of Aftab’s seven years ago = x – 7 years
Age of Aftab’s daughter seven years ago = y – 7 years
Algebraic equations of their ages before seven years
x-7 = 7(y-7)
x – 7 = 7y – 49
x – 7y = -49 + 7
x – 7y = -42…(1)
Age of Aftab’s three years after = x + 3 years
Age of Aftab’s daughter after three years = y +3 years
Algebraic equations of their ages after three years
x + 3 = 3(y + 3)
x + 3 = 3y + 9
x – 3y = 6 …(2)
Algabric equations are x – 7y = -42…(1)
x – 3y = 6 …(2)
To obtain the equivalent geometric representation, we find two solutions of each equation.
Solutions for equation (1), ie, x – 7y = -42
| X | 0 | -42 |
| y | 6 | 0 |
Solutions for equation 2, ie, x – 3y = 6
| x | 0 | 6 |
| y | -2 | 0 |
Plot these points in a graph. These straight lines are coincide to each other at one point.
(2) The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 3 more balls of the same kind for Rs 1300. Represent this situation algebraically and geometrically.
Ans-
Let price of one bat = Rs x
Price of one ball = Rs y
Algebraically representation of equation-
Cond 1 -The coach of a cricket team buys 3 bats and 6 balls for Rs 3900.
3x + 6y = Rs 3900
x + 2y = 1300…(1)
Cond 2 -She buys another bat and 3 more balls of the same kind for Rs 1300.
x + 3y = Rs 1300…(2)
To obtain the equivalent geometric representation, we find two solutions of each equation.
Solutions for equation (1), ie, x + 2y = Rs 1300
| X | 0 | 1300 |
| y | 650 | 0 |
Solutions for equation (2), ie, x + 3y = Rs 1300
| X | 0 | 1300 |
| y | 433.33 | 0 |
Plot these points in a graph. These straight lines are coincide to each other at one point.
(3) The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situation algebraically and geometrically.
Ans-
Let cost of 1kf apple = Rs x
Cost of 1 kg of grapes = Rs y
Algebraically representation of equation-
The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs 160.
2x + y = Rs 160…(1)
After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300.
4x + 2y = Rs 300…(2)
To obtain the equivalent geometric representation, we find two solutions of each equation.
Solutions for equation (1), ie, 2x + y = Rs 160
| X | 0 | 80 |
| y | 160 | 0 |
Solutions for equation (2), ie, 4x + 2y = Rs 300
| X | 0 | 75 |
| y | 150 | 0 |
Plot these points in a graph. These straight lines are coincide to each other at one point.
Exercise 3.2
(1) From the pair of linear equation in the following problems, and find their solutions graphically.
(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the umber of boys and girls who took part in the quiz.
(ii) 5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens together cost Rs 46. Find the cost of one pencil and that of one pen.
Ans-
(i) Let number of girls = x
Number of boys = y
Algebraically representation of equation-
x + y = 10…(1)
x + 4 = y
x – y = -4..(2)
To obtain the equivalent geometric representation, we find two solutions of each equation.
Solutions for equation (1), ie, x + y = 10
| x | 0 | 10 |
| y | 10 | 0 |
Solutions for equation (2), ie, x – y = -4
| x | 0 | -4 |
| y | 4 | 0 |
Plot these points in a graph. These straight lines are coincide to each other at one point.
Therefore, number of girls = 7 and number of boys = 3.
(ii) Let cost of one pencil = Rs x
Cost of one pen = Rs y
Algebraically representation of equation-
5x + 7y = 50…(1)
7x + 5y = 46…(2)
To obtain the equivalent geometric representation, we find two solutions of each equation.
Solutions for equation (1), ie, 5x + 7y = 50
| x | 0 | 10 |
| y | 7.1 | 0 |
Solutions for equation (2), ie, 7x + 5y = 46
| x | 0 | 6.5 |
| y | 9.2 | 0 |
Plot these points in a graph. These straight lines are coincide to each other at one point.
Therefore, cost of one pencil is Rs 3 and cost of one pen is Rs 5.
Ans-
(i) 3x + 2y = 5;
2x – 3y = 7
3x + 2y – 5 = 0;
2x – 3y – 7 = 0
From equations
a1 = 3, b1 = 2 and c1 = -5
a2 = 2, b2 = -3 and c2 = -7
(ii) 2x – 3y = 8; 4x – 6y = 9
2x – 3y = 8;
4x – 6y = 9
2x – 3y – 8 = 0;
4x – 6y – 9 = 0
From equations
a1 = 2, b1 = -3 and c1 = -8
a2 = 4, b2 = -6 and c2 = -9
Therefore, lines are parallels. So they have no common solutions. Hence, pair of lines are inconsistent.
Therefore, lines will intersect each other and has one common solution. Hence, lines are consistent.
(iv) 5x – 3y = 11; -10x + 6y = -22
5x – 3y = 11;
-10x + 6y = -22
5x – 3y – 11 = 0
-10x + 6y + 22 = 0
From equations
a1 = 5, b1 = -3 and c1 = -11
a2 = -10, b2 = 6 and c2 = 22
Therefore, lines are coincident and has common solutions. Hence, pair of lines are consistent.
(v) x + y = 8; 2x + 3y = 12
x + y = 8;
2x + 3y = 12
x + y – 8 = 0
2x + 3y -12 = 0
From equations
a1 = , b1 = 2 and c1 = -8
a2 = 2, b2 = 3 and c2 = -12
Therefore, lines are coincident and has common solutions. Hence, pair of lines are consistent.
(4) Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:
(i) x + y = 5, 2x + 2y = 10
(ii) x – y = 8, 3x – 3y = 16
(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0
Ans-
(i) x + y = 5, 2x + 2y = 10
x + y – 5 = 0
2x + 2y -10 = 0
From equations
a1 = 1, b1 = 1 and c1 = -5
a2 = 2, b2 = 2 and c2 = -10
Therefore, lines are coincident and has common solutions. Hence, pair of lines are consistent.
Solutions for equation (1), ie, x + y – 5 = 0
| x | 0 | 5 |
| y | 5 | 0 |
Solutions for equation (2), ie, 2x + 2y -10 = 0
| x | 0 | 5 |
| y | 5 | 0 |
Lines AB and CD coincides. Therefore, equations have infinitely many solutions. So, both lines are consistent.
(ii) x – y = 8, 3x – 3y = 16
x – y – 8 = 0
3x – 3y – 16 = 0
From equations
a1 = 1, b1 = -1 and c1 = -8
a2 = 3, b2 = -3 and c2 = -16
Therefore, lines are parallel and have no common solution. Hence, these lines are inconsistent.
Hence, pair of linear equations do not have any graphical solution.
(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
2x + y – 6 = 0
4x – 2y – 4 = 0
From equations
a1 = 2, b1 = 1 and c1 = -6
a2 = 4, b2 = -2 and c2 = -4
Therefore, lines will intersect and have one common solution. Hence pair of linear equations are consistent.
Solutions for equation (1), ie, 2x + y – 6 = 0
| x | 0 | 3 |
| y | 6 | 0 |
Solutions for equation (2), ie, 4x – 2y – 4 = 0
| x | 0 | 1 |
| y | -2 | 0 |
Therefore, lines will intersect and have one common solution. Hence pair of linear equations are consistent.
Hence, lines have a unique solution, ie, x = 2 and y = 2.
(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0
2x – 2y – 2 = 0,
4x – 4y – 5 = 0
From equations
a1 = 2, b1 = -2 and c1 = -2
a2 = 4, b2 = -4 and c2 = -5
Therefore, lines are parallels. So they have no common solutions. Hence, pair of lines are inconsistent.
(5) Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
Ans-
Let width of the rectangle = x m
Let length of the rectangle = y m
Perimeter of the rectangle
2(x + y) = 36(2)
x + y = 36 …(1)
y = 4 + x …(2)
Solutions for equation (1), ie, x + y = 36
| x | 0 | 36 |
| y | 36 | 0 |
Solutions for equation (1), ie, – x + y = 4
| x | 0 | -4 |
| y | 4 | 0 |
(6) Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines
Ans- Given the linear equation 2x + 3y – 8 = 0
(i) Intersecting lines
New linear equation 3x – 4y – 7 = 0
From equations
a1 = 2, b1 = 3 and c1 = -8
a2 = 3, b2 = -4 and c2 = -7
Therefore, equation is correct.
(ii) Parallel lines
Given the linear equation 2x + 3y – 8 = 0
New equation is 4x + 6y + 9 = 0
From equations
a1 = 2, b1 = 3 and c1 = -8
a2 = 4, b2 = 6 and c2 = 9
Therefore, equation is correct.
(iii) Coincident lines
Given the linear equation 2x + 3y – 8 = 0
New equation is 4x + 6y – 16 = 0
From equations
a1 = 2, b1 = 3 and c1 = -8
a2 = 4, b2 = 6 and c2 = -16
(7) Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0.Determine the coordinates of the vertices of the triangle formed by these lines and the x – axis, and shade the triangular region.
Ans-
Coordinates of the linear equation x – y + 1 = 0
| x | 0 | -1 | 1 | 2 |
| y | 1 | 0 | 2 | 3 |
Coordinates of the linear equation 3x + 2y – 12 = 0
| x | 0 | 4 | 2 | -2 |
| y | 6 | 0 | 3 | 9 |
Hence, BDF is required triangle. Coordinates of triangle are B(-1, 0), D(2, 3) and F(4, 0).
Ans-
(i) x + y = 14
x – y = 4
x + y = 14…(1)
x – y = 4 …(2)
x = 4 +y…(3) [From equation 2]
Put value of x in equation (1)
x + y = 14
4 + y + y = 14
2y = 10
y = 5
Put value of y in equation (3)
x = 4 + 5
x = 9
Hence value of x is 9 and value of y is 5.
6 + 2t + 3t = 6 6
5t = 36 – 6
5t = 30
t = 6
Put value of t in equation 3
s = 3 + t
s = 3 + 6
s = 9
Hence, value of s is 9 and value of t is 6.
(iii) 3x – y = 3
9x – 3y = 9
3x – y = 3 …(1)
9x – 3y = 9 …(2)
-y = 3 – 3x
y = 3x – 3 ..(3) From equation 1
Put value of y in equation 2
9x – 3(3x – 3) = 9
9x – 9x + 9 = 9
9 = 9
This statement is true for all values of x. We can not get specific values of y and x. This situation has arisen because both the given equations are same. Therefore, Equations (1) and Equation (2) have infinitely many solutions.
(iv) 0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3
0.2x + 0.3y = 1.3…(1)
0.4x + 0.5y = 2.3…(2)
0.2x = 1.3 – 0.3y
2(1.3 – 0.3y) + 0.5y = 2.3
2.6 – .6y + 0.5 y = 2.3
-0.1y = 2.3 – 2.6
= – 0.3
y = 3
Put value of y in equation 3
Hence, value of x is 2 and value of y is 3.
(v) √2 x + √3 y = 0
√3 x – √8 y = 0
√2 x + √3 y = 0…(1)
√3 x – √8y = 0..(2)
√3 x = √8 y
√2 ( √8 y) +√3 (√3 y) = 0
√16y + 3y = 0
4y + 3y = 0
7y = 0
y = 0
Put value of y in equation (2)
√3x – √8y = 0
√3x – √8(0) = 0
√3x – 0 = 0
x = 0
Hence, value of x is 0 and value of y is 0.
(2) Solve 2x + 3y = 11 and 2x – 4y = -24 and hence find the value of ‘m’ for which y = mx + 3.
Ans- 2x + 3y = 11…(1)
2x – 4y = -24…(2)
2x = -24 + 4y…(3)
Put value of x in equation (1)
-24 + 4y + 3y + 11
7y = 11 + 24
7y = 35
y = 5
Put value of y in equation 1
2x + 3(5) = 11
2x + 15 = 11
2x = 11 – 15
2x = – 4
x = -2
To find value of m put value of x and y in y = mx + 3
y = mx + 3
5 = m(-2) + 3
2m = 3 – 5
2m = -2
m = -1
Hence, value of m is – 1.
(3) From the pair of linear equations for the following problems and find their solution by substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.
(iv) The taxi charges in a city consist of a fixed charge together with he charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
(v) A fraction becomes 9/11 , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6 . Find the fraction.
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times of his son. What are their present ages?
Ans-
(i)Let x and y are two numbers.
x – y = 26 (From conditions)…(1)
x = 3y (from conditions)…(2)
Put value of x from equation 2 to equation (1)
x – y = 26
3y – y = 26
2y = 26
y = 13
Put value of y in equation (2)
x = 3(13)
= 39
Hence, value of x is 39 and value of y is 13.
(ii) Let larger angle of supplementary angle = x
Let smaller angle of supplementary angle = y
x = y + 18 (From Conditions)…(1)
x + y = 180 (Sum of supplementary angles)…(2)
Put value of x from equation (1) to equation (2)
(y + 18) + y = 180
2y = 180 – 18
2y = 162
y = 81
Put value of y in equation (1)
x = 81 + 18
x = 99
Hence, measures of angles are 81 and 99 in degrees.
(iii) Let cost of one bat = Rs x
Let cost of the one ball == Rs y
From conditions-
7x + 6y = 3800…(1)
3x + 5y = 1750…(2)
3x = 1750 – 5y
7(1750- 5y) + 3(6y) = 3(3800)
12250 – 35y + 18y = 11400
-17y = 11400 – 12250
-17y = -850
y = 50
Put value of y in equation (1)
7x + 6(50) = 3800
7x + 300 = 3800
7x = 3800 – 300
7x = 3500
x = 500
Hence, cost of one bat is Rs 500 and cost of one ball is Rs 50.
(iv) Let charge per km = Rs x
Let fix charge sof the taxi = Rs y
From conditions-
10x + y = Rs105…(1)
15x + y = Rs 155…(2)
y = 155 – 15x…(3)
Put value of y in equation (1)
10x + (155 – 15x) = 105
10x + 155 – 15x = 105
-5x = 105 – 155
-5x = -50
x = 10
Put value of x in equation 3
y = 155 – 15x
y = 155 – 15(10)
y = 155 – 150
y = 5
Payment of taxi for 25km = Fixed charges + charges for per km
= y + 25x
= 5 + 25(10)
= 5 + 250
= 255 Rs
Hence, fixed charges for taxi is Rs 5 and charges per km is Rs 10.
Total payment for 25 km is Rs 255.
From equation (1)
11(x + 2) = 9(y + 2) [By cross multiplication]
11x + 22 = 9y + 18
11x – 9y = 18 – 22
11x – 9y = -4…(3)
From equation (2)
6(x + 3) = 5(y + 3) [By cross multiplication]
6x + 18 = 5y + 15
6x – 5y = 15 – 18
6x – 5y = -3 …(4)
6x = -3 + 5y
-33 + y = -4(6)
y = -24 + 33
y = 9
Put value of y in equation (3)
11x – 9(9) = -4
11x – 81 = -4
11x = -4 + 81
11x = 77
x = 7
(vi) Let present age of Jacob = x years
Let present age of Jacobs son = y years
From conditions –
x + 5 = 3(y + 5)
x + 5 = 3y + 15
x – 3y = 15 – 5
x – 3y = 10 …(1)
x – 5 = 7(y – 5)
x – 5 = 7y – 35
x – 7y = -35 + 5
x – 7y = – 30 …(2)
x = -30 + 7y
Put value of x in equation (1)
(-30 + 7y) – 3y = 10
-30 + 7y – 3y = 10
4y = 10 + 30
4y = 40
y = 10
Put value of y in equation (1)
x – 3y = 10
x – 3(10) = 10
x – 30 = 10
x = 10 + 30
x = 40
Hence, present age of Jacob is 40 years and present age of Jacob’s son is 10 years.
Exercise 3.4
(1) Solve the following pair of linear equations by the elimination method and the substitution method:
(i) x + y = 5 and 2x – 3y = 4
(ii) 3x + 4y = 10 and 2x – 2y = 2
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
Ans-
(i) x + y = 5 …(1)
2x – 3y = 4 …(2)
Elimination Method –
x + y = 5 …(1)
2x – 3y = 4 …(2)
Multiply equation (1) by 2
2x + 2y = 10 …(3)
Subtract equation (2) from equation (3) to eliminate x
(2x + 2y) – (2x – 3y) = 10 – 4
2x + 2y – 2x + 3y = 6
5y = 6
Substitution method –
x + y = 5 …(1)
2x – 3y = 4 …(2)
From equation (1)
x = 5 – y
Put value of x in equation (2)
2(5 – y) – 3y = 4
10 – 2y – 3y = 4
-5y = 4 – 10
-5y = – 6
(ii) 3x + 4y = 10 …(1)
2x – 2y = 2 …(2)
Elimination Method –
3x + 4y = 10 …(1)
2x – 2y = 2 …(2)
Multiply equation (1) by 2
6x + 8y = 20 …(3)
Multiply equation (2) by 3
6x – 6y = 6 …(4)
Subtract equation (3) from equation (4) to eliminate x
(6x – 6y) – (6x + 8y) = 6 – 20
6x – 6y – 6x – 8y = 6 – 20
-14y = -14
y = 1
Put value of y in equation (1)
2x – 2(1) = 2
2x – 2 = 2
2x = 2 + 2
2x = 4
x = 2
Substitution method –
3x + 4y = 10 …(1)
2x – 2y = 2 …(2)
2x = 2 + 2y
6 + 14y = 2(10)
14y = 20 – 6
14y = 14
y = 1
Put value of y in equation (1)
3x + 4(1) = 10
3x + 4 = 10
3x = 10 – 4
3x = 6
x = 2
Hence, value of x is 2 and value of y is 1.
(iii) 3x – 5y – 4 = 0 …(1)
9x = 2y + 7…(2)
Elimination Method –
3x – 5y – 4 = 0
3x – 5y = 4…(1)
9x = 2y + 7
9x – 2y = 7…(2)
Multiply equation (1) by 2
6x – 10 y = 8…(3)
Multiply equation (2) by 5
45x – 10y = 35…(4)
Subtract equation (3) from equation (4)
(45x – 10y) – (6x – 10 y) = 35 – 8
45x – 10y – 6x + 10y = 27
39x = 27
Substitution method –
3x – 5y – 4 = 0
3x – 5y = 4…(1)
9x = 2y + 7
9x – 2y = 7…(2)
9x = 7 + 2y
2(-5y) = 6(5)
-10y = 30
y = -3
Put value of y in equation (2)
9 + 5y = 6(-1)
5y = -6 – 9
5y = – 15
y = -3
Put value of y in equation (2)
(2) From the pair of linear equations in the following problems, and their solutions (if they exist) by the elimination method:
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes if we only add 1 to the denominator. What is the fraction?
(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
(iii) The sum of the digits of a two – digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
(iv) Meena went to a bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received.
(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Ans-
2x = y + 1 [By cross multiplication]
2x – y =1…(4)
Subtract equation (4) from equation (3)
(x – y) – (2x – y) = -2 -1
x – y – 2x +y = -3
-x = -3
x = 3
Put value of x in equation (3)
x – y = -2
3 – y = -2
-y = -2 -3
-y = -5
y = 5
(ii)Let present age of Nuri = x years
Let present age of Sonu = y years
From conditions-
x -5 = 3(y – 5)
x – 5 = 3y – 15
x – 3y = -15 + 5
x – 3y = -10…(1)
x + 10 = 2(y + 10)
x + 10 = 2y + 20
x – 2y = 20 – 10
x – 2y = 10…(2)
Subtract equation (2) from equation (1)
(x – 3y) –(x – 2y) = -10 – 10
x – 3y – x + 2y = -20
-y = -20
y = 20
Put value of y in equation (1)
x -3y = -10
x – 3(20) = -10
x – 60= -10
x= -10 + 60
x = 50
Hence, age of Nuri is 50 years and age of Sonu is 20 years.
(iii) Let the unit of the number is x.
Let tens unit of number is y
Sum of the digits = x + y
From conditions
x + y = 9…(1)
The original number is 10y + x
The reversal number is y + 10x
According to given second condition-
9(10y + x) = 2(y + 10x)
90y + 9x = 2y + 20x
88y -11x = 0….(2)
From equation (1) x = 9 – y
Put value of x in equation (2)
88y -11(9 – y) = 0
88y – 99 + 11y = 0
99y = 99
y = 1
Put value of y in equation (1)
x = 9 – 1
x = 8
The required number is 10y + x
= 10(1) + 8
= 18
Hence, the required number is 18.
(iv) Let number of Rs 50 = x
Let number of Rs 100 = y
From conditions-
x + y = 25…(1)
50x + 100y = Rs 2000…(2)
Multiply equation (1) by 50
50x + 50y = 50(25)
50x + 50y = 1250…(3)
Subtract equation (3) from equation (2)
(50x + 100y) – (50x + 50y) = 2000 – 1250
50y = 750
y = 15
Put value of y in equation (1)
x + y = 25
x + 15 = 25
x = 25 – 15
x = 10
Hence, 10 notes of Rs 50 and 15 notes of Rs 100 is received by her.
(v) Let fixed charges for first three days for a book = Rs x
Let additional charge for each day = Rs y
From conditions-
Saritha paid for seven days.
She pays fixed charges for first three days and additional charges according to days for last four days.
Therefore,
x + 4y = Rs 27…(1)
Susy paid for five days.
She pays fixed charges for first three days and additional charges according to days for last two days.
x + 2y = Rs 21…(2)
Subtract equation (2) from equation (1)
(x + 4y) – (x + 2y) = 27 – 21
2y = 6
y = 3
Put value of y in eqution (1)
x + 4(3) = 27
x + 12 = 27
x = 27 – 12
x = 15
Hence, fixed charge for three days is Rs 15 and charge for extra day is Rs 3.
Exercise 3.5
(1) Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.
(i) x – 3y – 3 =0
3x – 9y – 2 = 0
(ii) 2x + y = 5
3x + 2y = 8
(iii) 3x – 5y = 20
6x – 10y = 40
(iv) x – 3y – 7 = 0
3x – 3y – 15 = 0
Ans-
(i) x – 3y – 3 =0
3x – 9y – 2 = 0
a1 = 1, b1 = -3, c1 = -3
a2 = 3, b2 = -9, c2 = -2
Hence, there is no solution.
(ii) 2x + y = 5
3x + 2y = 8
2x + y -5 = 0
3x + 2y – 8 = 0
a1 = 2, b1 = 1, c1 = -5
a2 = 3, b2 = 2, c2 = -8
Hence, there is a unique solution
(iii) 3x – 5y = 20
6x – 10y = 40
3x – 5y – 20 = 0
6x – 10y – 40 = 0
a1 = 3, b1 = -5, c1 = -20
a2 = 6, b2 = -10, c2 = -40
Hence, there are infinitely many solutions.
(iv) x – 3y – 7 = 0
3x – 3y – 15 = 0
a1 = 1, b1 = -3, c1 = -7
a2 = 3, b2 = -3, c2 = – 15
2(i) For which values of a and b does the following pair of linear equations have an infinite number of solutions?
2x + 3y = 7
(a – b)x + (a + b)y = 3a + b – 2
(ii) For which value of k will the following pair of linear equations have no solution?
3x + y = 1
(2k – 1)x + (k – 1)y = 2k + 1
Ans-
(i) 2x + 3y = 7
(a – b)x + (a + b)y = 3a + b – 2
2x + 3y – 7 = 0…(1)
(a – b)x + (a + b)y –( 3a + b – 2) = 0…(2)
a1 = 2, b1 = 3, c1 = -7
a2 = (a – b), b2 = (a + b), c2 = – (3a + b -2)
2(a + b) = 3(a – b) [By cross multiplication]
2a + 2b = 3a – 3b
2a – 3a + 2b + 3b = 0
-a + 5b = 0…(1)
a = 5b…(2)
2[- (3a + b -2)] = -7(a – b)
2(-3a – b + 2) = – 7a + 7b
-6a – 2b + 4 = – 7a + 7b
-6a + 7a – 2b – 7b + 4 = 0
a -9b + 4 = 0…(3)
Put value of a in equation (3)
5b – 9b + 4 = 0
-4b = -4
b = 1
Put value of b in equation (2)
a = 5b
a = 5(1)
a = 5
Hence, for value of a is 5 and value of b is 1 pair of linear equations have an infinite number of solutions.
(ii) 3x + y = 1
(2k – 1)x + (k – 1)y = 2k + 1
3x + y – 1 = 0…(1)
(2k – 1)x + (k – 1)y – (2k + 1) = 0…(2)
a1 = 3, b1 = 1, c1 = -1
a2 = (2k – 1), b2 = (k – 1), c2 = – (2k + 1)
3(k – 1) = 2k – 1
3k – 3 = 2k – 1
3k – 2k = -1 + 3
K = 2
Hence, for value of k = 2 following pair of linear equations have no solution.
(3) Solve the following pair of linear equations by the substitution and cross multiplication methods:
8x + 5y = 9
3x + 2y = 4
Ans-
Substitution method-
8x + 5y = 9…(1)
3x + 2y = 4…(2)
3x = 4 – 2y
32 – y = 9(3)
32 – y = 27
-y = 27 – 32
-y = – 5
y = 5
Put value of y in equation (3)
Cross multiplication method-
8x + 5y – 9 = 0…(1)
3x + 2y – 4 = 0…(2)
a1 = 8, b1 = 5, c1 = -9
a2 = 3, b2 =2, c2 = – 4
Hence, value of x is -2 and value of y is 5.
(4) From the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:
(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charges and the cost of food per day.
(ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes ¼ when 8 is added to its denominator. Find the fraction.
(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded, then Yash would have scored 50 marks. How many questions were there in the test?
(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they met in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
(v) The area of rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the legth by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.
Ans-
(i) Let monthly fixed hostel charges = Rs x
Let charges of food depending on number of days = Rs y
From conditions –
Monthly charges Paid by A
A has taken food for 20 days. So he will pay fixed charges + food charges.
x + 20y = Rs 1000…(1)
Monthly charges Paid by B
B has taken food for 26 days. So he will pay fixed charges + food charges.
x + 26y = Rs 1180…(2)
Subtract equation (1) from equation (2)
x + 26y – x – 20y = 1180 – 1000
6y = 180
y = 30
Put value of y in equation (1)
x + 20y = 1000
x + 20(30) = 1000
x = 1000 -600
x = 400
Hence, fixed charges for hostel is Rs 400 and charges for food per day is Rs 30.
4x = y + 8 (By cross multiplication)
4x – y = 8 …(2)
Subtract equation 1 from equation (2)
(4x –y) –(3x – y) = 8 – 3
4x – y – 3x + y = 5
x = 5
Put value of x in equation (1)
3(5) – y = 3
15 – y = 3
-y = 3 – 15
– y = -12
y = 12
Therefore, fraction is 5/12.
Hence, fraction is 5/12.
(iii)Let number of correct question =x
Let number of wrong question = y
From conditions –
3x – y = 40 ….(1)
-y = 40 – 3x
y = -40 + 3x
4x – 2y = 50….(2)
Put value of y in equation (2)
4x – 2(-40 + 3x) = 50
4x + 80 – 6x = 50
-2x = 50 – 80
-2x = – 30
x = 15
Put value of x in equation (1)
3(15) – y = 40
45 – y = 40
-y = 40 – 45
-y = -5
y = 5
Total number of question = x + y
= 15 + 5
= 20
Hence, total number of question are 20.
(iv)
Let the speed of the car starts from A = x km/h
Let the speed of the car starts from B = y km/h
After 5 hours cars are moving in same direction.
Relative speed –
x – y = 100/5
x – y = 20…(1)
When cars are moving in opposite direction-
Relative speed –
x + y = 100/1
x + y = 100….(2)
Add equation (1) and (2)
2x = 120
x = 60 km/h
Put value of x in equation (2)
x + y + 100
60 + y = 100
y = 40
Hence, speed of car A is 60 km/h and speed of car B is 40 km/h.
(v)Let length of the given rectangle = x units
Let breadth of the given rectangle = y units
Area of the rectangle = xy
If length is reduced by 5 units and breadth is increased by 3 units.
Then new length of rectangle = x – 5
New breadth of the rectangle = y + 3
Area of rectangle = (x – 5) (y + 3)
According to condition-
(x – 5) (y + 3) = xy – 9
xy + 3x -5y -15 = xy – 9
3x -5y = 6…(1)
If length is increased by 3 units and breadth is increased by 2 units.
Then new length of rectangle = x + 3
New breadth of the rectangle = y + 2
Area of rectangle = (x + 3) (y + 2)
(x + 3) (y + 2)= xy + 67
xy + 2x + 3y + 6 = xy + 67
2x + 3y = 61..(2)
Multiply equation (1) by 2 and multiply equation (2) by 3
6x – 10y = 12…(3)
6x + 9y = 183…(4)
Subtract equation 3 from Equation 4
6x + 9y – (6x – 10y) = 183 – 12
6x + 9y – 6x + 10y = 171
19y = 171
y = 9
Put value of y I equation (1)
3x -5y = 6
3x -5(9) = 6
3x – 45 = 6
3x = 51
x = 17
Hence, length of the rectangle = 17 units
And breadth of the rectangle = 9 units
Exercise 3.6
(1) Solve the following pairs of equations by reducing them to a pair linear equations:
Ans-
Now,
2p + 3q = 2…(3)
4p – 9q = – 1…(4)
Multiply equation (1) by 2
2(2p) + 2(3q) = 2(2)
4p + 6q = 4….(5)
Subtract equation 5 from equation (4)
[4p – 9q] – [4p + 6q] = -1 – 4
4p – 9q – 4p – 6q = -5
-9q – 6q = – 5
– 15q = -5
4p + 3y = 14….(3)
3p – 4y = 23….(4)
Multiply equation (3) by 3 and multiply equation (4) by 4
3(4p) +3(3y) = 3(14)
12p + 9y = 42….(5)
4(3p) – 4(4y) = 4(23)
12p – 16 y = 92…(6)
Subtract equation (6) from equation (5)
(12p + 9y) – (12p – 16 y) = 42 – 92
12p + 9y – 12 p + 16y = – 50
9y + 16 y = – 50
25y = – 50
y = -2
Put value of y in equation (3)
4p + 3(-2) = 14
4p – 6 = 14
4p = 14 + 6
4p = 20
p = 5
But,
Now,
5p + q = 2 …(3)
6p – 3q = 1…(4)
Multiply equation (3) by 6 and multiply equation (4) by 5
6(5p) + 6q = 6(2)
5(6p) – 5(3q) = 5(1)
30p + 6q = 12…..(5)
30p – 15q = 5…….(6)
Subtract equation (6) from equation (5)
(30p + 6q) – (30p – 15q) = 12 – 5
30p + 6q – 30p + 15q = 7
21p = 7
Put values of p and q
-2p + 7q = 5…(5)
7p + 8q = 15….(6)
Multiply equation (5) by 8 and multiply equation (6) by 7
8(-2)p + 8(7q) = 8(5)
-16p + 56q = 40……(7)
7(7p) + 7(8q)= 7(15)
49p + 56q = 105 …..(8)
Subtract equation (7) from equation (8)
(49p + 56q) – (-16p + 56q) = 105 – 40
49p + 56q + 16p – 56q = 65
49p + 16p = 65
65p = 65
p = 1
Put value of p in equation (5)
-2p + 7q = 5
-2(1) + 7q = 5
-2 + 7q = 5
7q = 7
q = 1
But,
(vi) 6x + 3y = 6xy
2x + 4y = 5xy
6x + 3y = 6xy
Now,
Equation (1) can be written as
6q + 3p = 6
Or,
3p + 6q = 6….(3)
Equation (2) can be written as
2q + 4p = 5
Or,
4p + 2q = 5….(4)
Multiply equation (3) by 4 and multiply equation (4) by 3
4(3p) +4( 6q) = 4(6)
12p + 24q = 24…..(5)
And
3(4p) + 3(2q) = 3(5)
12p + 6q = 15….(6)
Subtract equation (6) from equation (5)
(12p + 24q) – (12p + 6q) = 24 – 15
12p + 24q – 12p – 6q = 9
18q = 9
Now, we can write equation (1) as
10p + 2q = 4…..(3)
15p – 5q = -2……(4)
Multiply equation (3) by 3
30p + 6q = 12…..(5)
Multiply equation (4) by 2
30p – 10q = -4 ….(6)
Subtract equation (6) from equation (5)
(30p + 6q) – (30p – 10q) = 12 – (-4)
30p + 6q – 30p + 10q = 12 + 4
16q = 16
q = 1
Put value of q in equation (3)
10p + 2(1) = 4
10p + 2 = 4
10 p = 2
x – y = 1….(8)
Subtract equation (8)from equation (7)
(x + y) – (x – y) = 5 – 1
x +y – x + y = 4
2y = 4
y = 2
Put value of y in equation (8)
x – 2 = 1
x = 3
Hence, the value of x is 3 and value of y is 2.
(3x + y) – (3x – y) = -4 –(-8)
3x + y – 3x + y = -4 + 8
2y = 4
y = 2
Put value of y in equation (7)
3x + y = -4
3x + 2 = – 4
3x = -4 – 2
3x = -6
x = -2
Hence, value of x is -2 and value of y is 2.
(2) Formulate the following problems as a pair of equations, and hence find their solutions:
(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.
(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.
Ans-
(i) Let, speed of the boat in still water = x km/h
Let, speed of the stream = y km/h
Therefore,
Speed of the boat downstream = x + y km/h
Speed of the boat upstream = x – y km/h
x – y = 2 ….(2)
Subtract equation (2) from equation (1)
(x + y) – (x – y) = 10 – 2
x + y – x + y = 8
2y = 8
y = 4
Put value of y in equation (1)
x + y = 10
x + 4 = 10
x = 6
Hence, speed of the boat in still water = 6 km/h
Speed of the current = 4km /h
3(3p + 6q) = 1
9p + 18q = 1….(4)
Multiply equation (3) by 9
9(8p) + 9(12q) = 9
72p + 108q = 9…..(5)
Multiply equation (4) by 8
8(9p) + 8(18q) = 8
72p + 144q = 8…..(6)
Subtract equation (6) from equation (5)
(72p + 108q) – (72p + 144q) = 8 – 9
72p + 108q – 72p – 144q = – 1
-36q = -1
and,
One man can finish work = y days
= q
y = = 36 day
Hence, one woman can complete work in 12 days and a man can finish work in 36 days.
(iii) Let speed of the train = s1km/h
Let speed of the bus = s2 km/h
Let time taken by bus = t hours
Condition 1 –
Total distance travel by Roohi = 300 km
Distance travel by train = 60 km
Distance travel by bus = 300 – 60
= 240 km
Time taken to complete the journey= 4 hours
Time taken by train = time taken in total journey – time taken by bus
=(4 – t) hours
Condition 2-
If she travel 100 km by train,
Distance travel by bus = 300 – 100 = 200 km
Then,
Time taken to complete the journey = 4hours + 10 min
Exercise 3.7 (Optional)
(1) The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.
Ans-
Let age of Ani = x years
Let age of Biju = y years
Let age of Dharam = z years
Let age of Cathy = A year
Conditions-
The ages of two friends Ani and Biju differ by 3 years.
x – y = 3 …(1)
Ani’s father Dharam is twice as old as Ani
z = 2x …(2)
Biju is twice as old as his sister Cathy
y = 2A…(3)
The ages of Cathy and Dharam differ by 30 years
z – A = 30 …(4)
Put value of y in equation 1
x – y = 3
x = 3 + y
= 3 + 16
= 19
Hence age of Ani is 19 years and age of Biju is 16 years.
(2) One says,” Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is amount of their (respective) capital? [From the Bijaganita of Bhaskara II]
[Hint: x + 100 = 2(y – 100), y + 10 = 6(x – 10)].
Ans-
Let capital of first person = x rupees
Let capital of second person = y rupees
Condition-
One says,” Give me a hundred, friend! I shall then become twice as rich as you”.
x + 100 = 2(y – 100) …(1)
The other replies, ”If you give me ten, I shall be six times as rich as you”.
y + 10 = 6(x – 10) …(2)
y = 6x – 60 – 10
y = 6x – 70
Put value of y in equation (1)
x + 100 = 2(y – 100)
x + 100 = 2(6x – 70 – 100)
x + 100 = 12x – 140 – 200
-11x = -140 -200 – 100
-11x = – 440
x = 40 Rs
Put value of x in equation (1)
x + 100 = 2(y – 100)
40 + 100 = 2y – 200
140 + 200 = 2y
340 = 2y
y = 170 Rs
Hence, first person has 4o Rs and second person has 170 Rs.
(3) A train covered a certain distance at a uniform speed. If the train would have been 10km/h faster, it would have taken 2 hours less than the scheduled time. And,if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.
Ans –
Let the total distance covered by train = x km
Let the uniform speed of the train = y km/h
Let scheduled time = t hours
Distance = speed time
x = yt km …(1)
Conditions-
If the train would have been 10km/h faster, it would have taken 2 hours less than the scheduled time.
Time taken by train = t – 2 hours
Speed of the train = y + 10 km/h
Distance = speed × time
x = (y +10) × (t – 2)
x = yt – 2y + 10t – 20
yt = yt – 2y + 10t – 20 [From equation 1]
2y – 10t = -20 …(2)
If the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time
Time taken by train = t + 3
Speed taken by train = y – 10
Distance = speed time
x =(y -10) × (t + 3)
x = yt + 3y – 10t -30
yt = yt + 3y – 10t -30 (From equation 1)
-3y + 10t = -30…(3)
10t = -30 + 3y…(4)
Put value of 10t in equation 2
2y – (-30 + 3y) = -20
2y + 30 – 3y = -20
-y = -50
y = 50
Put value of y in equation 2
2(50) – 10t = -20
100 – 10t = -20
-10t = -20-100
-10t = -120
t = 12
Total distance = yt
= 50 (12)
= 600 km
Hence, train travelled 600 km.
(4) The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.
Ans-
Let number of students in class = x
Let number of rows = y
Let number of students in a row = z
Number of students in a class =Number of rows × Number of students in a row
x = yz ….(1)
Conditions-
If 3 students are extra in a row, there would be 1 row less.
Number of students in a row = z + 3
Number of rows = y – 1
Number of students in a class = (z + 3) (y – 1)
x = yz + 3y – z -3
yz = yz + 3y – z -3 (From equation 1)
-3y + z = -3…..(2)
If 3 students are less in a row, there would be 2 rows more.
Number of students in a class = (z – 3) (y + 2)
x = zy + 2z – 3y – 6
yz = zy + 2z – 3y -6 (From equation 1)
-2z + 3y = -6 …(3)
3y = -6 + 2z
Put value in equation 2
-(-6 + 2z) + z = -3
6 – 2z + z = -3
6 – z = -3
-z = -9
z = 9
Put value of z in equation 2
-3y + 9 = -3
-3y = -12
y = 4
Therefore,
Number of students in a class, x = yz
Number of students in a class, x = 4(9)
= 36
Hence, number of students in the class are 36.
(5) In a Δ ABC, ∠C = 3 ∠B = 2( ∠A + ∠ B). Find the three angles.
Ans-
Condition-
∠C = 3 ∠B = 2( ∠A + ∠B)…(1)
∠C = 3 ∠B…(2) (From equation 1)
3 ∠B = 2( ∠A + ∠B)
3 ∠B = 2 ∠A + 2 ∠B
∠B = 2 ∠A
Sum of the angles of ΔABC = 1800
∠A + ∠B + ∠C = 1800
And,
∠C = 3 ∠B
∠C = 3( 40)
∠C = 1200
Hence, angles of ΔABC are ∠A = 200, ∠B = 400 and ∠C = 1200.
(6) Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the co – ordinates of the vertices of the triangle formed by these lines and y – axis.
Ans-
To obtain the equivalent geometric representation, we find two solutions of each equation.
Solutions for equation (1), ie, 5x – y = 5
| x | 0 | -5 |
| y | 1 | 0 |
Solutions for equation (2), ie, 3x – y = 3
| x | 0 | – 3 |
| y | 1 | 0 |
So vertices are (0,1), (-3, 0) and (-5, 0).
(7)Solve the following pair of linear equations:
(i) px + qy = p – q
qx – py = p + q
(ii) ax + by = c
bx + ay = 1 + c
ax + by = a2 + b2
(iv) (a – b) x + (a + b)y = a2 – 2ab – b2
(a +b ) ( x + y) = a2 + b2
(v) 152x – 378y = -74
-378x + 152y = -604
Ans-
(i)
px + qy = p – q…(1)
qx – py = p + q…(2)
Multiply equation (1) by q
q(px) + q(qy) = q(p – q)
pqx + q2y = qp – q2…(3)
Multiply equation (2) by p
p(qx) – p(py) = p(p + q)
pqx – p2y = p2 + pq…(4)
Subtract equation 4 from equation 3
pqx + q2y – (pqx – p2y )= (qp – q2) – (p2 + pq)
pqx + q2y – pqx + p2y = qp – q2 – p2 – pq
(q2 + p2)y = -(q2 + p2)
y = -1
Put value of y in equation (1)
px + q(-1) = p – q
px –q = p – q
x = 1
Hence, value of x = 1 and y = -1.
(ii)
ax + by = c …(1)
bx + ay = 1 + c …(2)
Multiply equation (1) by b
b(ax) + b2y = bc
abx + b2y = bc….(3)
Multiply equation (2) by a
abx + a2y = a(1 + c)
abx + a2y = a + ac….(4)
Subtract equation 4 from equation 3
abx + b2y – (abx + a2y) = bc –(a + ac)
abx + b2y – abx – a2y = bc – a – ac
b2y – a2y = c(b – a) – a
y(b2 – a2) = c(b – a) – a
(iv)
(a – b) x + (a + b)y = a2 – 2ab – b2….(1)
(a +b ) ( x + y) = a2 + b2…(2)
(a – b) x + (a + b)y = a2 – 2ab – b2
ax –bx + ay + by = a2 – 2ab – b2…(3)
(a +b ) ( x + y) = a2 + b2
ax + bx + ay + by = a2 + b2…(4)
Subtract equation 4 from equation 3
ax –bx + ay + by – (ax + bx + ay + by) = a2 – 2ab – b2 – (a2 + b2)
ax –bx + ay + by- ax – bx –ay –by = a2 – 2ab – b2 – a2 – b2
-2bx = -2ab – 2b2
Divide equation by -2b
x = a + b …(5)
Put value of x in equation (1)
(a – b) x + (a + b)y = a2 – 2ab – b2
(a – b) (a + b) + (a + b)y = a2 – 2ab – b2
a2 – b2 + (a + b)y = a2 – 2ab – b2
(a + b)y = -2ab
(v)
152x – 378y = -74 …(1)
-378x + 152y = -604..(2)
Multiply equation (1) by 378
152 × 378 x – 378 × 378 y = -74 × 378
152 × 378 x = -74 × 378 + 378 × 378 y…(3)
Multiply equation (2) by 152
-378 × 152 x + 152 × 152 y = -604 × 152 …(4)
Put value of 152 × 378 x in equation (4)
-(-74 × 378 + 378 × 378 y) + 152 ×152 y = -604 × 152
74 × 378 – 378 × 378 y + 152 × 152 y = -604 × 152
– 378 × 378 y + 152 × 152 y = -604 × 152 – 74 × 378
(-142884 + 23104)y = -91808 – 27972
-119780y = – 119780
y = 1
Put value of y in equation (1)
152x – 378y = -74
152x – 378(1) = -74
152x – 378 = -74
152x = -74 + 378
152x = 304
x = 2
Hence, Value of x = 2 and y = 1.
(8) ABCD is a cyclic quadrilateral (fig given below).Find the angles of the cyclic quadrilateral.
Ans-
In ABCD, cyclic quadrilateral
∠A = 4y + 20
∠B = 3y – 5
∠C = -4x
∠D = -7x + 5
∠A + ∠C = 1800 …..(1)(sum of opposite angles of cyclic quadrilateral is 1800)
∠B + ∠D = 1800 ….(2) (sum of opposite angles of cyclic quadrilateral is 1800)
Put values of all angles in equation (1) and equation (2)
4y + 20 + (-4x)= 1800
4y – 4x = 1600…(3)
3y – 5 + (-7x + 5) = 1800
3y – 5 – 7x + 5 = 1800
3y – 7x = 1800…(4)
Multiply equation (3) by 3
4y × 3 – 4x × 3 = 1600 × 3
12y – 12 x = 4800 …(5)
Multiply equation (4) by 4
3y × 4 – 7x × 4 = 1800 × 4
12y – 28x = 7200…(6)
Subtract equation (5) from equation (6)
12y – 28x – (12y – 12 x )= 7200 – 4800
12y – 28x – 12y + 12 x = 7200 – 4800
-16x = 2400
x = – 150
Put value of x in equation (4)
3y – 7x = 1800
3y – 7(-15) = 180
3y + 105 = 180
3y = 75
y = 250
Therefore,
∠A = 4y + 20
∠A = 4 25+ 20
∠A = 100 + 20
∠A = 1200
And,
∠B = 3y – 5
∠B = 3 25 – 5
∠B = 75 – 5
∠B = 700
And,
∠C = -4x
∠C = -4 (-15)
∠C = 600
And,
∠D = -7x + 5
∠D = -7(-15) + 5
∠D = 105 + 5
∠D = 1100
Hence, ∠A = 1200, ∠B = 700, ∠C = 600 and ∠D = 1100
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