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Attempt:

Let $x, y \in \mathbb{R}$ and $4x + 3y = 6$. $$\Rightarrow y = \frac{6 - 4x}{3} = 2 - \frac{4x}{3}.$$ Thus, we have found a real number $y$ for every real number $x$ s.t $4x + 3y = 6$ and conclude that $\forall x \in \mathbb{R}, \exists y \in \mathbb{R}, \ 4x + 3y = 6$.

This seems too simple to be right. Am I on the right track?

Faith Alone
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    To prove $\forall x , \exists y \dots$, start by assuming that $x$ is given. Then state what $y$ is or how you can find it. Finally prove the property that is supposed to hold for $x$ and $y$. It's like a tennis game: Somebody serves you the $x$, show that you can always return the serve (find a $y$ with the desired property.) Your attempt has all the right formulae, but it does not follow this logical scheme. – Hans Engler Sep 04 '22 at 01:19
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    You can prove by solving, I suppose. The real issue is how were the real numbers defined for you. To my mind, part of the definition of $\mathbb R$ is that $\mathbb R$ is a field (do you know what that means?). As such any $ax + by =c$ will have solutions (unless both $a,b=0$ and $c$ doesn't> – fleablood Sep 04 '22 at 01:22
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    To be logically impeccable, plug this value in for $y$ and verify that it is a solution. – GEdgar Sep 04 '22 at 01:46

1 Answers1

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It is this simple, since addition and multiplication preserve elements in $\mathbb{R}$ (unless you are dividing by zero). So $y$ is a real number.

Kamal Saleh
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