nbsp-eg no-5 chapter 4 -lesson 3-it is given that nbsp- nbsp-x nbsp- 3- -y nbsp- 1- - nbsp-xy nbsp- 15x nbsp-y nbsp- 1- - 3 nbsp-y nbsp- 1- - nbsp-xy nbsp- 15xy nbsp- nbsp-x nbsp- 3y nbsp- 3 - nbsp-xy nbsp- 15- nbsp-x nbsp- 3y nbsp- 3 - - 15- nbsp-x nbsp- 3y nbsp- 12but shouldnt the ans nbsp-have been x-3y-12- nbsp-

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Concept of Inequality

nbsp;eg no-5 ...

Question

eg no-5 chapter 4 ,lesson 3,it is given that (x + 3) (y – 1) = xy – 15
x (y – 1) + 3 (y – 1) = xy – 15
xy – x + 3y – 3 = xy – 15
– x + 3y – 3 = – 15
∴ x – 3y = 12
but shouldnt the ans have been x+3y=12.

Open in App

Solution

Suggest Corrections

Similar questions

\frac{x^{- 3} - y^{- 3}}{x^{- 3} y^{- 1} + (x y)^{- 2} + y^{- 3} x^{- 1}}

\frac{1}{2 x} + \frac{1}{3 y} = 2; \frac{1}{3 x} + \frac{1}{2 y} = \frac{13}{6}

\frac{2}{\sqrt{x}} + \frac{3}{\sqrt{y}} = 2; \frac{4}{\sqrt{x}} - \frac{9}{\sqrt{y}} = - 1

\frac{4}{x} + 3 y = 14; \frac{3}{x} - 4 y = 23

\frac{5}{x - 1} + \frac{1}{y - 2} = 2; \frac{6}{x - 1} - \frac{3}{y - 2} = 1

\frac{7 x - 2 y}{x y} = 5; \frac{8 x + 7 y}{x y} = 15

6 x + 3 y = 6 x y; 2 x + 4 y = 5 x y

\frac{10}{x + y} + \frac{2}{x - y} = 4; \frac{15}{x + y} - \frac{5}{x - y} = - 2

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

Factorize 15x + 3y + xy + 45 and give its value for x = 2 and y = 3.
[3 MARKS]

x^{2} + y^{2} + x y = 1

x, y \in R,

x^{3} y + x y^{3} + 4

x^{2} - x y + y^{2} - x + y + 3

- x^{2} + 3 y^{2} - 4 x y + 1

Join BYJU'S Learning Program