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}$
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 $\angle D = (4x -7)$ Solution
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
70^{0},53^{0},110^{0},127 ^{0}
b) $\angle A = (2x - 1)$, $\angle B = (y + 5)$, $\angle C = (2y + 15)$ and $\angle D = (4x -7)$
Solving similarly, we get
65^{0}, 55^{0}, 115^{0}, 125^{0}
Question 4.
Given below are three equations. Two of them have infinite solutions and two have a unique solution. State the two pairs:
$3x - 2y = 4$
$6x + 2y = 4$
$9x -6y = 12$ Solution
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:
$2x + 3y = 7$
$2ax + (a + b)y = 28$ Solution
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:
$2x -3y = 7$
$(a + b)x - (a + b- 3)y = 4a + b$ Solution
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. Solution
In a triangle,sum of angles is equal to 360^{0}
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. Solution
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. Solution
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. Solution
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 $3x + 2y - 5 = 0$? Solution
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. Solution
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.
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