If 6x-y-36 and 3x-y-729- then x2-y2-

Given: #160;6x y=36and #160;3x+y=729 #160;6x y=36 #8658; #160;6x y=62 #8658;x y=2 #160; #160; #160; #160; #160; #160; #160; #160; ...

You visited us 78 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard VII

Mathematics

Types of Brackets

If 6x-y=36 an...

Question

If 6^x–y = 36 and 3^x+y= 729, then x² – y² = _________.

Open in App

Solution

$Given : 6^{x - y} = 36 and 3^{x + y} = 729 6^{x - y} = 36 \Rightarrow 6^{x - y} = 6^{2} \Rightarrow x - y = 2 . . . (1) 3^{x + y} = 729 \Rightarrow 3^{x + y} = 3^{6} \Rightarrow x + y = 6 . . . (2) Adding (1) and (2), we get 2 x = 8 \Rightarrow x = 4 Substituting the value of x in (2), we get 4 + y = 6 \Rightarrow y = 2 Now, x^{2} - y^{2} = 4^{2} - 2^{2} = 16 - 4 = 12$

Hence, x² – y² = 12.

Suggest Corrections

Similar questions

Let $f (x, y) = \sqrt{x^{2} + y^{2}} + \sqrt{x^{2} + y^{2} - 2 x + 1} + \sqrt{x^{2} + y^{2} - 2 y + 1} + \sqrt{x^{2} + y^{2} - 6 x - 8 y + 25} \forall x, y ϵ R, t h e n$

Q. If

(x, y)

is a solution of

x^{2} + y^{2} = 160

and

y = 3 x

, then

y^{2}

is equal to

Q. If

(x^{2} + 1) (y^{2} + 1) + 9 = 6 (x + y)

(where

x, y \in R

), then the value of

(x^{2} + y^{2})

is,

Q. If

\frac{4}{x + y} + \frac{2}{x - y} = 2

then prove that

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

Q. If

{(x + y)}^{3} - z^{2} = 4

{(y + z)}^{2} - x^{2} = 9

and

{(z + x)}^{2} - y^{2} = 36

, then find the value of

x + y + z

Join BYJU'S Learning Program

If 6x–y = 36 and 3x+y = 729, then x2 – y2 = _________.

Given: 6x-y=36and 3x+y=729 6x-y=36⇒ 6x-y=62⇒x-y=2 ...13x+y=729⇒3x+y=36⇒x+y=6 ...2Adding 1 and 2, we get2x=8⇒x=4Substituting the value of x in 2, we get4+y=6⇒y=2Now,x2-y2=42-22 =16-4 =12 Hence, x2 – y2 = 12.

If 6^x–y = 36 and 3^x+y= 729, then x² – y² = _________.