How do I solve simultaneous equations with squared terms and fractions?

[nirvana1990]

[SOLVED] Simultaneous equations

Homework Statement

Solve: x^2+y^2=10 and 1/x+1/y=4/3

Homework Equations

er... I've tred substituting y=mx into eliminate y

The Attempt at a Solution

I tried squaring the2nd equation to give: 1/x^2+1/y^2+2/xy=16/9
Then I substituted y^2=10-x^2 into give: 1/x^2+(1/10-x^2)+2/xy=16/9

 

[EnumaElish]

I would've tried to obtain one of the variables in terms of the other from the 2nd eq., then substitute that into the 1st.

 

[arildno]

Note that from the 2. equation, you have:
$$2xy=\frac{3}{2}(x+y)$$

Add this to your first equation, yielding:
$$(x+y)^{2}=10+\frac{3}{2}(x+y)$$

This is a quadratic equation in the unknown (x+y) that is easily solved.

 

[dynamicsolo]

Homework Equations

er... I've tred substituting y=mx into eliminate y

This relation won't help you because it is not supported by either of the equations you are given. There is nothing that tells us that x and y are related by a linear equation.

The Attempt at a Solution

I tried squaring the2nd equation to give: 1/x^2+1/y^2+2/xy=16/9
Then I substituted y^2=10-x^2 into give: 1/x^2+(1/10-x^2)+2/xy=16/9

It's good to see that you are wrestling with this. Unhappily, there is no general method for solving systems of *non-linear* equations: you have to look for approaches that suit the particular types of relations that appear in the system. So stumbling around is something everyone does; if you do enough of these sorts of problems, you will gain experience in what works in different situations.

I'd say here that the key thing is to avoid having to deal with square roots or mixed terms containing both x and y, because these create some of the more awkward algebraic equations to solve. You found that squaring the second equation still leaves an *unsquared* x and y, so substitution from the first equation remains problematic. OK, that route won't help, so let's try something else.

Solve the second equation for (1/y). You now have the variables separated (which will be *essential* in getting anywhere with this) and the expression involving x is a difference of two fractions. Put those over a common denominator, so you have (1/y) expressed as a rational function in x. Now you can take the reciprocal to get *y* expressed as a rational function in x. You now have something you can substitute into the first equation. (I didn't say the algebra would be pretty, but you will at least now have something which can be reduced to solving for the roots of a polynomial equation in x. I'll work this through to see how bad it gets...)

 

[nirvana1990]

Thanks for the reply! I've now rearranged the 2nd eq. : 1/x=4/3-1/y and I've inversed it all to get: x=-3/(3y-4), I then substituted this into the 1st eq. and rearranged to get: 9y^4-24y^3-74y^2+240y-151=0 is this right??

 

[dynamicsolo]

Note that from the 2. equation, you have:
$$2xy=\frac{3}{2}(x+y)$$

Add this to your first equation, yielding:
$$(x+y)^{2}=10+\frac{3}{2}(x+y)$$

This is a quadratic equation in the unknown (x+y) that is easily solved.

Well, I guess this illustrates what I was saying about "stumbling": the approach I proposed does not "tidy up" for this problem and leaves a fourth-degree polynomial to find zeroes for. I regret not spotting this approach and it will work cleanly (with a *little* work). Thanks, arildno.

 

[dynamicsolo]

Thanks for the reply! I've now rearranged the 2nd eq. : 1/x=4/3-1/y and I've inversed it all to get: x=-3/(3y-4),

I believe this should be x = 3y/(y-4) . But you *will* end up with a quartic equation to solve. While you can get approximations of zeroes with graphing software nowadays, this is not going to be what mathematicians like and won't be available at an exam.

arildno's method is much better for this problem; my apologies...

 

[nirvana1990]

Oh ok so using what dynamicsolo says I got: 1/y=(4x-3)/3x
Taking the reciprocal I get: y=3x/(4x-3)
Sub this in eq.1: x^2+9x^2/(4x^2-24x+9)=10 is this right so far?
Can I then multiply up to get a quartic equation and (somehow) solve that?!

Thanks for the help!

 

[nirvana1990]

Note that from the 2. equation, you have:
$$2xy=\frac{3}{2}(x+y)$$

Add this to your first equation, yielding:
$$(x+y)^{2}=10+\frac{3}{2}(x+y)$$

This is a quadratic equation in the unknown (x+y) that is easily solved.

Thanks for the help! I've rearranged the 2nd eq. ad divided by 2 to get (like you say) 3/2(x+y)=2xy but I don't understand how this has gone into the 1st eq. and what we're substituting it for?

 

[arildno]

Thanks for the help! I've rearranged the 2nd eq. ad divided by 2 to get (like you say) 3/2(x+y)=2xy but I don't understand how this has gone into the 1st eq. and what we're substituting it for?

Note that 3/2(x+y) is the SAME number as 2xy.
That is what the equality sign between them means!

Therefore, since you can add the same number to each side of an equation, add 2xy to the left hand side of eq. 1, and 3/2(x+y) to your right hand side.

Your left hand side of modified 1 can now be rewritten by the identity
$$(x+y)^{2}=x^{2}+2xy+y^{2}$$
for all x and y.

 

[nirvana1990]

Oh yay I get it now! Thank you sooo much!

 

[cristo]

Marked thread as solved to test.