Question

hey

$\sqrt{x} + y = 7 \\ x + \sqrt{y} = 11$
find the values of x and y​

Answer 1

x+√y=7

√x+y=11

from eq 1...

√y=7-x

y=(7-x)^2

from eq 2...

√x=11-y

substitute (7-x)^2 for y

√x=11-(7-x)^2

square both sides

x=121-22(7-x)^2+(7-x)^4

(7-x)^4=(x-7)^2*(x-7)^2=(x^2-14x+49)(x^2-14x-49)

x^4-28x^3+294x^2-1372x+2401

x=121-22x^2+308x-1078+x^4-28x^3+294x^2-1372x+2401

0=-x+121-22x^2+308x-1078+x^4-28x^3+294x^2-1372x+2401

x^4-28x^3+272x^2-1065x+1444=0

from the rational root theorem, possible roots are factors of 1444

4 is a factor of 1444 and a root of the equation...

256-1792+4352-4260+1444=0

-1536+4352-4260+1444=0

2816-4260+1444=0

-1444+1444=0

0=0

x=4
x+√y=7
4+√y=7
√y=7-4
√y=3
square both sides
y=9
x=4 and y=9

Hope it helps u..

Answer 2

Answer:

x = 9 and b = 4

Step-by-step explanation:

Given that-

$\sf \sqrt{x} + y = 7 \\ \sf x + \sqrt{y} = 11$

Let us consider, √x = a and √y = b,

$\sf a + b^{2} = 7 \: \: ... (i)\\ \\ \sf a {}^{2} + b = 11 \: \:...( ii)$

Multiplying equation (i) by a² and equation (ii) by a, we get -

$\sf {a}^{2} (a + {b}^{2} ) = {a}^{2} \times 7 \\ \\ \implies \sf {a}^{3} + {a}^{2} {b}^{2} = 7 {a}^{2} \: \: ....( iii)$

And,

$\sf a( {a}^{2} + b) = a \times 11 \\ \\ \implies \sf {a}^{3} + ab = 11a \: \: \:....( iv)$

Subtracting equation (iii) from (iv)-

$\sf {a}^{3} + {a}^{2} {b}^{2} - ({a}^{3} + ab) = 7 {a}^{2} - 11a \\ \\ \implies \sf {a}^{3} + {a}^{2} {b}^{2} - {a}^{3} - ab= 7 {a}^{2} - 11a \\ \\ \implies \sf {a}^{2} {b}^{2} - ab = 7 {a}^{2} - 11a \\ \\ \implies \sf ab(ab - 1) = a(7a - 11) \\ \\ \implies \sf b(ab - 1) = 7a - 11 \\ \\ \implies \sf ab {}^{2} - b = 7a - 11 \\ \\ \implies \sf - b = 7a - ab {}^{2} - 11 \\ \\ \implies \sf - b = a(7 - b {}^{2} ) - 11 \\ \\ \implies \sf a(7 - {b}^{2} ) = 11 - b \\ \\ \implies \sf a = \frac{11 - b}{7 - {b}^{2} }$

By using hit and trial method, we will find the value of b.

Let b = 1,2...

When b = 2,

$\implies \sf a = \frac{11 - b}{7 - {b}^{2} } \\ \\ \implies \sf a = \frac{11 - 2}{7 - 2 {}^{2} } \\ \\ \implies \sf a = \frac{9}{3} \\ \\ \implies \sf a = 3$

Now, find x and y:

$\implies \sf \sqrt{x} = a \\ \\ \implies \sf \sqrt{x} = 3 \\ \\ \bf squaring \: both \: sides : \\ \\ \boxed{ \therefore \: \bf x = 9} \\ \\ \implies \sf \sqrt{y} = b \\ \\ \implies \sf \sqrt{y} = 2 \\ \\ \bf squaring \: both \: sides : \\ \\ \boxed{ \therefore \: \bf y= 4}$

Hence, the value of x is 9 and y is 4.