# Solve each of the following systems of equations by using the method of cross multiplication: We have, …(i) …(ii)

Let 1/(x + y) = p and 1/(x - y) = q. Now,

From equation (i), 5p – 2q + 1 = 0 …(iii)

From equation (ii), 15p + 7q – 10 = 0 …(iv)

From equation (iii), we get a1 = 5, b1 = - 2 and c1 = 1

And from equation (iv), we get a2 = 15, b2 = 7 and c2 = - 10

Using cross multiplication,      and  and p = 1/5 and q = 1 and [ p = 1/(x + y) and q = 1/(x - y)]

To solve these, we need to take reciprocal of these equations. By taking reciprocal, we get

x + y = 5 and x – y = 1

Rearranging them again,

x + y – 5 = 0 …(v)

x – y – 1 = 0 …(vi)

From equation (v), we get a1 = 1, b1 = 1 and c1 = - 5

And from equation (vi), we get a2 = 1, b2 = - 1 and c2 = - 1

Using cross multiplication,      and  and x = 3 and y = 2

Thus, x = 3, y = 2

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos  Dealing With the Real Life Problems53 mins  Quiz | Solution of Linear Equations53 mins  Champ Quiz | Consistency and Inconsistency of Solutions36 mins  Pair of Linear Equations in Two Variables46 mins  Smart Revision | Important Word Problems37 mins  Dealing with the Real Life Problems54 mins  Quiz | Real Life Problems Through Linear Equations56 mins  Elimination (quicker than quickest)44 mins  Bonus on Applications of Linear Equations in Two Variables43 mins  NCERT I Quiz on Solution of Linear Equations in Two Variables49 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses 