# Trying to evaluate $xz$ [closed]

$$x = 11-y$$

$$z = y-3$$

• Evaluate $xz$

I've tried to multiply

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

However, there will be no exact solution from here.

Regards

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Xander Henderson, Taroccoesbrocco, Leucippus, Sil
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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
• 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
• 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
• @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 $(-\infty,16)$.