Three Equations But Two Unknowns

Algebra Level 4

$\begin{cases} x^4- y^4 = 24 \\ x^2 + y^2 = 6 \\ x + y = 4 \end{cases}$

If $x$ and $y$ satisfy the above system of equations, find $x-y$ .

-2 1 -1 2 None of these choices

1 solution

Paul Ryan Longhas
Feb 15, 2016

Clearly, $x^4 - y^4 = 24 \implies (x-y)(x+y)(x^2 + y^2) = 24 \implies (x-y)(4)(6) = 24$ .

Thus, $x-y = 1$ . But if $x-y = 1$ and $x+ y = 4 \implies x = 5/2$ and $y = 3/2$ .

So, $x^2 + y^2 = (5/2)^2 + (3/2)^2 = 17/2 \neq 6$ which is a contradiction. Thus, none of the choices.

If the question hadnt been lvl 4 I would have put 1 as the answer for sure

Shreyash Rai - 5 years, 3 months ago

Lol i did the question when it was unrated and got it wrong :-)

Department 8 - 5 years, 3 months ago

Wow! What a confusing question! It will help me in my finals.

Department 8 - 5 years, 4 months ago

Good question!.

A Former Brilliant Member - 5 years, 4 months ago

Is there any solution to this problem and if yes what is it ?

Raven Herd - 5 years, 4 months ago

Paul's solution shows that there is no solution to this system of equations.

A priori, there is no reason why a list of random equations must yield a (unique) solution.

Calvin Lin Staff - 5 years, 3 months ago

Let there be a system of equation or just one equation (a random one) with some variables and it does yield some finite value( maybe real , maybe complex) so in order to be sure that it is indeed the solution we must put that value and then compare the L.H.S and R.H.S .Is that what you say? And there is another thing. If we have an equation something like 1-1+1-1+1-1 ......forever that an absurd solution to this problem is 1/2 so how can we put that value into the equation because there seems to be no practical method to solve that and yes what about infinite surds?

Raven Herd - 5 years, 3 months ago

@Raven Herd – When you manipulate an equation, you have to think if you possibly introduced extraneous solutions. The most common case when this occurs (and most people are aware of it), is when we square an equation. For example, the solutions to $x = 1$ are very different from the solutions to $x ^2 = 1$ .

The instance here when additional solutions were introduced, was when we divided the two equations. The solutions to $\{ f(x,y) = 1, g(x,y) = 1$ is a subset of the solutions to $\frac{ f(x,y) } { g(x,y) } = 1$ . As such, any solutions that we obtain to the latter system have to be checked against the initial system. Otherwise, you are solving a completely different problem.

Calvin Lin Staff - 5 years, 3 months ago

I get that why the answer is non of the choices, but can you please tell me what went wrong in the equation , as by the rule, the answer should be 1

Yatharth Chowdhury - 5 years, 3 months ago

Three Equations But Two Unknowns

1 solution

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