Pair of Linear Equations in Two Variables
Mathematics Class Tenth
NCERT Exercise 3.3-(2)
Question: 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 are `x` and `y`
Thus, according to question,
`x - y = 26` -----------(i)
And, `x = 3y` -----------(ii)
Now, by substituting the value of `x` from equation (ii) in equation (i), we get
`3y - y = 26`
`=> 2y = 26`
`:. y = 26/2 = 13`
Now, by substituting the value of `y` in equation (ii), we get
`x = 3 xx 13 = 39`
Thus, `x = 39` and `y = 13` Answer
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Solution:
Let, given supplementary angles are `x` and `y`
Thus, according to question,
`x+y = 180` ------------------(i)
[∵ as given angles are supplementary which is equal to `180^0`]
And, `x - y = 18` ------------(ii)
Now, from equation, (ii), we get
`x = 18+y` ---------(iii)
Now, by substituting the value of `x` from equation (iii) in equation (i), we get
`18+y + y = 180`
`=> 18 +2y = 180`
`=> 2y = 180-18 = 162`
`:. y = 162/2 = 81`
Now, by substituting the value of `y` in equation (ii), we get
`x = 18 + 81 = 99`
Thus, angles are `99^0` and `81^0` Answer
(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.
Solution:
Let, cost of one bat = Rs `x`
And cost of one ball = Rs `y`
Thus, according to question,
`7x + 6 y = 3800` ------------(i)
`3x + 5y = 1750` --------------(ii)
`=> 3x = 1750 - 5y`
`=> x = (1750 - 5y)/3` ------------(iii)
Now, by substituting the value of `x` in equation (i), we get
`7 (1750 - 5y)/3 + 6y = 3800`
`=> (12250-35y)/3+6y = 2800`
`=> ((12250-35y) + 18y)/3 = 3800`
`=> 12250 - 35y +18y = 3800xx3`
`=> 12250 +17y = 11400`
`=> -17 y = 11400 - 12250`
`=> -17 y = -850`
`=> y = (-850)/(-17)=50`
Now, by substituting the value of `y` in equation (ii), we get
`3x+5xx50 = 1750`
`=> 3x+250 = 1750`
`=> 3x = 1750-250`
`=>3x = 1500`
`:. x = 1500/3 = 500`
Thus, cost of one bat = Rs 500 and cost of one ball = Rs 50. Answer
(iv) The taxi charges in a city consist of a fixed charge together with the charges for this distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 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?
Solution:
Let the fixed charge = Rs `x`
And charge per km = Rs `y`
Therefore, according to question,
Charge paid for 10 km
`x + 10 y = 105` ------------(i)
And, charge paid for 15 km
`x+15 y = 155` ----------(ii)
`=> x = 155 - 15y` -------------(iii)
Now, by substituting the value of `x` in equation (i), we get
`155-15y +10y = 105`
`=> 155 - 5 y = 105`
`=> -5y = 105 - 155``=> -5y = -50`
`=> y = (-50)/(-5) = 10`
Now, by substituting the value of `y` in equation (i), we get
`x + 10 xx 10 = 105`
`=> x +100 = 105`
`=> x = 105 - 100 = 5`
Thus, fixed charge is Rs 5 and charge per km is equal to Rs 10.
Now, charge for 25 km
= fixed charge + charge per km × 25
`= 5 + 10 xx 25`
`= 5+250 = 255`
Thus,
Fixed charge = Rs 5, charge per km = Rs 10 and charge for 25 km = Rs 255 Answer
(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, fraction is `x/y`
Thus, according to question,
After addition of `2` in numerator and denominator
`(x+2)/(y+2) = 9/11`
After cross multiplication
`=> 11(x+2) = 9(y+2)`
`=> 11x + 22 = 9y + 18`
`=> 11x - 9y =18-22`
`=> 11x - 9y = -4` -----------(i)
And according to question, After addition of `3` to numerator and denominator,
`(x+3)/(y+3) = 5/6`
After cross multiplication,
`=> 6(x+3) = 5(y+3)`
`=> 6x+18 = 5y +15`
`=> 6x-5y = 15-18`
`=> 6x-5y = -3` ----------(ii)
`=> 6x = -3 + 5y`
`=> x = (-3+5y)/6` ---------(iii)
Now, by substituting, the value of `x` from equation (iii) in equation (i), we get
`11xx(-3+5y)/6 - 9y = -4`
`=> (-33+55y)/6 -9y = -4`
`=> (-33+55y-54y)6=-4`
`=> -33 + y = -24`
`=> y = -24 + 33 = 9`
Now, by substituting the value of `y` in equation (iii), we get
`x = (-3+5xx9)/6`
`=> x = (-3+45)/6`
`=>x = 42/6 = 7`
Thus, `x = 7` and `y = 9`
And thus, fraction is `7/9` Answer
(vi) Five year 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, present age of Jacob `= j` years
And present age of his son `= s` years
Thus, according to question, five years hence
`j+5 = 3(s+5)`
`=> j + 5 = 3s+15`
`=> j -3s = 15-5`
`=> j-3s = 10` ----------(i)
And according to question, five years ago
`j-5 = 7(s-5)`
`=> j-5= 7s - 35`
`=> j - 7s = -35+5 = -30` ---------(ii)
Now,
`=> j = -30 +7s` ------ (iii)
Now, by substituting the value `j` in equation (i), we get
`-30 + 7s - 3s = 10`
`=> -30 +4s =10`
`=> 4s = 10+30 = 40`
`=> s = 40/4 = 10`
Now, by substituting the value of `s` in equation (i), we get
`j - 3xx10 = 10`
`=> j - 30 = 10`
`=> j = 10+30 = 40`
Thus, present age of Jacob = 40 years and present age of his son = 10 years. Answer
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