UP Board Solutions for Class 10 Maths Chapter 3 Exercise 3.3 Pair Of Linear Equations In Two Variables
Exercise 3.3
1.Solve the following pair of linear equations by the substitution method :
Sol. (i) The given system of equations is x +3 = 14 …(1)
and, x-y = 4 …(2)
Substituting y = 14-x in (2), we get x – (14 -x) = 14 => x-14 x = 4 => 2x = 4 + 14 => 2x = 18 => x= 9 Putting x = 9 in (1), we get 9+y = 14 => y = 14 – 9 => y =5 Hence, the solution of the given system of equations is
x= 9, y = 5
(ii) The given system of equations is
s-t = 3 …(1)
From (1), s =3+1 Substituting 8 = 3 + t in (2), we get
=> 2(3+t) +3t =36
=> 6 + 2t + 3t = 36 => 5t = 30 =› t = 6 Putting t = 6 in (1), we get s – 6 = 3 => s = 3 + 6 = 9 Hence, the solution of the given system of equations is s= 9,t= 6 (iii) The given system of equations is 3x – y = 3 …(1) and, 9x – 3y = 9 … (2) From (1), y = 3x – 3 Substituting y = 3x – 3 in (2), we get 9x – 3(3x – 3) = 9 => 9x – 9x + 9 = 9 => 9 = 9 This statement is true for all values of x. However, we do not get a specific value of x as a solution. So, we cannot obtain a specific value of y. This situation has arisen because both the given equations are the same. * Equations (1) and (2) have infinitely many solutions.
(iv) The given system of equations is 0.2x + 0.3y = 1.3 => 2x + 3y = 13 …(1) and, 0.4x + 0.5y = 2.3 => 4x + 5y = 23 …(2)
(vi) The given system of equations is
2.Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m‘ for which y = mx + 3.
Sol. The given system of equations is 2x + 3y = 11 …(1) and, 2x – 4y = – 24 …(2) From (1), 3y = 11 – 2x = >
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. (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 for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each ball. (iv) The taxi change in a city consist of a fixed change together with the change for the distance covered .For a distance of 10km ,the charge paid is Rs 105 and for a journey of 15km ,the change paid is Rs 155. what are the fixed charges and the 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 , if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator, it becomes 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?
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Hence, the cost of one bat is Rs 500 and the cost of one ball is Rs 50. (iv) Let the fixed charges of taxi be Rs ~ per km and the running charges be Rs y km/hr. According to the given condition, we have x + 10y = 105 …(1)
x+ 15y = 155 …(2) From (1), X = 105 – 10y Substituting x = 105 – 10y in (2), we get
105 – 10y + 15y = 155 = > 105 + 5y = 155 = > 5y = 155 – 105 = > 5y = 50 => y = 10 Putting y = 10 in (1), we get
x + 10 × 10 = 105 => x = 105 – 100 => х = 5 Total charges for travelling a distance of 25 km = x + 25y = Rs (5 + 25 × 10) => Rs =255 Hence, the fixed charge is Rs 5, the charge per km is Rs 10 and the total charge for travelling a distance of 25 km is Rs 255. (v) Let the fraction be Then, according to the given conditions, we have
(vi)Let the present age of Jacob be x years and the present age of his son be y years. Five years hence, Jacob’s age = (x + 5) years Son’s age = (y +5) years Five years ago, Jacob’s age = (x – 5) years
, if 2 is added to both
Find the fraction
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