Math, asked by Sid1509cool, 9 months ago

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Answers

Answered by MANTUMEHER
1

Answer:

x=a^2. y=b^2

Step-by-step explanation:

 \frac{x}{a}  +  \frac{y}{b}  = a + b \\  \frac{xb + ya}{ab}  = (a + b) \\ bx + ay = (a + b)ab \: ...........1 \\  \\  \frac{x}{ {a}^{2} }  +  \frac{y}{ {b}^{2} }  = 2 \\  \frac{x {b}^{2}  + y {a}^{2} }{ {a}^{2} {b}^{2}  }  = 2 \\ x {b}^{2}  + y {a}^{2}  = 2 {a}^{2}  {b}^{2}  \: .............2 \\ now \: solving \: the \: eqs\: we \: have \\ eqs \: 1 \times a = abx +  {a}^{2} y = (a + b) {a}^{2} b \: \\  subtracting \: eqs \: 2 \: x {b}^{2}  + y {a}^{2}  = 2 {a}^{2}  {b}^{2}  \\ so \: we \: get \: abx - x {b}^{2}  = (a + b) {a}^{2} b - 2 {a}^{2}  {b}^{2}  \\ xb(a - b) =  {a}^{2} b(a + b - 2b) \\ x =  \frac{ {a}^{2} b(a - b)}{(a - b)b}  =  {a}^{2}  \\  \\ now \: putting \: the \: value \: of \: x \: \: in \: eqs \: 2 we \: get \\  {a}^{2}  {b}^{2}  + y {a}^{2}  = 2 {a}^{2}  {b}^{2}  \\ y {a}^{2}  =  {a}^{2}  {b}^{2}  \\ y =  \frac{ {a}^{2} {b}^{2}  }{ {a}^{2} }  =  {b}^{2}

Answered by vgehlot224
0

Answer:

x=a^2. y=b^2

Step-by-step explanation:

\begin{lgathered}\frac{x}{a} + \frac{y}{b} = a + b \\ \frac{xb + ya}{ab} = (a + b) \\ bx + ay = (a + b)ab \: ...........1 \\ \\ \frac{x}{ {a}^{2} } + \frac{y}{ {b}^{2} } = 2 \\ \frac{x {b}^{2} + y {a}^{2} }{ {a}^{2} {b}^{2} } = 2 \\ x {b}^{2} + y {a}^{2} = 2 {a}^{2} {b}^{2} \: .............2 \\ now \: solving \: the \: eqs\: we \: have \\ eqs \: 1 \times a = abx + {a}^{2} y = (a + b) {a}^{2} b \: \\ subtracting \: eqs \: 2 \: x {b}^{2} + y {a}^{2} = 2 {a}^{2} {b}^{2} \\ so \: we \: get \: abx - x {b}^{2} = (a + b) {a}^{2} b - 2 {a}^{2} {b}^{2} \\ xb(a - b) = {a}^{2} b(a + b - 2b) \\ x = \frac{ {a}^{2} b(a - b)}{(a - b)b} = {a}^{2} \\ \\ now \: putting \: the \: value \: of \: x \: \: in \: eqs \: 2 we \: get \\ {a}^{2} {b}^{2} + y {a}^{2} = 2 {a}^{2} {b}^{2} \\ y {a}^{2} = {a}^{2} {b}^{2} \\ y = \frac{ {a}^{2} {b}^{2} }{ {a}^{2} } = {b}^{2}\end{lgathered}

a

x

+

b

y

=a+b

ab

xb+ya

=(a+b)

bx+ay=(a+b)ab...........1

a

2

x

+

b

2

y

=2

a

2

b

2

xb

2

+ya

2

=2

xb

2

+ya

2

=2a

2

b

2

.............2

nowsolvingtheeqswehave

eqs1×a=abx+a

2

y=(a+b)a

2

b

subtractingeqs2xb

2

+ya

2

=2a

2

b

2

sowegetabx−xb

2

=(a+b)a

2

b−2a

2

b

2

xb(a−b)=a

2

b(a+b−2b)

x=

(a−b)b

a

2

b(a−b)

=a

2

nowputtingthevalueofxineqs2weget

a

2

b

2

+ya

2

=2a

2

b

2

ya

2

=a

2

b

2

y=

a

2

a

2

b

2

=b

2

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