SOLUTIONS:
1. Solve the following pair of linear equations by the elimination method and the substitution method :
(i) ${x + y = 5}$ and ${2x – 3y = 4}$
Solution
Elimination method.
${x + y = 5}$ ......................(1)
${2x – 3y = 4}$ .....................(2)
Multiply eqn (1) by 2. Eqn (1) becomes:
i.e ${2x + 2y = 10}$ ................(3)
Subracting eqn (3) with eqn (2)
${(2x + 2y)-(2x – 3y) = 10 - 4}$
i.e $5y=6$ => $y=6/5$
Substitution the value of $y$ in eqn (1).
${x + y = 5}$ => ${x + 6/5 = 5}$ => ${x =5 -6/5}$
${x=19/5}$
Ans: Therefore ${x=19/5}$ and $y=6/5$
Substitution method.
${x + y = 5}$ => ${x= 5 - y}$ ..........(4)
Substitute $x$ in eqn (2)
${2x – 3y = 4}$ => ${2(5-y) – 3y = 4}$
i.e $10-2y -3y = 4 $
i.e $5y=6$
i.e $y=6/5$
for find the value of $x$ , substitute the value of $y$ in eqn (4)
$x=5-6/5$= $19/5$
Ans:Therefore ${x=19/5}$ and $y=6/5$
(ii) ${3x + 4y = 10}$ and ${2x – 2y = 2}$
Solution
Elimination method.
${3x + 4y = 10}$ ......................(1)
${2x – 2y = 2}$ .....................(2)
Multiply eqn (2) by 2. Eqn (2) becomes:
i.e ${4x - 4y = 4}$ ................(3)
Adding eqn (3) with eqn (1)
${(3x + 4y)+(4x - 4y) = 10 + 4}$
i.e $7x=14$ => $x=2$
Substitution the value of $x$ in eqn (1).
${3x + 4y = 10}$ => ${3(2) + 4y = 10}$ => $4y = 4$
${y=1}$
Ans: Therefore $x=2$ and ${y=1}$
Substitution method.
${3x + 4y = 10}$ ......................(1)
${2x – 2y = 2}$ .....................(2)
i.e $x=y+1$
Substitute $x$ in eqn (1)
${3x + 4y = 10}$ => ${3(y+1) + 4y = 10}$
i.e $3y+3+4y=10$
i.e $7y=7$
i.e $y=1$
and $x=y+1$ => $x=2$
Ans: Therefore $x=2$ and ${y=1}$
(iii) ${3x – 5y – 4 = 0}$ and ${9x = 2y + 7}$
Solution
Elimination method.
${3x – 5y – 4 = 0}$ i.e ${3x – 5y = 4 }$ ......................(1)
${9x = 2y + 7}$ i.e ${9x - 2y =7 }$ .....................(2)
Multiply eqn (1) by 3. Eqn (1) becomes:
i.e ${9x-15y= 12 }$ ................(3)
subracting eqn (3) with eqn (1)
${(9x-15y)-(9x - 2y) = 12 - 7}$
i.e $-13y=5$ => $y={-5/13}$
Substitution the value of $y$ in eqn (1).
${3x – 5y = 4}$ => ${3x – 5(-5/13) = 4}$ => $3x=27/13$
${x=9/13}$
Ans: Therefore ${x=9/13}$ and $y={-5/13}$
Substitution method.
${3x – 5y – 4 = 0}$ i.e ${3x – 5y = 4 }$ ......................(1)
${9x = 2y + 7}$ i.e ${9x - 2y =7 }$ .....................(2)
From eqn (1), ${3x = 5y+ 4 }$ => $x=(5y+4)/3$ ..........(3)
Substitute $x$ in eqn (2)
${9x - 2y =7 }$ => ${9((5y+4)/3) - 2y =7 }$
$15y+12 -2y = 7$
$13y = 7-12$
$y = -5/13$
For find the value of $x$, put the value of $y$ in eqn (3)
$x=(5y+4)/3$ => $x=(5(-5/13)+4)/3$=> $x= (-25+52)/(13×3)$
$x= 27/{13×3}$ = $9/13$
Ans: Therefore ${x=9/13}$ and $y={-5/13}$
(iv) ${x/2 +2/3y = -1}$ and ${x-y/3=3}$
Solution
Elimination method.
${x/2 +2/3y = -1}$ .................(1)
${x-y/3=3}$ .................(2)
Multiplying eqn (1) with 2
$2× x/2 +2 ×2/3y = 2×(-1)$
$x + 2/3y= -2$ .............. (3)
Subracting eqn (2) and (3)
${(x + 2/3y) - (x-y/3)= -2 -3 }$
$2/3y +y/3 = -5 $
$y=-5$
utting the value of $y$ in eqn (2)
i.e ${x-y/3=3}$ => ${x-(-5)/3=3}$
$x= (3-5/3)$ => $(9-5)/3$
$x=4/3$
2. 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?
Solution:
LET the fraction be $x/y$
Given: If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1
i.e ${x+1}/{y-1}=1$ => $x+1=y-1$ => $x-y=-2$ ..............(1)
Also given: It becomes ${1/2}$ if we only add 1 to the denominator
i.e $x/{y+1} = 1/2$
$2x=y+1 $ => $2x-y=1$ ...................(2)
Subracting eqn(1) and eqn (2), we get
$(2x-y)-(x-y) = 1 - (-2)$
$x=3$
Putting the $x=3$ in eqn (1)
$x-y=-2$ => $3-y=-2$
$y=5$
Ans: Fraction is $3/5$
(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?
Solution
Let Nuri present age be $x$
Let Sonu present age be $y$
given: Five years ago, Nuri was thrice as old as Sonu
i.e $x-5=3(y-5)$
i.e $x-5=3y-15$
$x-3y=-10$ .............(1)
Also given: Ten years later, Nuri will be twice as old as Sonu
i.e $x+10= 2(y+10)$
$x+10=2y+20$
$x-2y=10$ ................(2)
Subracting eqn (1) and eqn (2), we get
$(x-2y)-(x-3y)= 10-(-10)$
$y=20$
Putting the value of $y$ in eqn (2), we get
$x-2(20)=10$
$x=50$
Ans: So, present age of Nuri is 50 years and Sonu is 20 years
(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.
Solution:
Let the digit in 10th place be $x$ and once's place be $y$.
So , the number is given as= $10x+y$
and Number after reversing the order is $10y+x$
Given: sum of the digits is 9; ie $x+y=9$
and nine times this number is twice the number obtained by reversing the order of the digits.
i.e 9($10x+y$) = $2(10y+x)$
$90x + 9y = 20 y + 2x$
$88x-11y=0$ ..............(2)
Multiply eqn (1) with 11 and Adding eqn (1) and (2) , we get,
$11x+11y=99$
$88x-11y=0$
Therefore: $99x=99$ => $x=1$
and $y=9-x$ => $y=8$
Ans: So the number is $18$
(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.
Solution
Let total number of ₹ 100 note be $x$
and total number of ₹ 50 note be $y$
Given: $x+y=25$ ..............(1)
and $100x+50y=2000$ i.e $2x+y=40$ .................(2)
Subracting eqn (1) and eqn(2), we get,
$x=15$ and $y=10$
Ans: Total number of ₹ 100 note is 15 and total number of ₹ 50 note is 10
(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 and the charge for each extra day.
Solution:
Let $c$ be the fixed charge for first 3 days and $x$ be the extra charge for each day.
Given: Saritha paid ₹ 27 for a book kept for seven days
i.e $27= 3c+ 4x$ .............(1)
and Susy paid ₹ 21 for the book she kept for five days.
i.e $21= 3c+ 2x$ ..............(2)
Solving eqn (1) and (2) , we get
$6=2x$
$x=3$
and fixed charge is $27= 3c+ 4x$ => $27= 3c+ 4(3)$
$3c=27-12=15$
$c=5$
Ans: So, fixed charge per days is ₹ 5 and charge for each extra day is ₹ 3