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
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)
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}
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.
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
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
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
For unique solution
$\frac {a_1}{a_2} \neq \frac {b_1}{b_2}$
Subsituting the values
$\frac {1}{5} \neq \frac {2}{k}$
$k\neq10$
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
For unique solution
$\frac {a_1}{a_2} \neq \frac {b_1}{b_2}$
Subsituting the values
$\frac {1}{2} \neq \frac {2}{k}$
$k\neq4$
3(5) +2(-5) -5=0
So x=5 and y=-5 is solution of the equation
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
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.