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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.

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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.

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

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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$ $\forall x,y \in \mathbb{R}$.

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