Rejecting values in multivariable equations

Question

Say we have a system of equations involving more than one variable, for eg: $${x}^2 - {y}^2 = 48$$ $$x - y = 8$$ This can be solved as follows: $$(x-y)(x+y) = 48$$ Since $x-y=8$ , the equation becomes $$8(x+y)=48$$ $$x+y=6$$ Adding ($x-y=8$) and ($x+y=6$) to get $$2x = 14$$ $$x=7$$

Till here, there is no problem. But when we put the value of $x=7$ in the very first equation we get two values for y(+1 and -1). But only one of them i.e., $y=-1$ is consistent with both the equations. The other value of y gets rejected. So my question is:

Why do we get two values of $y$ if only one of them is consistent?

I know it's absurd, but this question has been gnawing at my brain for a week and I haven't been able to come up with a decent answer.

Thanks in advance!

Answer 1

It is a general fact when you solve equations by implications. If \begin{equation} \text{System A}\Longrightarrow\text{System B} \end{equation} then it means that System B has more solutions than System A, that is to say every solution of System A is a solution of System B but the converse is not necessarily true.

In your case, System B is $\left\{\begin{align}y^2=1\cr x=7\end{align}\right.$.

Of course if you replace $x$ by $7$ in the second equation ($x-y=8$), you obtain directly that $y=-1$.

To elaborate a little more on this matter, note that if $\text{System A}\Longrightarrow\text{System B}$, then \begin{equation} \text{System A}\Longleftrightarrow \left\{\begin{array}[l]\cr \text{System A}\cr \text{System B} \end{array}\right. \end{equation} So in your case, this entails \begin{equation} \left\{\begin{array}[l]\cr x^2-y^2=48\cr x - y = 8 \end{array}\right. \Longleftrightarrow \left\{\begin{array}[l]\cr x^2-y^2=48\cr x-y=8\cr x=7 \end{array}\right. \Longleftrightarrow \left\{\begin{array}[l]\cr y^2=1\cr y=-1\cr x=7 \end{array}\right. \Longleftrightarrow \left\{\begin{array}[l]\cr y=-1\cr x=7 \end{array}\right. \end{equation}

Answer 2

While solving, you have reason by equivalences, hence always maintain 2 equations. Your initial system implies what you wrote after, but is not equivalent. Written rigorously, your argument leads to the (unique) solution : $$[(x-y)(x+y)=48\text{ and }x-y=8]$$ $$\Leftrightarrow[8(x+y)=48\text{ and }x-y=8]$$ $$\Leftrightarrow[x+y=6\text{ and }x-y=8]$$ $$\Leftrightarrow[x=7\text{ and }x-y=8]$$ $$\Leftrightarrow[x=7\text{ and }y=-1].$$

Answer 3

Way 1:

For a given value of $x$ the quadratic equation will return 2 roots while the linear equation will return 1 root. So, while the linear equation will give me the correct value of $y$ that satisfies both, the quadratic equation will give me 2 values instead and while one is a crew-mate other must be the imposter.
So, instead put $x=7$ in the equations of the line to get the required value.

An interesting analogy is that suppose A who is an engineer is finding his partner who is also an engineer to work on a project. A's partner lives in a rented shared room. So, when A is standing out his partner's room and calls for him, 2 people come out as both of their names are B, one is engineer and the other a doctor. A recognizes his partner and goes on to work with him.

Way 2:

As you can see from the above pic, there are 3 things in play. Once you have found out that $x=7$, now you must find the common point between the 3 equations.