## radical syst. of eqn's: xy +sqrt(xy)=90 ; x + y =30

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Luke53
Posts: 54
Joined: Sun Mar 13, 2011 9:46 am
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### radical syst. of eqn's: xy +sqrt(xy)=90 ; x + y =30

I'm stuck again with this system:
xy + sqrt(xy) = 90 (1)
x + y = 30 (2) => x = 30 - y
Substituting x = 30 - y into the first eqn. gives: (30 - y) * y + sqrt(( 30 - y) * y) = 90
-(y^2) + 30y + sqrt(-((y^2) + 30y)) = 90
This looks very messy to work out.
Is there an easier way to do this?
Luke.

Luke53
Posts: 54
Joined: Sun Mar 13, 2011 9:46 am
Contact:

### Re: radical syst. of eqn's: xy +sqrt(xy)=90 ; x + y =30

xy + sqrt(xy) = 90 (1)
x + y = 30 (2) => x = 30 - y
Substituting x = 30 - y into the first eqn. gives: (30 - y) * y + sqrt(( 30 - y) * y) = 90
-(y^2) + 30y + sqrt(-((y^2) + 30y)) = 90
This looks very messy to work out.
Is there an easier way to do this?
Luke.
Luke53

Berichten: 18
Geregistreerd op: zo maart 13, 2011 10:46 am
Think I found the least difficult way to solve this system. Change the first eqn. to:
sqrt(xy) = 90 - x * y Squaring both sides gives: x * y = x² * y² - 180 * x * y + 8100 => x² * y² - 181 *x * y + 8100 = 0
Knowing that x = 30 - y and substituting this into the previous eqn.
((30 - y)²) * y² - 181y * (30 - y) + 8100 = 0 ; working out gives : y^4 - 60 y^3 + 1081y² - 5430y + 8100 = 0
Solving this gives four roots witch are: y = 3 or y = 27 ; y = 3.8197 or y = 26.1803.
So if y = 3 then x must be 27 (out of eqn (2)) and vice versa.
Putting these results into eqn (1) gives: 81 + sqrt(81) = 81 + 9 = 90 (OK!).
Luke.