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 16x=p,16y=q and then solve in p& q and then find x and y
ii. (1 ,3)
Hint: Take 1x=p,1y=q and then solve in p& q and then find x and y
iii. (8,3)
Hint: Take 1x+y=p,1x−y=q and then solve in p& q and then find x and y
iv. (3,2)
Hint: Take 1x+y=p,1x−y=q and then solve in p& q and then find x and y
v. ab,bc
vi. (2,1)
Hint: Take 12x+3y=p,13x−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
7y−2x=5
8y+7y=15
Take 1x=p,1y=q and then solve in p& q and then find x and y
ix. (a,b)
x. (11, 8)
a. ∠A=(2x+4), ∠B=(y+3), ∠C=(2y+10), ∠D=(4x−5)
In a cyclic quadrilateral, Opposite angles are supplementary.
∠A+∠C=1800 and ∠B+∠D=1800
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) ∠A=(2x−1), ∠B=(y+5), ∠C=(2y+15) and ∠D=(4x−7)
Solving similarly, we get
650, 550, 1150, 1250
For Infinite solution:
We have a1a2=b1b2=c1c2 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
a1a2=b1b2=c1c2
22a=3a+b=728
Solving this , we get a=4 and b=8
We have
a1a2=b1b2=c1c2
2a+b=−3a+b−3=74a+b
2a+b=−3a+b−3
a+b+6=0 -(1)
−3a+b−3=74a+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 3600
So
x+(3x−2)+y=180 or 4x+y=182 -(1)
Also given
∠C−∠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
a1a2≠b1b2
Subsituting the values
15≠2k
k≠10
For parallel lines
a1a2=b1b2=c1c2
2a−1=3a+1=73a−1
or a=5
For unique solution
a1a2≠b1b2
Subsituting the values
12≠2k
k≠4
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.
Go back to Class 10 Main Page using below links