# Hard simultaneous equations

1. Jan 28, 2010

### benf.stokes

Hi

How do you solve this equations analyticaly without solving a 4th degree polynomial
sqrt(x)+y=11; x+sqrt(y)=7

Thanks

2. Jan 28, 2010

### Kyouran

I'll give a general outline on how to solve this:

First of all, use new variables. a = sqrt(x) and b = sqrt(y). That will remove those ugly square roots for now.

Then, add the 2 equations and form a new equation a + a² + b + b² = 18. In this equation, 18 can be split up into 2 unknown values c and d to form 2 quadratic equations. Of these 2 values, you only need to know 1 to calculate the other, that is d = 18 - c. The roots of these 2 quadratic equations can be written in terms of these 2 values. Keeping in mind that the solutions can only be positive, that should lead you to the right answer

Last edited: Jan 28, 2010
3. Jan 28, 2010

### HallsofIvy

Staff Emeritus
I may be misunderstanding but you seem to be saying that if you have $a+ a^2+ b+ b^2= 18= c+ d$ then you must have $a+ a^2= c$ and $b+ b^2= d$. That is not true.

Last edited: Jan 28, 2010
4. Jan 28, 2010

### Kyouran

Do u want to explain why this is not true? If I define c as being a+a², and d as being b+b², then a + a² + b + b² = c + d = 18

Edit: nvm, you probably did misunderstand me^^

5. May 8, 2010

### farook

can you plees put full steps for this equation plees

6. Sep 24, 2010

### mgpodbury

Hello I'm 14 years old so i don't know what a 4th degree polynomial is... however i agree with the guy named Kyouran. This is how i did it:

sqrt(x) + y = 11 ...(1)
sqrt(y) + x = 7 ...(2)

Let sqrt(x) = a, and sqrt(y) = b.

The eqs will then be:
a + b^2 = 11 ...(3)
b + a^2 = 7 ...(4)

Add them together and you get:

a + a^2 + b + b^2 = 18

The way i solved it is just by doing guess and check. This only takes common sense:

I first tried a = 2 and b = 3 which when substituted into the equation equals 18:

2 + 2^2 + 3 + 3^2 = 18

However when subbing them into eqs 3 and 4, they weren't correct:

So i then switched the values around saying a = 3, and b = 2 which when subbed into eqs 3 and 4 were correct!

And so, as said before...

Since I let sqrt(x) = a, and the sqrt(y) = b:

sqrt (x) = 3
and so therefore x = 9

sqrt (y) = 2
and so therefore y = 4

therefore x = 9, y = 4 ;)

Now you will find, that when subbing these x and y values into your original eqs (1) and (2), they are sure to be correct!

Last edited: Sep 24, 2010