$$x = 11-y$$

$$z = y-3$$

  • Evaluate $xz$

I've tried to multiply

$$xz = 11-y(y-3)$$

However, there will be no exact solution from here.


closed as off-topic by Andrei, Xander Henderson, Taroccoesbrocco, Leucippus, Sil Aug 11 at 9:45

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  • If you have two variables, both of which are functions of only $y$, quite obviously the product will also be a function of $y$. – Matti P. Aug 10 at 10:10
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    Also, remember the brackets in the result of the multiplication. – Matti P. Aug 10 at 10:11
  • @MattiP. I couldn't get your hint. – Hamilton Aug 10 at 10:12
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    What is the difficulty you are facing? If $x= 11-y$ and $z=y-3$, then indeed $xy = (11-y)(y-3)$ as you correctly intended to write. And what do you mean with exact solution? You didn't ask anything in your post, so it's a bit tricky to figure out what you exactly want. – Matti P. Aug 10 at 10:17
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    @Hamilton The point is, we don't know what you mean by "solution".$x$ and $z$ are algebraic expressions in $y$, so their product will also be an algebraic expression in $y$, and this product should qualify as the "solution", to the question "evaluate $xz$". – астон вілла олоф мэллбэрг Aug 10 at 11:22

So $$xz = \underbrace{-y^2+14y-33}_{f(y)} =-(y-7)^2+16\leq 16$$ So since $f$ is continous function we have $xz$ can take any value in $(-\infty,16)$.

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