# Is this a valid expression? x/x-1 = 1/x-1 It would appear that x=1 but that would...

#### dogleg

##### New member
x/x-1 = 1/x-1 It would appear that x=1 but that would give us an undefined quantity. Explain please.

#### tkhunny

##### Moderator
Staff member
x/x-1 = 1/x-1 It would appear that x=1 but that would give us an undefined quantity. Explain please.
x = 1 is fine AS WRITTEN.

If you mean this: $$\displaystyle \dfrac{x}{x-1} = \dfrac{1}{x-1}$$ then x = 1 is NOT fine.

Use parentheses to clarify intent.

#### Dr.Peterson

##### Elite Member
x/(x-1) = 1/(x-1) It would appear that x=1 but that would give us an undefined quantity. Explain please.
Presumably you solved by multiplying both sides by (x-1), or something equivalent. Then you get x = 1. But in doing this, you are assuming that you are not multiplying by zero, which would produce a non-equivalent equation; and in fact, when x = 1, x-1 = 0.

So x = 1 is an extraneous solution, because, as you observed on checking, the equation is undefined in that case. Such a check is mandatory.

And the final answer is that there is no solution, since the only possible solution is not a solution.

#### Denis

##### Senior Member
Is this a classroom assignment?
If so, please show how far you got.

#### dogleg

##### New member
no it is just a personal exercise I got to: -x squared+y squared = 10 Stuck there as far as what x y= ? Thanks.

#### Dr.Peterson

##### Elite Member
Thank you for your help with this. Also may I ask about this one: if x^2 + xy = 20 and y^2 + xy = 30 what would xy = ?
You can just use the quadratic formula to solve one equation for one variable, and substitute that into the other.

I don't see a really nice way to directly isolate xy, but an alternative method I used is to add the two equations, and also subtract them as you did, and then use some factoring to combine them so that it eventually becomes possible to isolate either variable. In doing that, I was just trying things out without a specific goal, just to see what might happen.

#### tkhunny

##### Moderator
Staff member
xy = 30 - y^2 = 20 - x^2

What can you tell us from there?

x(x+y) = 30
y(x+y) = 20

x+y = 30/x = 20/y

Anything strike you, yet?

#### JeffM

##### Elite Member
If there is a solution to this system of equations, it can be found by substitution.

$$\displaystyle \text {Given: } x^2 + xy = 20 \text { and } y^2 + xy = 30.$$

$$\displaystyle x = 0 \implies x^2 + xy = 0 + 0 = 0 \ne 20 \implies \text { a contradiction.}$$

$$\displaystyle \therefore x \ne 0.\text { And so, } x^2 + xy = 20 \implies y = \dfrac{20 - x^2}{x}.$$

$$\displaystyle \therefore 30 = y^2 + xy = \left ( \dfrac{20 - x^2}{x} \right )^2 + x * \dfrac{20 - x^2}{x} \implies$$

$$\displaystyle 30 = \dfrac{400 - 40x^2 + x^4}{x^2} + 20 - x^2 = \dfrac{400 - 20x^2}{x^2} \implies$$

$$\displaystyle 30x^2 = 400 - 20x^2 \implies 50x^2 = 400 \implies x^2 = 8 = 4 * 2 \implies x = 2\sqrt{2} \text { or } -\ 2\sqrt{2}.$$

$$\displaystyle \therefore y = \dfrac{20 - 8}{2\sqrt{2}} = \dfrac{6}{\sqrt{2}} = \dfrac{6\sqrt{2}}{2} = 3\sqrt{2} \text { or}$$

$$\displaystyle y = -\ 3\sqrt{2}.$$

NOW CHECK.

$$\displaystyle (2 \sqrt{2})^2 + (2 \sqrt{2}) (3 \sqrt{2}) = 4 * 2 + 6 * 2 = 8 + 12 = 20.$$

$$\displaystyle (3\sqrt{2})^2 + (2 \sqrt{2}) (3 \sqrt{2} = 9 * 2 + 6 * 2 = 18 + 12 = 30.$$

So $$\displaystyle x = 2\sqrt{2} \text { and } y = 3\sqrt{2}$$ is a valid answer.

And $$\displaystyle (-\ x)^2 + (-\ x)(-\ y) = x^2 + xy \text { and }(-\ y)^2 + (-\ x)(-\ y) = y^2 + xy.$$

So $$\displaystyle x = -\ 2\sqrt{2} \text { and } y = -\ 3\sqrt{2}$$ is also a valid answer.

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#### Jomo

##### Elite Member
x/x-1 = 1/x-1 It would appear that x=1 but that would give us an undefined quantity. Explain please.
An algebraic expression does not have an equal sign, that is by design it does not have an answer like x= 1 or x = 3. You can simplify an expression by combining like terms and/or factoring. What you gave us is an equation, since it has an equal sign, and one may or may not be able to solve it. In your case, as many have pointed out, this equation has no solution.

#### Subhotosh Khan

##### Super Moderator
Staff member
xy = 30 - y^2 = 20 - x^2

What can you tell us from there?

x(x+y) = 20 ....................(1)
y(x+y) = 30 ....................(2)

x+y = 30/x = 20/y

Anything strike you, yet?
(x^2 + xy) + (y^2 +xy) = 50

x+ y = $$\displaystyle \pm$$√50 = $$\displaystyle \pm$$5√2 ................. (3)

Divide (1) by (2).............. y = 1.5 * x...........(4)

using (4) in (3)

2.5*x = $$\displaystyle \pm$$5√2

x = $$\displaystyle \pm$$2√2

y = (3/2) * ($$\displaystyle \pm$$2√2) = $$\displaystyle \pm$$(3)*√2 .................... [edited]

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#### Dr.Peterson

##### Elite Member
(x^2 + xy) + (y^2 +xy) = 50

x+ y = √50 = $$\displaystyle \pm$$5√2 ................. (3)

Divide (1) by (2).............. x = 1.5 * y...........(4)

using (4) in (3)

2.5*x = $$\displaystyle \pm$$5√2

x = $$\displaystyle \pm$$2√2

y = 2/3 * $$\displaystyle \pm$$2√2 = $$\displaystyle \pm$$(4/3)*√2
Unfortunately, equations (1) and (2) are wrong!

But this is essentially one of the ways I was suggesting; I can see many approaches that will work.

But what is xy? No one has finished yet ...

#### Jomo

##### Elite Member
(x^2 + xy) + (y^2 +xy) = 50

x+ y = √50 = $$\displaystyle \pm$$5√2 ................. (3)

Divide (1) by (2).............. x = 1.5 * y...........(4)

using (4) in (3)

2.5*y = $$\displaystyle \pm$$5√2

y = $$\displaystyle \pm$$2√2

x = 2/3 * $$\displaystyle \pm$$2√2 = $$\displaystyle \pm$$(4/3)*√2

#### Subhotosh Khan

##### Super Moderator
Staff member
Yes ... I did - instead of starting from my own work I tried to steal somebody else's work ...

Sitting in the corner for 12 hrs |@

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#### Denis

##### Senior Member
......I tried to steal somebody else's work ...
Shame, shame, double shame...

#### JeffM

##### Elite Member
Unfortunately, equations (1) and (2) are wrong!

But this is essentially one of the ways I was suggesting; I can see many approaches that will work.

But what is xy? No one has finished yet ...

It is true that there are multiple methods that will work, but, as I explained before, substitution is a very general method that easily works for this problem. Moreover, it is generally considered adequate to find x and y rather than their product. Nevertheless, xy = 12 is explcitly calculated twice my prior post in this thread.

Would you please be so kind as to explain what is incorrect or incomplete in my prior post? Thank you.

#### Dr.Peterson

##### Elite Member

It is true that there are multiple methods that will work, but, as I explained before, substitution is a very general method that easily works for this problem. Moreover, it is generally considered adequate to find x and y rather than their product. Nevertheless, xy = 12 is explcitly calculated twice my prior post in this thread.

Would you please be so kind as to explain what is incorrect or incomplete in my prior post? Thank you.
Maybe I'm not seeing something, but the problem asks "what would xy = ?", and I don't think anyone ever concluded their answer by saying the answer to that question. If I were grading your answer on a test, I would take off something for not having explicitly answered the question asked. (I've been assuming that everyone, like me, was just leaving the last step for the OP to do, but I thought it was worth pointing this out.)

Also, I never said that anything you did was wrong; only that, when the question is about xy rather than x or y, I tend to wonder if there is a direct way to get that without first solving for x and y separately -- some contest-type problems have such shortcuts. And I haven't found a quick, elegant way to that answer. Often, substitution is the long way to a solution, so looking for other ways is a good habit, anyway.

#### JeffM

##### Elite Member
Maybe I'm not seeing something, but the problem asks "what would xy = ?", and I don't think anyone ever concluded their answer by saying the answer to that question. If I were grading your answer on a test, I would take off something for not having explicitly answered the question asked. (I've been assuming that everyone, like me, was just leaving the last step for the OP to do, but I thought it was worth pointing this out.)
So it does. I am so used to solving problems for the unknowns that habit took over. Thank you.

And it is interesting that the solution for xy is unique whereas the solutions for x and y are not.

Also, I never said that anything you did was wrong; only that, when the question is about xy rather than x or y, I tend to wonder if there is a direct way to get that without first solving for x and y separately -- some contest-type problems have such shortcuts. And I haven't found a quick, elegant way to that answer. Often, substitution is the long way to a solution, so looking for other ways is a good habit, anyway.
Now this represents a pedagogical difference of opinion. I like to start students off on systems of simultaneous equations with two general ideas, namely that there may not be any solution and that substitution is a very general method of solution. I agree that substitution is not always a quick method, but it stops students from saying that they "have no idea where to start."

Perhaps one reason for this approach is that I start off in word problems by identifying what is not known. This introduces systems of equations as soon as word problems are introduced. The substitutions arising from word problems are usually very simple.

#### dogleg

##### New member
If there is a solution to this system of equations, it can be found by substitution.

$$\displaystyle \text {Given: } x^2 + xy = 20 \text { and } y^2 + xy = 30.$$

$$\displaystyle x = 0 \implies x^2 + xy = 0 + 0 = 0 \ne 20 \implies \text { a contradiction.}$$

$$\displaystyle \therefore x \ne 0.\text { And so, } x^2 + xy = 20 \implies y = \dfrac{20 - x^2}{x}.$$

$$\displaystyle \therefore 30 = y^2 + xy = \left ( \dfrac{20 - x^2}{x} \right )^2 + x * \dfrac{20 - x^2}{x} \implies$$

$$\displaystyle 30 = \dfrac{400 - 40x^2 + x^4}{x^2} + 20 - x^2 = \dfrac{400 - 20x^2}{x^2} \implies$$

$$\displaystyle 30x^2 = 400 - 20x^2 \implies 50x^2 = 400 \implies x^2 = 8 = 4 * 2 \implies x = 2\sqrt{2} \text { or } -\ 2\sqrt{2}.$$

$$\displaystyle \therefore y = \dfrac{20 - 8}{2\sqrt{2}} = \dfrac{6}{\sqrt{2}} = \dfrac{6\sqrt{2}}{2} = 3\sqrt{2} \text { or}$$

$$\displaystyle y = -\ 3\sqrt{2}.$$

NOW CHECK.

$$\displaystyle (2 \sqrt{2})^2 + (2 \sqrt{2}) (3 \sqrt{2}) = 4 * 2 + 6 * 2 = 8 + 12 = 20.$$

$$\displaystyle (3\sqrt{2})^2 + (2 \sqrt{2}) (3 \sqrt{2} = 9 * 2 + 6 * 2 = 18 + 12 = 30.$$

So $$\displaystyle x = 2\sqrt{2} \text { and } y = 3\sqrt{2}$$ is a valid answer.

And $$\displaystyle (-\ x)^2 + (-\ x)(-\ y) = x^2 + xy \text { and }(-\ y)^2 + (-\ x)(-\ y) = y^2 + xy.$$

So $$\displaystyle x = -\ 2\sqrt{2} \text { and } y = -\ 3\sqrt{2}$$ is also a valid answer.

Thanks to all for your kind help. You have shown that there is more than one way to skin a cat.

#### Dr.Peterson

##### Elite Member
You can just use the quadratic formula to solve one equation for one variable, and substitute that into the other. ...
Now this represents a pedagogical difference of opinion. I like to start students off on systems of simultaneous equations with two general ideas, namely that there may not be any solution and that substitution is a very general method of solution. I agree that substitution is not always a quick method, but it stops students from saying that they"have no idea where to start."
I'm not sure there's any difference of opinion. After all, I did start by suggesting substitution, at least as a fallback (though I didn't actually do it, so I missed the fact that the quadratic formula is never needed).

What I like to encourage is, after solving one way, to look for others, which stops students from falling into ruts, and keeps them flexible. In this case, the oddity of the question about xy led me to look in other directions first. And, not knowing the context, perhaps this was a contest-type problem in which that is the point. But I don't think it was. In fact, the real problem may have been worded as "what can x and y be?"