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; s/3+t/2=6`
iii) ` 3x-y=3; 9x-3y=9`
iv) `0.2x+0.3y=1.3; 0.4x+0.5y=2.3`
v) `sqrt2x+ sqrt3y=0; sqrt3x-sqrt8y=0`
vi) `(3x)/2-(5y)/3=-2; x/3+y/2=13/6`
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=10/2`
`y=5`
Put in equation (3)
`x=14-5`
`x=9`
Answer :
`x=9, y=5`.
ii)
`s-t=3`___________(1)
`s/3+t/2=6`_______(2)
from equation (1)
`s=t+3`__________(3)
Substituting this value in equation (2)
`(t+3)/3+t/2=6`
`(2(t+3)+3t)/6=6`
`2t+6+3t=36`
`5t=36-6`
`5t=30`
`t=30/5`
`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=(13-3y)/2`________(3)
Substituting this value in equation (2)
`4((13-3y)/2)+5y=23`
`2((13-3y)/2)+5y=23`
`26-6y+5y=23`
`-6y+5y=23-26`
`-y=-3`
`y=3`
Put in equation (3)
`x=(13-3(3))/2`
`x=(13-9)/2`
`x=4/2`
`x=2`
Answer :
`x=2, y=3`
v)
`sqrt2x+ sqrt3y=0`_________(1)
`sqrt3x-sqrt8y=0`__________(2)
from equation (1)
`x=-(sqrt3y)/sqrt2`_____________(3)
Sustituting this value in equation (2)
`sqrt3(-(sqrt3y)/sqrt2)-sqrt8y=0`
`-(3y)/sqrt2-2sqrt2y=0`
`-3y-4y=0`
`-7y=0`
`y=0/-7`
`y=0`
Put in equation (3)
`x=-(sqrt3(0))/sqrt2`
`x=0`
Answer :
`x=0, y=0`
vi)
`(3x)/2-(5y)/3=-2`
`9x-10y=-12`__________(1)
`x/3+y/2=13/6`
Multiply `6` both sides
`2x+3y=13`____________(2)
from equation (1)
`x=(10y-12)/9`_________(3)
Substituting this value in equation (2)
`2((10y-12)/9)+3y=13`
`20y-24+27y=117`
`20y+27y=117+24`
`47y=141`
`y=141/47`
`y=3`
Put in equation (3)
`x=(10(3)-12)/9`
`x=(30-12)/9`
`x=18/9`
`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=(11-3y)/2`_______(3)
Substituting this value in equation (2)
`2((11-3y)/2)-4y=-24`
`11-3y-4y=-24`
`-7y=-24-11`
`-7y=-35`
`y=35/7`
`y=5`
Put in equation (3)
`x=(11-3(5))/2`
`x=(11-15)/2`
`x=(-4)/2`
`x=-2`
Put `x=-2` and `y=5` in equation (A)
`5=m(-2)+3`
`5-3=-2m`
`2=-2m`
`m=(-2)/2`
`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 `15 km,` 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 `25 km`?
v) A fraction becomes `9/11,` if `2` is added to both the numerator. If, `3` is added to both the numerator and the denominator it becomes `5/6`. 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=26/2`
`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 `=180^0.`
`x+y=180^0`__________(1)
and
`x=y+18^0`__________(2)
Substituting `x=y+18^0` in equation (1)
`y+18^0+y=180^0`
`2y=180^0-18^0`
`2y=162^0`
`y=162^0/2`
`y=81^0`
Put in equation (1)
`x=81^0+18^0`
`x=99^0`
Answer :
`x=99^0, y=81^0`.
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=(3800-6y)/7`________(3)
Substituting this value in equation (2)
`3((3800-6y)/7)+5y=1750`
`11400-18y+35y=12250`
`17y=12250-11400`
`17y=850`
`y=850/17`
`y=50`
Put in equation (3)
`x=(3800-6(50))/7`
`x=(3800-300)/7`
`x=3500/7`
`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=50/5`
`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+2)/(y+2)=9/11`
`11(x+2)=9(y+2)`
`11x+22=9y+18`
`11x-9y=18-22`
`11x-9y=-4`_______(1)
and
`(x+3)/(y+3)=5/6`
`6(x+3)=5(y+3)`
`6x+18=5y+15`
`6x-5y=15-18`
`6x-5y=-3`_______(2)
from equation (1)
`x=(-4+9y)/11`____(3)
Substituting this value in equation (2)
`6(-4+9y)/11-5y=-3`
`-24+54y-55y=-33`
`-y=-33+24`
`-y=-9`
`y=9`
Put in equation (3)
`x=(-4+9(9))/11`
`x=(-4+81)/11`
`x=77/11`
`x=7`
Answer :
`x=7, y=9` and fraction `7/9`.
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`
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