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Solve for x,ySquareroot x + squareroot y=squareroot 2 + 1 x-y=1

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singup | Student, Grade 11 | eNoter

Posted January 24, 2012 at 5:49 PM via web

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Solve for x,y

Squareroot x + squareroot y=squareroot 2 + 1

x-y=1

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted January 24, 2012 at 6:04 PM (Answer #1)

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You need to solve the system of simultaneous equations:

`sqrtx + sqrty = sqrt2 + 1`

`` `x - y = 1`

You may use substitution to solve the system, hence you may use the second equation to write x in terms of y such that:

`x= y+1`

Plugging `x = y+1`  in the first equation yields:

`sqrt(y+1) + sqrt y = sqrt 2+ 1`

You need to raise to square both sides such that:

`y + 1 + 2sqrt(y(y+1)) + y = 2 + 2sqrt2 + 1`

Reducing like terms both sides yields:

`2y + 2sqrt(y(y+1)) = 2 + 2sqrt2`

Reducing by 2 both sides yields:

`y + sqrt(y^2 + y) = 1 + sqrt2`

Keeping the square root to the left side yields:

`sqrt(y^2 + y) = 1 + sqrt2 - y`

Raising to square both sides yields:

`y^2 + y = (1+sqrt2)^2 - 2y(1+sqrt2) + y^2`

Reducing `y^2`  both sides yields:

`y = 1 + 2sqrt2 + 2 - 2y - 2sqrt2*y`

`` `y + 2y + 2sqrt2*y = 3 + 2sqrt2`

`3y + 2sqrt2*y = 3 + 2sqrt2`

Factoring out y to the left side yields:

`y(3 + 2sqrt2) = 3 + 2sqrt2 =gt y = 1`

`x = y + 1 =gt x = 2`

Hence, the solution to the system of equations is `x = 2 ; y = 1.`

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