The values of x and y satisfying (x + 3) (y – 5) = xy + 39 and (x �

1.	The values of x and y satisfying (x + 3) (y – 5) = xy + 39 and (x – 2) (y + 3) = xy – 40 respectively are : (a) 6 and 8 (b) 6 and –8 (c) –6 and 8 (d) –6 and –8
Answer» (c) –6 and 8 (x + 3)(y – 5) = xy + 39 \(\Rightarrow\) xy + 3y – 5x – 15 = xy + 39 \(\Rightarrow\) –5x + 3y = 54  5x – 3y = –54 ..........(i) (x – 2) (y + 3) = xy – 40 \(\Rightarrow\) xy – 2y + 3x – 6 = xy – 40 \(\Rightarrow\) –2y + 3x = –34 \(\Rightarrow\) 3x – 2y = – 34 .............(ii) Multiply (i) by 3 and (ii) by 5 and then subtract eqn (ii) from eqn (i). so that we get solution.

The values of x and y satisfying (x + 3) (y – 5) = xy + 39 and (x – 2) (y + 3) = xy – 40 respectively are : (a) 6 and 8 (b) 6 and –8 (c) –6 and 8 (d) –6 and –8

Answer»

(c) –6 and 8

(x + 3)(y – 5) = xy + 39

\(\Rightarrow\) xy + 3y – 5x – 15 = xy + 39

\(\Rightarrow\) –5x + 3y = 54  5x – 3y = –54 ..........(i)

(x – 2) (y + 3) = xy – 40

\(\Rightarrow\) xy – 2y + 3x – 6 = xy – 40

\(\Rightarrow\) –2y + 3x = –34 \(\Rightarrow\) 3x – 2y = – 34 .............(ii)

Multiply (i) by 3 and (ii) by 5 and then subtract eqn (ii) from eqn (i).

so that we get solution.