Math, asked by rakshit814, 1 year ago

Solve for x and y, using cross-multiplication method:
bx/a + ay/b = a^2 + b^2 ; x+y=2ab

Answers

Answered by BrainlyQueen01
21
hey dear friend here's ur answer :-)
 \frac{bx}{a}  +  \frac{ay}{b}  = a ^{2}  + b^{2} .......eq(1) \\ x + y = 2ab........eq(2)
now... \\  \\ from \: eq.(1) \\  \frac{b ^{2} x + a ^{2}y }{ab}  = a {}^{2}  + b {}^{2}  \\  \\  b {}^{2} x+ a ^{2}y  = a {}^{3} b + b {}^{3} a \\  \\ b {}^{2} x -  b{}^{3} a = a {}^{3} b - a {}^{2} y \\  \\ b {}^{2} (x - ab) = a {}^{2} (ab - y) \\  \\ now \: from \: eq \: (2) \\  \\ b {}^{2} ( \frac{x - x}{2}  +  \frac{y}{2} ) = a {}^{2} ( \frac{x}{2}  +  \frac{y - y}{2} ) \\  \\ b {}^{2} ( \frac{x}{2}  -  \frac{y}{2} ) = a {}^{2} ( \frac{x}{2}  -  \frac{y}{2} ) \\  \\ b {}^{2} x - b {}^{2} y = a {}^{2} x - a {}^{2}y \\  \\ x(b {}^{2}  - a {}^{2} ) = y(b {}^{2}  - a {}^{2} ) \\  \\ x = y \\  \\ now \: put \: the \: value \: of \: x \: in \: eq \: (2) \\ x + y = 2ab \\ y + y = 2b \\ 2y = 2ab \\  \\ hope \: it \: helps \: u \:

BrainlyQueen01: plzz mark it as branliest
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Answered by harshfoujdar
4

Answer:

hey dear friend here's ur answer :-)

\begin{lgathered}\frac{bx}{a} + \frac{ay}{b} = a ^{2} + b^{2} .......eq(1) \\ x + y = 2ab........eq(2)\end{lgathered}

a

bx

+

b

ay

=a

2

+b

2

.......eq(1)

x+y=2ab........eq(2)

\begin{lgathered}now... \\ \\ from \: eq.(1) \\ \frac{b ^{2} x + a ^{2}y }{ab} = a {}^{2} + b {}^{2} \\ \\ b {}^{2} x+ a ^{2}y = a {}^{3} b + b {}^{3} a \\ \\ b {}^{2} x - b{}^{3} a = a {}^{3} b - a {}^{2} y \\ \\ b {}^{2} (x - ab) = a {}^{2} (ab - y) \\ \\ now \: from \: eq \: (2) \\ \\ b {}^{2} ( \frac{x - x}{2} + \frac{y}{2} ) = a {}^{2} ( \frac{x}{2} + \frac{y - y}{2} ) \\ \\ b {}^{2} ( \frac{x}{2} - \frac{y}{2} ) = a {}^{2} ( \frac{x}{2} - \frac{y}{2} ) \\ \\ b {}^{2} x - b {}^{2} y = a {}^{2} x - a {}^{2}y \\ \\ x(b {}^{2} - a {}^{2} ) = y(b {}^{2} - a {}^{2} ) \\ \\ x = y \\ \\ now \: put \: the \: value \: of \: x \: in \: eq \: (2) \\ x + y = 2ab \\ y + y = 2b \\ 2y = 2ab \\ \\ hope \: it \: helps \: u \:\end{lgathered}

now...

fromeq.(1)

ab

b

2

x+a

2

y

=a

2

+b

2

b

2

x+a

2

y=a

3

b+b

3

a

b

2

x−b

3

a=a

3

b−a

2

y

b

2

(x−ab)=a

2

(ab−y)

nowfromeq(2)

b

2

(

2

x−x

+

2

y

)=a

2

(

2

x

+

2

y−y

)

b

2

(

2

x

2

y

)=a

2

(

2

x

2

y

)

b

2

x−b

2

y=a

2

x−a

2

y

x(b

2

−a

2

)=y(b

2

−a

2

)

x=y

nowputthevalueofxineq(2)

x+y=2ab

y+y=2b

2y=2ab

hopeithelpsu

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