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Q2.    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.

Let the speed of Ritu in still water be x and speed of current be y, Let's solve this problem by using relative motion concept, the relative speed when they are going in the same direction (downstream)= x +y  the relative speed when they are going in the opposite direction (upstream)= x - y Now, As we know, Relative distance = Relative speed * time . So, According to the question, And, Now,...

Q1.    Solve the following pairs of equations by reducing them to a pair of linear equations:

(viii)    $\\\frac{1}{3x + y} + \frac{1}{3x -y} = \frac{3}{4}\\ \frac{1}{2(3x+y)} - \frac{1}{2(3x -y)} = \frac{-1}{8}$

Given Equations, Let,  Now, our equation becomes And Now, Adding (1) and (2), we get Putting this value in (1) Now, And Now, Adding (3) and (4), we get Putting this value in (3), Hence,

Q1.    Solve the following pairs of equations by reducing them to a pair of linear equations:

(vii)    $\\\frac{10}{x + y} + \frac{2}{x - y}= 4\\ \frac{15}{x+y} - \frac{5}{x - y} = -2$

Given Equations, Let,  Now, our equation becomes And By Cross Multiplication method, Now, And, Adding (3) and (4) we get, Putting this value in (3) we get, And Hence,

Q1.    Solve the following pairs of equations by reducing them to a pair of linear equations:

(vi)    $\\6x + 3y = 6xy\\ 2x + 4y = 5 xy$

Given Equations, Let,  Now, our equation becomes And By Cross Multiplication method, And Hence,

Q1.    Solve the following pairs of equations by reducing them to a pair of linear equations:

(v)    $\\\frac{7x - 2y}{xy} = 5\\ \frac{8x + 7y}{xy} = 15$

Given Equations, Let,  Now, our equation becomes And By Cross Multiplication method, And Hence,

Q1.    Solve the following pairs of equations by reducing them to a pair of linear equations:

(iv)    $\\\frac{5}{x - 1} + \frac{1}{y -2} = 2\\ \frac{6}{x-1} - \frac{3}{y -2} =1$

Given Equations, Let,  Now, our equation becomes And Multiplying (1) by 3 we get Now, adding (2) and (3) we get Putting this in (2) Now, Hence,

Q1.    Solve the following pairs of equations by reducing them to a pair of linear equations:

(iii)    $\\\frac{4}{x} + 3y = 14\\ \frac{3}{x} - 4y = 23$

Given Equations, Let,  Now, our equation becomes And By Cross Multiplication method, And Hence,

Q1.    Solve the following pairs of equations by reducing them to a pair of linear equations:

(ii)    $\\ \frac{2}{\sqrt x} + \frac{3}{\sqrt y} = 2\\ \frac{4}{\sqrt x} - \frac{9}{\sqrt y} = -1$

Given Equations, Let,  Now, our equation becomes And By Cross Multiplication method, So, . And hence

Q1.    Solve the following pairs of equations by reducing them to a pair of linear equations:

(i)    $\\\frac{1}{2x} +\frac{1}{3y} = 2\\ \frac{1}{3x} + \frac{1}{2y} = \frac{13}{6}$

Given Equations, Let,  Now, our equation becomes And By Cross Multiplication method, And Hence,
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