10th Maths 3.4
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 3.4
Q1. Solve the following pair of linear equations by the elimination method and the substitution method :
i) `x+y=5` and `2x-3y=4`
ii) `3x+4y=10` and `2x-2y=2`
iii) `3x-5y-4=0` and `9x=2y+7`
iv) `x/2+(2y)/3=-1` and `x-y/3=3`
Sol. :
i)
By Elimination Method :
`x+y=5`___________(1)
`2x-3y=4`__________(2)
Multiply equation (1) by `3` and equation (2) by `1`,
`3x+3y=15`________(3)
`2x-3y=4`__________(4)
By Equation (3) + (4)
`3x+3y+2x-3y=15+4`
`5x=19`
`x=19/5`
Put in equation (1)
`19/5+y=5`
`y=5-19/5`
`y=(25-19)/5`
`y=6/5`
By Substitution Method :
from equation (1)
`y=5-x`__________(5)
Substituting this value in equation (2)
`2x-3(5-x)=4`
`2x-15+3x=4`
`5x=4+15`
`5x=19`
`x=19/5`
put in equation (5)
`y=5-19/5`
`y=(25-19)/5`
`y=6/5`
Answer :
`x=19/5, y=6/5`
ii)
By Elimination Methiod :
`3x+4y=10`__________(1)
`2x-2y=2`___________(2)
Multiplying equation (1) by `2` and equation (2) by `3`
`6x+8y=20`__________(3)
`6x-6y=6`___________(4)
By Equation (3) - (4)
`6x+8y-(6x-6y)=20-6`
`6x+8y-6x+6y=14`
`14y=14`
`y=14/14`
`y=1`
Put in equation (1)
`3x+4(1)=10`
`3x+4=10`
`3x=10-4`
`3x=6`
`x=6/3`
`x=2`
By Substituting Method :
from equation (1)
`x=(10-4y)/3`________(5)
Substituting this value in equation (2)
`(2(10-4y))/3-2y=2`
`20-8y-6y=6`
`-14y=6-20`
`-14y=-14`
`y=(-14)/(-14)`
`y=1`
Answer :
`x=2, y=1`
iii)
By Elimination Method :
`3x-5y-4=0`__________(1)
`9x-2y-7=0`__________(2)
Multiply equation (1) by `9` and equation (2) by `3`
`27x-45y-36=0`_______(3)
`27x-6y-21=0`_________(4)
By Equation (3) - (4)
`27x-45y-36-(27x-6y-21)=0`
`27x-45y-36-27x+6y+21=0`
`-39y-15=0`
`y=-15/39`
`y=-5/13`
Put in equation (1)
`3x-5(-5/13)-4=0`
`3x+25/13-4=0`
`3x+(25-52)/13=0`
`39x-27=0`
`x=27/39`
`x=9/13`
By Substituting Method :
from equation (1)
`x=(5y+4)/3`_______(5)
Substituting this in equation (2)
`(9(5y+4))/3-2y-7=0`
`3(5y+4)-2y-7=0`
`15y+12-2y-7=0`
`13y+5=0`
`y=-5/13`
Put in equation (5)
`x=(5(-5/13)+4)/3`
`x=(-25/13+4)/3`
`x=(-25+52)/(3(13))`
`x=27/(3(13))`
`x=9/13`
Answer :
`x=9/13, y=-5/13`.
iv)
By Elimination Method
`x/2+(2y)/3=-1`
`3x+4y=-6`_________(1)
`x-y/3=3`
`3x-y=9`___________(2)
By Equation (1) - (2)
`3x+4y-(3x-y)=-6-9`
`3x+4y-3x+y=-15`
`5y=-15`
`y=-15/5`
`y=-3`
Put in equation (1)
`3x+4(-3)=-6`
`3x-12=-6`
`3x=-6+12`
`3x=6`
`x=6/3`
`x=2`
By Substituting Method
from equation (2)
`y=3x-9`_________(3)
Substituting this value in equation (1)
`3x+4(3x-9)=-6`
`3x+12x-36=-6`
`15x=30`
`x=30/15`
`x=2`
Put in equation (3)
`y=3(2)-9`
`y=6-9`
`y=-3`
Answer :
`x=2, y=-3`
Q2. Form the pair of linear equations in the following problems, and find their solutions ( if they exist) by the elimination method :
i) If we add `1` to the numerator and subtract `1` from the denominator, a fraction reduces to 1. It becomes `1/2` if we only add `1` to the denominator, What is the fraction?
ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
iii) The sum of the digits of a two-digit number is `9`. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
iv) Meena went to a bank to withdraw ₹2000. She asked the cashier to give her ₹50 and ₹100 notes only. Meena got ₹25 notes in all. Find how many notes of `₹50` and `₹100 she received.
v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid 27 for a book kept for seven days, while Susy paid 21 for the book she kept for five days. Find the fixed charge for each extra day.
Sol. :
i)
By Elimination Method
Let the numerator is `x`,
and the denominator is `y`.
then the fraction is `x/y`
According to question
`(x+1)/(y-1)=1`
`x+1=y-1`
`x-y=-1-1`
`x-y=-2`____________(1)
and
`x/(y+1)=1/2`
`2x=y+1`
`2x-y=1`____________(2)
By Equation (1) - (2)
`x-y-(2x-y)=-2-1`
`x-y-2x+y=-3`
`-x=-3`
`x=3`
Put in equation (1)
`3-y=-2`
`y=3+2`
`y=5`
Answer :
`x=3, y=5` and fraction is `3/5`.
ii)
By Elimination Method
Let the age of Nuri is `x` years,
and the age of Sonu is `y` years.
According to question
`(x-5)=3(y-5)`
`x-5=3y-15`
`x-3y=-15+5`
`x-3y=-10`___________(1)
and
`(x+10)=2(y+10)`
`x+10=2y+20`
`x-2y=20-10`
`x-2y=10`__________(2)
By Equation (1) - (2)
`x-3y-(x-2y)=-10-10`
`x-3y-x+2y=-20`
`-y=-20`
`y=20`
Put in equation (1)
`x-3(20)=-10`
`x-60=-10`
`x=60-10`
`x=50`
Answer :
Age of Nuri `(x)=50` years, Age of Sonu `(y)=20` years.
iii)
By Elimination Method
Let the ten's digit is `x`,
and the one's digit is `y`.
The Original number is `10x+y`
According to question
`x+y=9`_____________(1)
and
`9(10x+y)=2(10y+x)`
`90x+9y=20y+2x`
`90x-2x+9y-20y=0`
`88x-11y=0`
`8x-y=0`____________(2)
By Equation (1) + (2)
`x+y+8x-y=9`
`x+8x=9`
`9x=9`
`x=9/9`
`x=1`
Put in equation (1)
`1+y=9`
`y=9-1`
`y=8`
Answer :
`x=1, y=8` and number is `18`.
iv)
By Elimination Method
Let Meena gets `x` notes of `₹50`,
and Meena gets `y` notes of `₹100`
According to question
`x+y=25`___________(1)
and
`50x+100y=2000`
Divide by `50` both the sides.
`x+2y=40`__________(2)
By Equation (1) - (2)
`x+y-x-2y=25-40`
`-y=-15`
`y=15`
Put in equation (1)
`x+15=25`
`x=25-15`
`x=10`
Answer :
`x=10, y=15`.
v)
By Elimination Method
Let the fixed charge is `₹ x`,
and the extra charge is `₹ y`.
According to question
`x+(7-3)y=27`
`x+4y=27`_________(1)
and
`x+(5-3)y=21`
`x+2y=21`_________(2)
By Equation (1) - (2)
`x+4y-(x+2y)=27-21`
`x+4y-x-2y=6`
`2y=6`
`y=6/2`
`y=3`
Put in equation (1)
`x+2(3)=21`
`x=21-6`
`x=15`
Answer :
`x=15, y=3`.
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