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Worksheets for Linear equations in two variable




Given below are the linear equations in two variables class 10 worksheet with solutions


Question 1.
Solve these linear equation in two variable ( x and y)
a. $37x + 41y = 70$
$41x + 37y = 86$

b. $99x + 101y = 499$
$101x + 99y = 501$

c. $23x - 29y = 98$
$29x - 23y =110$

d. $ax + by = a - b$
$bx - ay = a + b$

e. $x + y = a + b$
$ax - by = a^2 - b^2$

f. $(a - b) x + (a + b) y = a^2 - 2ab - b^2$
$(a + b) (x + y) = a^2 - b^2$

g. $8x - 3y = 5xy$
$5y = -2xy$

h. $3(2x + y) = 7xy$
$3(x + 3y) = 11xy$

i. $49x + 51y = 499$
$51x + 49y = 501$

j. $217x + 131y = 913$
$131x + 217y = 827$

Answer

1.
i. (3, 2)
ii. (3, -1)
iii. (8, 3)
iv. (1,-1)
v. (a, b)
vi. (a,-b)
vii. (0, 0) ,(22/31, 11/23)
viii. (0,0) (1,3/2)
ix. (1 ½ , 9/2)
x. 3,2


Qustion 2
Solve these linear equation in two variable ( x and y)
i. $ \frac {1}{2x} + \frac {1}{3y} = 2$
$ \frac {1}{3x} + \frac {1}{2y} = \frac {13}{6}$

ii) $ \frac {2}{x} + \frac {3}{y} = \frac {9}{xy}$
$ \frac {4}{x} + \frac {9}{y} = \frac {21}{xy}$
Where $x \ne 0,y \ne 0$

iii. $ \frac {22}{x+y} + \frac {15}{x-y} = 5$
$ \frac {55}{x+y} + \frac {45}{x-y} = 14$

iv.$ \frac {5}{x+y} - \frac {2}{x-y} = -1$
$ \frac {15}{x+y} + \frac {7}{x-y} = 10$

v. $bx + cy=a+b$
$ ax( \frac {1}{a-b} - \frac {1}{a+b}) + cy ( \frac {1}{b-a} - \frac {1}{b+a})= \frac {2a}{a+b}$

vi)$ \frac {1}{2(2x+3y)} + \frac {1}{7(3x-2y)} = \frac {17}{20}$
$ \frac {7}{(2x+3y)} - \frac {1}{(3x-2y)} = -\frac {28}{5}$

vii.$ \frac {x+1}{2} - \frac {y +4}{11} = 2$
$ \frac {x+3}{2} + \frac {2y+3}{17} = 5$

viii.$ \frac {7x -2y}{xy} =5$
$ \frac {8x+7y}{xy} =15$

ix. $ \frac {x}{a} + \frac {y}{b} =2$
$ax -by = a^2 - b^2$

x.$ \frac {57}{x+y} + \frac {6}{x-y} = 5$
$ \frac {38}{x+y} + \frac {21}{x-y} = 9$

Answer

i. (1/2, 1/3)
Hint: Take $\frac {1}{6x}=p , \frac {1}{6y} =q$ and then solve in p& q and then find x and y

ii. (1 ,3)
Hint: Take $\frac {1}{x}=p , \frac {1}{y} =q$ and then solve in p& q and then find x and y

iii. (8,3)
Hint: Take $\frac {1}{x+y}=p , \frac {1}{x-y} =q$ and then solve in p& q and then find x and y

iv. (3,2)
Hint: Take $\frac {1}{x+y}=p , \frac {1}{x-y} =q$ and then solve in p& q and then find x and y

v. $ \frac {a}{b},\frac {b}{c}$ vi. (2,1)
Hint: Take $\frac {1}{2x+3y}=p , \frac {1}{3x-2y} =q$ and then solve in p& q and then find x and y

vii. 5, 7
viii. (1, 1)
Hint:
Convert into these forms
$ \frac {7}{y} - \frac {2}{x} =5$
$ \frac {8}{y} + \frac {7}{y} =15$
Take $\frac {1}{x}=p , \frac {1}{y} =q$ and then solve in p& q and then find x and y

ix. (a,b)
x. (11, 8)



Question 3.
In a cyclic quadrilateral ABCD,Find the four angles.
a. $\angle A = (2x + 4)$, $\angle B = (y + 3)$, $\angle C = (2y + 10)$, $\angle D = (4x -5)$.
b. $\angle A = (2x - 1)$, $\angle B = (y + 5)$, $\angle C = (2y + 15)$ and $\angle D = (4x -7)$

Answer

a. $\angle A = (2x + 4)$, $\angle B = (y + 3)$, $\angle C = (2y + 10)$, $\angle D = (4x -5)$
In a cyclic quadrilateral, Opposite angles are supplementary.
$\angle A + \angle C = 180^0$ and $\angle B + \angle D = 180^0$
So $2x+4+2y+10=180$ or $x+y=83$
$y+3+4x-5=180$ or $y+4x=182$
Solving the above equation by Substitution method
x=33 and y=50
So Angles are

700,530,1100,127 0
b) $\angle A = (2x - 1)$, $\angle B = (y + 5)$, $\angle C = (2y + 15)$ and $\angle D = (4x -7)$
Solving similarly, we get
650, 550, 1150, 1250


Question 4.
Given below are three equations. Two of them have infinite solutions and two have a unique solution. State the two pairs:
$3x - 2y = 4$
$6x + 2y = 4$
$9x -6y = 12$

Answer

For Infinite solution:

We have $\frac{a_1}{a_2} = \frac {b_1}{b_2} = \frac {c_1}{c_2}$ is the equation for infinite solution.
Equation 1 and 3 is satisfying the condition
$3x-2y=4$; $9x-6y=12$
3/9 = -2/-6 = 4/12

This also means equation 1 and equation 3 are same

Unique solution:
We have a1/a2 is not equal to b1/b2 is the equation for the unique solution.
Equation 1 and 2 and Equation 2 and 3 is satisfying the condition
3x-2y=4; 6x+2y=4 & 6x+2y=4; 9x-6y=1
3/6 is not equal to -2/2 & 6/9 is not equal to 2/-6.


Question 5
Find the values of a and b for which the following system of linear equations has infinite number of solutions:
$2x + 3y = 7$
$2ax + (a + b)y = 28$

Answer

We have
$\frac{a_1}{a_2} = \frac {b_1}{b_2} = \frac {c_1}{c_2}$
$\frac{2}{2a} = \frac {3}{a+b} = \frac {7}{28}$
Solving this , we get a=4 and b=8


Question 6
Find the values of a and b for which the following system of linear equations has infinite number of solutions:
$2x -3y = 7$
$(a + b)x - (a + b- 3)y = 4a + b$

Answer

We have
$\frac{a_1}{a_2} = \frac {b_1}{b_2} = \frac {c_1}{c_2}$
$\frac{2}{a+b} = \frac {-3}{a+b-3} = \frac {7}{4a+b}$
$\frac{2}{a+b} = \frac {-3}{a+b-3}$
a+b+6=0 -(1)
$\frac {-3}{a+b-3} = \frac {7}{4a+b}$
5a-4b+21=0 -(2)
Solving (1) and (2) by substitution method we get
a=-5,b=-1


Question 7.
In a ABC, $\angle A = x$, $\angle B = (3x -2)$, $\angle C = y$ Also $\angle C - \angle B = 9$. Find the three angles.

Answer

In a triangle,sum of angles is equal to 3600
So
$x+ (3x-2) + y=180$ or $4x+y=182$ -(1)
Also given
$\angle C - \angle B = 9$ or $y-(3x-2)=9$ or $y-3x=7$ or $y=7+3x$
Substituting this in (1)
$4x+ 7+3x=182$
$7x=175$
$x=25$
So angles are
25,73, 82


Question 8.
Find the value of k for which the system of equations x + 2y -3 = 0 and ky + 5x + 7 = 0 has a unique solution.

Answer

For unique solution
$\frac {a_1}{a_2} \neq \frac {b_1}{b_2}$
Subsituting the values
$\frac {1}{5} \neq \frac {2}{k}$
$k\neq10$


Question 9
For what value of a the system of linear equations $2x + 3y = 7$ and $(a -1)x + (a + 1)y = 3a - 1$ represent parallel lines.

Answer

For parallel lines
$\frac{a_1}{a_2} = \frac {b_1}{b_2} = \frac {c_1}{c_2}$
$\frac{2}{a-1} = \frac {3}{a+1} = \frac {7}{3a-1}$
or a=5


Question 10
If the lines $x + 2y + 7 = 0$ and $2x + ky + 18 = 0$ intersect at a point, then find the value of k.

Answer

For unique solution
$\frac {a_1}{a_2} \neq \frac {b_1}{b_2}$
Subsituting the values
$\frac {1}{2} \neq \frac {2}{k}$
$k\neq4$


Question 11
Is $x = 5, y = -5$ a solution of the linear equation $3x + 2y - 5 = 0$?

Answer

3(5) +2(-5) -5=0
So x=5 and y=-5 is solution of the equation


Question 12.
The father’s age is six times his son’s age. Four years hence, the age of the father will be four times his son’s age. Find the present ages of the son and the father.

Answer

let Father age is y and son age is x
First condition
$y=6x$
Second condition
$y+4=4(x+4)$
So we have
$6x+4=4x+16$
x=6 and y=36



Summary

This linear equations in two variables worksheet class 10 with solutions is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feedback on the mail.



Also Read




Go back to Class 10 Main Page using below links
Class 10 Maths Class 10 Science

Practice Question

Question 1 What is $1 - \sqrt {3}$ ?
A) Non terminating repeating
B) Non terminating non repeating
C) Terminating
D) None of the above
Question 2 The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is?
A) 19.4 cm3
B) 12 cm3
C) 78.6 cm3
D) 58.2 cm3
Question 3 The sum of the first three terms of an AP is 33. If the product of the first and the third term exceeds the second term by 29, the AP is ?
A) 2 ,21,11
B) 1,10,19
C) -1 ,8,17
D) 2 ,11,20



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