Question

(x) ^1/2 +y = 7 and x + (y) ^1/2 = 11 solve for x and y.

Answer 1

(x)^½ + y = 7
x^½ = 7-y
Squaring on both sides
(x^½)² = (7-y)²
x = 7²+y²-2(7)(y)
x = y²-14y+49

Put x = y²-14y+49 in equation : x + y^½ = 11

y²-14y+49 + (y^½) = 11
Let a = y^½ then y = a²
a^4 - 14a² + 49 + a = 11
a^4 - 14a² + a + 38 = 0
(a-2)(a³+2a²-10a-19) = 0
a-2 = 0
a = 2

Then y = a² = (2)² = 4
x = y²-14y+49 = 4²-14(4)+49
= 16-56+49
= 63-56
= 9

Therefore, x = 9 and y = 4

Hope it helps

Answer 2

Hi ,

√x + y = 7 ----( 1 )

x + √y = 11 ----( 2 )

Subtract equation ( 1 ) from equation

( 2 ),

x + √y - √x - y = 11 - 7

( x - y ) - ( √x - √y ) = 4

[ (√x)² - (√y )² ] - ( √x - √y ) = 4

[( √x + √y ) ( √x - √y )] - ( √x - √y )= 4

( √x - √y ) ( √x + √y - 1 ) = 1 × 4

√x - √y = 1 ------( 3 )

√x + √y - 1 = 4

√x + √y = 4 + 1

√x + √y = 5 ----( 4 )

Add equation ( 1 ) and ( 2 ) we get

2√x = 6

√x = 6 / 2

√x = 3---( 5 )

Do the square of both sides of the

Equation

x = 9

Put √x = 3 in equation ( 1 )

3 + y = 7

y = 7 - 3

y = 4

Therefore ,

x = 9 and y = 4

I hope this helps you.

:)