10th Maths 3.3
NCERT Class 10th solution of Exercise 3.1
NCERT Class 10th solution of Exercise 3.5
NCERT Class 10th solution of Exercise 3.6
Exercise 3.3
Q1. Solve the following pair of linear equations by the substitution method :
i) x+y=14;x-y=4
ii) s-t=3;s3+t2=6
iii) 3x-y=3;9x-3y=9
iv) 0.2x+0.3y=1.3;0.4x+0.5y=2.3
v) √2x+√3y=0;√3x-√8y=0
vi) 3x2-5y3=-2;x3+y2=136
Sol. :
i)
x+y=14________(1)
x-y=4 _________(2)
from equation (2)
x=y+4_________(3)
Substituting this value in equation (1)
y+4+y=14
2y=14-4
2y=10
y=102
y=5
Put in equation (3)
x=14-5
x=9
Answer :
x=9,y=5.
ii)
s-t=3___________(1)
s3+t2=6_______(2)
from equation (1)
s=t+3__________(3)
Substituting this value in equation (2)
t+33+t2=6
2(t+3)+3t6=6
2t+6+3t=36
5t=36-6
5t=30
t=305
t=6
Put in equation (3)
s=6+3
s=9
Answer :
s=9,t=6
iii)
3x-y=3___________(1)
9x-3y=9___________(2)
from equation (1)
y=3x-3____________(3)
Substituting this value in equation (2)
9x-3(3x-3)=9
9x-9x+9=9
9=9
Answer :
y=3x-3
where x can take any value, i.e., infinitely many solutions.
iv)
0.2x+0.3y=1.3
Multiply 10 both sides.
2x+3y=13__________(1)
0.4x+0.5y=2.3
Multiply 10 both sides.
4x+5y=23__________(2)
from equation (1)
x=13-3y2________(3)
Substituting this value in equation (2)
4(13-3y2)+5y=23
2(13-3y2)+5y=23
26-6y+5y=23
-6y+5y=23-26
-y=-3
y=3
Put in equation (3)
x=13-3(3)2
x=13-92
x=42
x=2
Answer :
x=2,y=3
v)
√2x+√3y=0_________(1)
√3x-√8y=0__________(2)
from equation (1)
x=-√3y√2_____________(3)
Sustituting this value in equation (2)
√3(-√3y√2)-√8y=0
-3y√2-2√2y=0
-3y-4y=0
-7y=0
y=0-7
y=0
Put in equation (3)
x=-√3(0)√2
x=0
Answer :
x=0,y=0
vi)
3x2-5y3=-2
9x-10y=-12__________(1)
x3+y2=136
Multiply 6 both sides
2x+3y=13____________(2)
from equation (1)
x=10y-129_________(3)
Substituting this value in equation (2)
2(10y-129)+3y=13
20y-24+27y=117
20y+27y=117+24
47y=141
y=14147
y=3
Put in equation (3)
x=10(3)-129
x=30-129
x=189
x=2
Answer :
x=2,y=3.
Q2. Solve 2x+3y=11 and 2x-4y=-24 and hence find the value of ′m′ for which y=mx+3.
Sol. :
2x+3y=11_________(1)
2x-4y=-24_________(2)
y=mx+3___________(A)
from equation (1)
x=11-3y2_______(3)
Substituting this value in equation (2)
2(11-3y2)-4y=-24
11-3y-4y=-24
-7y=-24-11
-7y=-35
y=357
y=5
Put in equation (3)
x=11-3(5)2
x=11-152
x=-42
x=-2
Put x=-2 and y=5 in equation (A)
5=m(-2)+3
5-3=-2m
2=-2m
m=-22
m=-1
Answer :
x=-2,y=5, and m=-1
Q3. Form 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 ₹3800. Later, she buys 3 bats and 5 balls for₹1750. Find the cost of each bat and each ball.
iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10km, the charge paid is ₹105 and for a journey of 15km, the charge paid is ₹155. what are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25km?
v) A fraction becomes 911, if 2 is added to both the numerator. If, 3 is added to both the numerator and the denominator it becomes 56. 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 that of his son. What are their present ages?
Sol. :
i)
Let the first number is x,
and the second number is y.
According to question
x-y=26________(1)
x=3y__________(2)
Substituting x=3y in equation (1)
3y-y=26
2y=26
y=262
y=13
Put in equation (2)
x=3(13)
x=39
Answer :
x=39,y=13
ii)
Let the first angle is x,
and the second angle is y.
According to question
Sum of supplymentry angles =1800.
x+y=1800__________(1)
and
x=y+180__________(2)
Substituting x=y+180 in equation (1)
y+180+y=1800
2y=1800-180
2y=1620
y=16202
y=810
Put in equation (1)
x=810+180
x=990
Answer :
x=990,y=810.
iii)
Let the cost of each bat ₹x,
and the cost of each ball ₹y.
According to question
7x+6y=3800__________(1)
3x+5y=1750__________(2)
from equation (1)
x=3800-6y7________(3)
Substituting this value in equation (2)
3(3800-6y7)+5y=1750
11400-18y+35y=12250
17y=12250-11400
17y=850
y=85017
y=50
Put in equation (3)
x=3800-6(50)7
x=3800-3007
x=35007
x=500
Answer :
x=₹500,y=₹50.
iv)
Let the fixed charge is ₹x,
and the per km fare is ₹y.
According to question
x+10y=105__________(1)
x+15y=155__________(2)
from equation (1)
x=105-10y__________(3)
Substituting this value in equation (2)
105-10y+15y=155
5y=155-105
5y=50
y=505
y=10
Put in equation (3)
x=105-10(10)
x=105-100
x=5
Answer :
x=₹5,y=₹10.
v)
Let the numerator is x
and the denominator is y
According to question
x+2y+2=911
11(x+2)=9(y+2)
11x+22=9y+18
11x-9y=18-22
11x-9y=-4_______(1)
and
x+3y+3=56
6(x+3)=5(y+3)
6x+18=5y+15
6x-5y=15-18
6x-5y=-3_______(2)
from equation (1)
x=-4+9y11____(3)
Substituting this value in equation (2)
6-4+9y11-5y=-3
-24+54y-55y=-33
-y=-33+24
-y=-9
y=9
Put in equation (3)
x=-4+9(9)11
x=-4+8111
x=7711
x=7
Answer :
x=7,y=9 and fraction 79.
vi)
Let the present age of Jacob is x years,
and the present age of his son is y years.
According to question
(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)
from equation (1)
x=10+3y________(3)
Substutiting this value in equation (2)
10+3y-7y=-30
-4y=-30-10
-4y=-40
y=-40-4
y=10
Put in equation (3)
x=10+3(10)
x=10+30
x=40
Answer :
x=40,y=10
Comments