» Unlabelled » NCERT Class 10 Maths Solutions Chapter - 3 Pairs of Linear Equations in Two Variables, Ex-3.6

NCERT Class 10 Maths Solutions Chapter - 3 Pairs of Linear Equations in Two Variables, Ex-3.6

Ex - 3.6

Question 1.  Solve the following pairs of equations by reducing them to a pair of linear equations:


Solution


Using cross-multiplication method, we obtain:
Concept insight: Here, the two given equations are not linear as the variables x and y both are in the denominator. By substituting  given equation reduces to 

linear equation in variables p and q. Note here that we are assuming the variables x and y to be non-zero, because only then the given equations will make sense. The linear equations can be solved by using any suitable algebraic method. Also, do not forget here that we need to find the values of x and y. So, put the values of p and q to find x and y.

Multiplying equation (1) by 3, we obtain:
6p + 9q = 6       (3)
Adding equation (2) and (3), we obtain:

Concept insight: Here, the two given equations are not linear as it involves  x-1/2  and y-1/2  If we substitute  in the equations, then they will reduce to 

linear form in variables p and q. Note here that we are assuming the variables x and y to be non negative real numbers. The linear equations can be solved by using any suitable algebraic method. Also, do not forget here that we need to find the values of x and y. So, put the values of p and q to find x and y.

Concept insight: Here, the two given equations are not linear as the variable x is in the denominator. If we substitute   in the equations, then they will reduce to linear form in 

variables p and y. Note here that we are assuming the variable x to be non-zero. The linear equations can be solved by using any suitable algebraic method. Also, do not forget here that we need to find the values of x and y. So, put the value of p to find x.

Concept insight: Here, the two given equations are not linear as the expressions on the LHS of both the equations are not linear. If we substitute   in the 

equations, then they will reduce to linear form in variables p and q. Note here that we are assuming x  1 and y  2. The linear equations can be solved by using any suitable algebraic method. Also, do not forget here that we need to find the values of x and y. So, put the values of p and q to find x and y.


Concept insight: Here, the two given equations are not linear as the expressions on the LHS of both the equations are not linear. So, here our first step will be to break the expression on the LHS. Thereafter, if we will substitute  in the equations, then they will 

reduce to linear form in variables p and q. Note here that we are assuming x  0 and y  0. The linear equations can be solved by using any suitable algebraic method. Also, do not forget here that we need to find the values of x and y. So, put the values of p and q to find x and y.

Concept insight: Here, the two given equations are not linear as the expressions on the LHS of both the equations are not linear. So, first step will be to divide both LHS and RHS by xy. Thereafter, substitute  in the equations, then they will reduce to linear 

form in variables p and q. Note here that we are assuming x  0 and y  0. The linear equations can be solved by using any suitable algebraic method. Also, do not forget here that we need to find the values of x and y. So, put the values of p and q to find x and y.

Concept insight: Here, the two given equations are not linear as the expressions on the LHS are not linear. If we substitute  in the equations, then they will 

reduce to linear form in variables p and q. Note here that we are assuming x + y  0 and x - y  0, otherwise  the equations will not make sense. The linear equations can then be solved by using any suitable algebraic method. Also, do not forget here that we need to find the values of x and y. So, put the values of p and q and two linear equations x + y = 5 and x - y = 1 will be obtained. And these equations can be solved easily by elimination method.

The given equations reduce to:
Concept insight: Here, the two given equations are not linear as the expressions on the LHS are not linear. If we substitute  in the equations, then they will 

reduce to linear form in variables p and q. Note here that we are assuming 3x + y  0 and 3x - y  0. The linear equations can be solved by using any suitable algebraic method. Also, do not forget here that we need to find the values of x and y. So, put the values of p and q to obtain two linear equations 
3x + y = 4 and 3x - y = 2. These equations can then be solved easily by elimination method.

Question 2.  Formulate the following problems as a pair of equations, and hence find their solutions:
 (i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
 (ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.
 (iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

Solution

(i)  Let the speed of Ritu in still water and the speed of stream be x          km/h and y km/h respectively.
Speed of Ritu while rowing upstream = (x-y) km/h
Speed of Ritu while rowing downstream = (x + y) km /h
According to the question,
Adding equations (1) and (2), we obtain:
2x = 12
x = 6
Putting the value of x in equation (1), we obtain:
y = 4
Thus, Ritu’s speed in still water is 6 km/h and the speed of the current is 4 km/h.
Concept insight: Two unknown quantities are speed in still water and the speed of the current, which can be represented by variables x and y respectively. Then using the given conditions, two linear equations can be formed. Remember the speed while rowing upstream will be x + y and while rowing downstream, it will be x – y. Now, the two equations can be solved easily by the substitution method.

(ii)        Let the number of days taken by a woman and a man to finish the work be x and y respectively.
Work done by a woman in 1 day = 
Work done by a man in 1 day =    
According to the question,
(iii)     Let the speed of train and bus be u km/h and v km/h respectively.

Concept insight: In this problem it is mentioned that a girl Roohi travels to her home partly by train and partly by bus. So, we represent the speed of the train and bus by variables u and v respectively. Here, to answer this problem, remember the fact that speed is the ratio of distance and time. Using this and the given conditions, formulate pair of linear equations. Observe that the equations obtained are not linear. So, first step will be to convert them into linear form by suitable substitution and then solve it using an appropriate algebraic method.


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Ex - 3.6

Question 1.  Solve the following pairs of equations by reducing them to a pair of linear equations:


Solution


Using cross-multiplication method, we obtain:
Concept insight: Here, the two given equations are not linear as the variables x and y both are in the denominator. By substituting  given equation reduces to 

linear equation in variables p and q. Note here that we are assuming the variables x and y to be non-zero, because only then the given equations will make sense. The linear equations can be solved by using any suitable algebraic method. Also, do not forget here that we need to find the values of x and y. So, put the values of p and q to find x and y.

Multiplying equation (1) by 3, we obtain:
6p + 9q = 6       (3)
Adding equation (2) and (3), we obtain:

Concept insight: Here, the two given equations are not linear as it involves  x-1/2  and y-1/2  If we substitute  in the equations, then they will reduce to 

linear form in variables p and q. Note here that we are assuming the variables x and y to be non negative real numbers. The linear equations can be solved by using any suitable algebraic method. Also, do not forget here that we need to find the values of x and y. So, put the values of p and q to find x and y.

Concept insight: Here, the two given equations are not linear as the variable x is in the denominator. If we substitute   in the equations, then they will reduce to linear form in 

variables p and y. Note here that we are assuming the variable x to be non-zero. The linear equations can be solved by using any suitable algebraic method. Also, do not forget here that we need to find the values of x and y. So, put the value of p to find x.

Concept insight: Here, the two given equations are not linear as the expressions on the LHS of both the equations are not linear. If we substitute   in the 

equations, then they will reduce to linear form in variables p and q. Note here that we are assuming x  1 and y  2. The linear equations can be solved by using any suitable algebraic method. Also, do not forget here that we need to find the values of x and y. So, put the values of p and q to find x and y.


Concept insight: Here, the two given equations are not linear as the expressions on the LHS of both the equations are not linear. So, here our first step will be to break the expression on the LHS. Thereafter, if we will substitute  in the equations, then they will 

reduce to linear form in variables p and q. Note here that we are assuming x  0 and y  0. The linear equations can be solved by using any suitable algebraic method. Also, do not forget here that we need to find the values of x and y. So, put the values of p and q to find x and y.

Concept insight: Here, the two given equations are not linear as the expressions on the LHS of both the equations are not linear. So, first step will be to divide both LHS and RHS by xy. Thereafter, substitute  in the equations, then they will reduce to linear 

form in variables p and q. Note here that we are assuming x  0 and y  0. The linear equations can be solved by using any suitable algebraic method. Also, do not forget here that we need to find the values of x and y. So, put the values of p and q to find x and y.

Concept insight: Here, the two given equations are not linear as the expressions on the LHS are not linear. If we substitute  in the equations, then they will 

reduce to linear form in variables p and q. Note here that we are assuming x + y  0 and x - y  0, otherwise  the equations will not make sense. The linear equations can then be solved by using any suitable algebraic method. Also, do not forget here that we need to find the values of x and y. So, put the values of p and q and two linear equations x + y = 5 and x - y = 1 will be obtained. And these equations can be solved easily by elimination method.

The given equations reduce to:
Concept insight: Here, the two given equations are not linear as the expressions on the LHS are not linear. If we substitute  in the equations, then they will 

reduce to linear form in variables p and q. Note here that we are assuming 3x + y  0 and 3x - y  0. The linear equations can be solved by using any suitable algebraic method. Also, do not forget here that we need to find the values of x and y. So, put the values of p and q to obtain two linear equations 
3x + y = 4 and 3x - y = 2. These equations can then be solved easily by elimination method.

Question 2.  Formulate the following problems as a pair of equations, and hence find their solutions:
 (i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
 (ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.
 (iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

Solution

(i)  Let the speed of Ritu in still water and the speed of stream be x          km/h and y km/h respectively.
Speed of Ritu while rowing upstream = (x-y) km/h
Speed of Ritu while rowing downstream = (x + y) km /h
According to the question,
Adding equations (1) and (2), we obtain:
2x = 12
x = 6
Putting the value of x in equation (1), we obtain:
y = 4
Thus, Ritu’s speed in still water is 6 km/h and the speed of the current is 4 km/h.
Concept insight: Two unknown quantities are speed in still water and the speed of the current, which can be represented by variables x and y respectively. Then using the given conditions, two linear equations can be formed. Remember the speed while rowing upstream will be x + y and while rowing downstream, it will be x – y. Now, the two equations can be solved easily by the substitution method.

(ii)        Let the number of days taken by a woman and a man to finish the work be x and y respectively.
Work done by a woman in 1 day = 
Work done by a man in 1 day =    
According to the question,
(iii)     Let the speed of train and bus be u km/h and v km/h respectively.

Concept insight: In this problem it is mentioned that a girl Roohi travels to her home partly by train and partly by bus. So, we represent the speed of the train and bus by variables u and v respectively. Here, to answer this problem, remember the fact that speed is the ratio of distance and time. Using this and the given conditions, formulate pair of linear equations. Observe that the equations obtained are not linear. So, first step will be to convert them into linear form by suitable substitution and then solve it using an appropriate algebraic method.


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