Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Basically what I have is that $x,y \in \mathbb{R}$ and that $3x + 2y \leq 5 $, so what I need to prove is that $x > 1 \rightarrow y < 1$

How would you prove this?

In some way i know that if I make $y \leq \frac{5-3x}{2}$, I would find an expression in which I certainly know that all the values that the variable $y$ is going to take will be less than $1$ because one of the premise is that $x > 1$, but I'm not sure if this is enough to make a correct proof.

share|cite|improve this question

If $x\gt 1$ then $3x\gt 3$, so $-3x\lt -3$, hence $5-3x\lt 2$.

Therefore, $2y\leq 5-3x\lt 2$, so $2y\lt 2$, hence $y\lt 1$ must hold.

share|cite|improve this answer

$3x+2y\le5$ and $x>1\implies 2y < 5-3\implies~y<1$.

share|cite|improve this answer

Let's see it as a system of inequalities: \begin{cases} 3x+2y\le5 \\ x > 1 \end{cases}

\begin{cases} x\le\frac{5-2y}{3} \\ x > 1 \end{cases}

$$ 1 < x \le \frac{5-2y}{3} $$

$$ 1 < \frac{5-2y}{3} $$

$$ 3 < 5-2y $$

$$ 0 < 2-2y $$

$$ 2y < 2 $$

$$ y < 1 $$

So, $ 3x+2y\le5 $ when $ x > 1$ will imply $ y < 1$ $\forall x,y \in \mathbb{R}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.