Question

Solve the equations
$\sf \: \sqrt{ {x}^{2} + 12y } + \sqrt{ {y}^{2} + 12x} = 33$

and

$\sf \: x + y = 23$

Spammers please far away.

Brainly Mods, Brainly stars or best user or genius. Please answer.

Don't co py and pa ste from Bro wser!!!!​

Answer 1

Given :-

• $\sf \: \sqrt{ {x}^{2} + 12y } + \sqrt{ {y}^{2} + 12x} = 33$

• $\sf x + y = 23$

To find :-

• Value of a and b

Solution :-

$\sf \implies\: \sqrt{ {x}^{2} + 12y } + \sqrt{ {y}^{2} + 12x} = 33$

$\sf \implies\: \sqrt{ {x}^{2} + 12y } + \sqrt{ {y}^{2} + 12x} =10 + 23$

$\sf \implies\: \sqrt{ {x}^{2} + 12y } + \sqrt{ {y}^{2} + 12x} =10 + (x + y)$

Squaring both sides

$\sf \implies\:( \sqrt{ {x}^{2} + 12y } + \sqrt{ {y}^{2} + 12x})^{2} =(10 + x + y)^{2}$

❒ Apply identity :-

• $\tt(A + B)^2 = A^2 + B^2+ 2AB$

• $\tt(A+B+C)^2 = A^2 +B^2 + C^2+ 2AB + 2BC + 2CA$

we get . . .

$\sf \implies\: {x}^{2} + 12y + {y}^{2} + 12x + 2 \Big( \sqrt{( {x}^{2} + 12y)( {y}^{2} + 12x) }\Big)=x^2+y^2+100+2xy+20y+20x$

$\sf \implies\: 2 \Big( \sqrt{( {x}^{2} + 12y)( {y}^{2} + 12x) }\Big)=x^2+y^2+100+2xy+20y+20x - {x}^{2} - {y}^{2} - 12y - 12x$

$\sf \implies\: 2 \Big( \sqrt{( {x}^{2} + 12y)( {y}^{2} + 12x) }\Big) = 8x + 8y + 100 + 2xy$

$\sf \implies\: 2 \Big( \sqrt{( {x}^{2} + 12y)( {y}^{2} + 12x) }\Big) = 8(x + y) + 100 + 2xy$

Divide 2 both sides

$\sf \implies\: \sqrt{( {x}^{2} + 12y)( {y}^{2} + 12x) }= 4(x + y) + 50 + xy$

Substitute value of x+y=23

$\sf \implies\: \sqrt{( {x}^{2} + 12y)( {y}^{2} + 12x) }= 4(23) + 50 + xy$

$\sf \implies\: \sqrt{( {x}^{2} + 12y)( {y}^{2} + 12x) }= 92 + 50 + xy$

$\sf \implies\: \sqrt{( {x}^{2} + 12y)( {y}^{2} + 12x) }= 142 + xy$

Again squaring both sides

$\sf \implies\: ( {x}^{2} + 12y)( {y}^{2} + 12x)= (142+ xy)^{2}$

❒ Apply identity :-

• $\tt (A+B)^2=A^2+2AB+B^2$

$\sf \implies\: {x}^{2} {y}^{2} + 12 {x}^{3} + 12 {y}^{3} + 144xy = {x}^{2} {y}^{2} + 20164 + 284xy$

$\sf \implies\: 12 {x}^{3} + 12 {y}^{3} + 144xy - 284xy- 20164 = 0$

$\sf \implies\: 12 {x}^{3} + 12 {y}^{3} - 140xy- 20164 = 0$

Divide whole equation by 12

$\sf \implies\: {x}^{3} + {y}^{3} - \dfrac{35}{3} xy- \dfrac{5041}{3} = 0$

❒ We know that :-

• $\tt(A+B)^3 = A^3 + B^3 + 3AB (A+B)$

We derived a new equation from this :-

• $\tt A^3 + B^3 = (A+B)^3 - 3AB(A+B)$

Using this we get ::

$\sf \implies\: {\bf{(x + y)^{3} - 3xy(x + y) }} - \dfrac{35}{3} xy- \dfrac{5041}{3} = 0$

Substitute value of ( x + y ) = 23

$\sf \implies\: (23)^{3} - 3xy(23)- \dfrac{35}{3} xy- \dfrac{5041}{3} = 0$

$\sf \implies\:12167 - 69xy- \dfrac{35xy}{3}- \dfrac{5041}{3} = 0$

$\sf \implies \: \dfrac{36501 - 5041}{3} - \dfrac{207xy - 35xy}{3} = 0$

$\sf \implies \: \dfrac{31460}{3} - \dfrac{242xy}{3} = 0$

$\sf \implies \: \dfrac{31460}{3} = \dfrac{242xy}{3}$

$\sf \implies \: 31460 = 242xy$

$\sf \implies \: \dfrac{ 31460}{242} = xy$

$\sf \implies\: 130 = xy$

‎ ‎ ‎

❒ Now we have :-

$\begin{cases} \tt \: x + y = 23 \\ \\ \tt \: xy = 130 \end{cases}$

By using hit and trial, we get :-

Either x = 13 and y = 10 OR x = 10 and y = 13.

This is the required result.