1. Solve the following pair of linear equations by the substitution method.
(i) ${x + y = 14}$ & ${x – y = 4}$
Solution:
${x + y = 14}$ ...........(1)
${x – y = 4}$ i.e $x=y+4$ ...........(2)
Subsituting the value of $x$ in eqn (1)
$(y+4) + y= 14$
$2y=10$
$y=5$
and $x=y+4=9$
Ans: $x$ = 9 and $y$ = 5
(ii) ${s – t = 3}$ & ${s/3 +t/2 = 6}$
Solution:
Given: ${s – t = 3}$ i.e $s=t+3$ ............(1)
${s/3 +t/2 = 6}$...............(2)
Subsituting the value of $s$ in eqn (2)
$(t+3)/3 +t/2 = 6$
$(2t+6+3t)/6 =6$
$5t+6=36$
$5t=30$
$t=6$
Since $s=t+3$ , therefore $s=6+3 = 9$
Ans: s=9 and t = 6
(iii) ${3x – y = 3}$ & ${9x – 3y = 9}$
Solution:
${3x – y = 3}$ i.e ${y = 3x -3}$ ........(1)
${9x – 3y = 9}$ ..........................(2)
Subsituting the value of $y$ in eqn (2).
${9x – 3(3x -3) = 9}$
${9x – 9x +9) = 9}$
${9x – 9x = 9-9}$
Therefore the pair of linear equations cannot be solved and has many solutions.
(iv) ${0.2x + 0.3y = 1.3}$ & ${0.4x + 0.5y = 2.3}$
Solution:
${0.2x + 0.3y = 1.3}$ ...........(1)
${0.4x + 0.5y = 2.3}$ ............(2)
Multiply eqn (1) by 2 and subsitute the value of $x$ in eqn (2)
${0.4x + 0.6y = 2.6}$ i.e ${0.4x = 2.6 - 0.6y}$
${0.4x + 0.5y = 2.3}$ becomes ${2.6 - 0.6y + 0.5y = 2.3}$
i.e $0.1 y = 0.3$ i.e $y=3$
and ${0.4x = 2.6 - 0.6y}$ => ${0.4x = 2.6 - 0.6(3)}$ => ${0.4x = 2.6 - 1.8}$
=> $0.4x = 0.8$ => $x=2$
Ans: $x=2$ and $y=3$
(v) ${√2x+√3y = 0}$ & ${√3x-√8y=0}$
Solution:
${√2x+√3y = 0}$ => ${√2x = -√3y }$ => $x= -√{3/2}y$ ......(1)
${√3x-√8y=0}$ ..........(2)
Subsituting the value of $x$ in eqn (2)
${√3x-√8y=0}$ => ${-√3(√{3/2})y-√8y=0}$ => $√{(9+16)/2} y = 0$.
i.e $y=0$
Ans: $y=0$ and $x=0$
(vi) ${3/2x -5/3y = -2 }$ & ${x/3+y/2 = 13/6}$
Solution:
Given ${3/2x -5/3y = -2 }$ => ${3/2x =5/3y -2 }$ => ${x =2/3(5/3y -2) }$ => ${x =(10/9y -4/3) }$ .........eqn(1)
${x/3+y/2 = 13/6}$ ................eqn(2)
Subsituting the value of $x$ in eqn (2)
${x/3+y/2 = 13/6}$ => ${({(10/9y -4/3)})/3+y/2 = 13/6}$
=> ${10/27}y - 4/9 + y/2 = 13/6 $
=> ${10/27}y + y/2 = 13/6 + 4/9 $
=> $(47/54)y = 47/18 $
=> $y = 3$
${x =(10/9y -4/3) }$ =>${(10/9{3} -4/3) }$ => ${10/3}-4/3$ => $6/3$=> $2$
Ans: $x$ = 2 and $y$ = 3
2. Solve ${2x + 3y = 11}$ and ${2x – 4y = – 24}$ and hence find the value of ${‘m’}$ for which
${y = mx + 3}$.
Solution
Given : ${2x + 3y = 11}$ => ${2x = 11 - 3y}$ ......(1)
${2x – 4y = – 24}$ ........................(2)
Substituting the value of $2x$ in eqn (1)
${2x – 4y = – 24}$ => ${(11 - 3y) – 4y = – 24}$
=> $7y=35$
$y=5$
and ${2x = 11 - 3y}$ => ${2x = 11 - 3(5)}$
i.e $x=-2$
Given: ${y = mx + 3}$. Therefore ${5 = m(-2) + 3}$.
$2m=-2$
$m=-1$
Ans: $x=-2$; $y=5$ ; $m=-1$
3. 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.
Solution:
Let the numbers be $x$ and $y$; where $x>y$
Given: difference between two numbers is 26
i.e $x-y=26$ ..............(1)
and , one number is three times the other
i.e $x=3y$ ................(2)
Substituting the value of $x$ in eqn (1)
therefore, $x-y=26$ => $3y-y=26$ => $2y=26$; i.e $y=13$
and $x=3y$ => $x=39$
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Solution
Since the angles are supplementary angles, their sum is 180 degree.
Let the first angle be $x$ degree.
Let the second angle be $y$ degree.
Given: $x+y = 180$ ..................(1)
and $x=y+18$ ...................(2)
Substituting the value of $x$ in eqn (1)
$x+y = 180$ => $(y+18)+y = 180$
$2y=172$
$y=86$
and $x = 86+18 $ i.e $x$ = 104
Ans: Angles are 104 degree and 86 degree.
(iii) The coach of a cricket team buys 7 bats and 6 balls for ₹ 3800.
Later, she buys 3 bats and 5 balls for ₹ 1750. Find the cost of each bat and each ball.
Solution:
Let the cost of 1 bat be ₹ $x$
Let the cost of 1 ball be ₹ $y$
Given:7 bats and 6 balls for ₹ 3800
i.e $7x+6y=3800$
i.e ${x=(3800-6y)/7}$ ..............(1)
and 3 bats and 5 balls for ₹ 1750
i.e $3x+5y=1750$ ...................(2)
Substituting the value of $x$ in eqn (2),
$3x+5y=1750$ => ${3(3800-6y)}/7+5y=1750$
$11400 - 18y + 35y = 12250$
$17y = 850$
$y = 50$
and ${x=(3800-6y)/7}$ => ${x=(3800-300)/7}$
$x=3500/7$ => $x=500$
Ans: Cost of 1 bat is ₹ 500 and 1 ball is ₹ 50.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered.
For a distance of 10 km, 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 25km?
Solution:
Let the fixed charges be ₹ $c$
Let the charges per km be ₹ $x$
Given:For a distance of 10 km, the charge paid is ₹ 105
i.e $105 = c + 10 x$ i.e $c= 105 - 10x$ .............(1)
and for a journey of 15 km, the charge paid is ₹ 155
i.e $155 = c + 15 x$ ....................(2)
Substitute the value of $c$ in eqn (2)
$155 = c + 15 x$ => $155 = 105 - 10x + 15 x$
$50 = 5x$ i.e $x=10$
$c= 105 - 10x$ => $c= 105 - 10(10)$ ; ie $c = 5$
So, the fixed charges is ₹ 5 and the charges per km is ₹ 10.
Charges for travelling 25 km = 5+ 25(10)=5+250 = ₹ 255
(v) A fraction becomes ${9/11}$ if 2 is added to both the numerator and the denominator.
If, 3 is added to both the numerator and the denominator it becomes ${5/6}$. Find the fraction.
Solution:
Let the fraction be $x/y$
Case 1: ${x+2}/{y+2} = 9/11 $ => $11x-9y+4=0$ ............(1)
Case 2: ${x+3}/{y+3} = 5/6 $ => $6x-5y+3 = 0$ .................(2)
Solving eqn (1) and (2) with substitution method.
$11x-9y+4=0$ => $11x=9y-4$ => $x={9y-4}/11$
$6x-5y+3 = 0$ => $6({9y-4}/11)-5y+3 = 0$
${54y-24}/11 -5y + 3 = 0 $
$54y-24-55y + 33 = 0 $
$y=9$
$x={9y-4}/11$ => $x={81-4}/11$ => $x=7$
Ans: Therefore the fraction is $7/9$
(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?
Solution
Let Jacob present age be $x$
and His son present age be $y$
Given: Five years hence, the age of Jacob will be three times that of his son.
i.e $x+5$ = $3(y+5)$
$x+5$ = $3y+15$
$x - 3y = 10$
i.e $x=3y+10 $ .......................(1)
Given: Five years ago, Jacob’s age was seven times that of his son.
i.e $x-5 = 7(y-5)$
i.e $x-5=7y-35$ ........................(2)
Substituting $x$ in eqn (2)
$x-5=7y-35$ => $3y+10-5=7y-35$
i.e $7y-3y = 5+35$
i.e $4y = 40$
i.e $y = 10$
Since $x=3y+10 $ i.e $x=3(10)+10 $
i.e $x=40$
Ans: Present age of Jacob is 40 years and hi son is 10 years.
SOLUTIONS: