If $3(x +7) - y(2x+9)$ is the same for all values of $x$, what number must $y$ be? If we let $x = 0$.
\begin{align*} 
3(0+7)-y(2(0)+9) \\
21-9y \\
\end{align*}
Then $9y$ should always equal $21$?
Solving for $y$ finds $\frac{7}{3}$.
But $3(x+7)-\frac{7}{3}(2(x)+9)$ does not have the same result for diffrent values of $x$.
Where am I going wrong?
 A: $3 (x + 7) - y (2 x + 9)=21-9y$ for $x=0$ and
$3 (x + 7) - y (2 x + 9)=24 - 11 y$ for $x=1$
They are the same, so must be
$$21-9y=24-11y\to y=\frac32$$
A: The easiest way to think about this is to realise you need the expression to be independent of the value of $x$. Which means the coefficient of the $x$ term has to vanish, leaving only a constant term.
That means $3-2y = 0 \implies y = \frac 32$.
A: If we let $x=0$ then we do indeed find that the expression equals $21-9y$. But this does not mean that $21-9y$ must equal $0$; it only means that $21-9y$ is a constant. If, as Raffaele has suggested, we let $x=1$, then we find that the expression equals $24-11y$. Hence, $21-9y$ and $24-11y$ must be equal to the same constant, and so solving the problem boils down to solving the equation $21-9y=24-11y$:
\begin{align}
21 - 9y &= 24 - 11y &&\text{Subtract $21$ from both sides}\\
-9y &= 3 - 11y &&\text{Add $11y$ to both sides}\\
2y &= 3 &&\text{Divide both sides by $2$}\\
y &= \frac{3}{2} \, .
\end{align}
A: You have the function:
$$ f \left( x, y \right) = 3 \left( x + 7 \right) - y \left( 2 x + 9 \right) $$
Since the function is constant for $ x $ then it means:
$$ \nabla_{x} f \left( x, y \right) = 0 = 3 - 2 y \Rightarrow y = \frac{3}{2} $$
Another way thinking about it, is by definition the function with respect to $ x $ must be constant (Specific case of Linear).
Hence its coefficients (Which are given by the derivative since it's a specific case of Linear Function) must vanish (As seen in @Deepak answer).
