Question on shifts in Supply and Demand Curves I have a super basic question on the shift in supply and demand curves ( I haven't seen this stuff since high school). More generally, this is about solving systems of simultaneous equations.

We have two curves, an IS curve and an LM curve. We know from theory
that the IS curve is downward sloping in $i-y$ space and the LM
curve is upward sloping in $i-y$ space. Let's take a basic example.
The IS curve is represented by
$$
i=-y
$$
and the $LM$ curve is represented by 
$$
i=y
$$
 Equilibrium would imply values of $y$ and $i$ that solve these
equations simultaneously, or $i=y=0.$ Now suppose government activity
increased, shifting the $IS$ curve to the right. In this framework,
this would result in a horizontal translation of $2$ units of the
$IS$ curve. 
$$
i=-y+2
$$
 while the $LM$ curve remains unchanged. The new equilibrium is given
by $y=1$ and $1.$ Notice that the initial increase of $2$ in the
IS curve resutlts in equilibrium increasing by only $1.$ Intuitively,
this is because as $y$ increases, so does $i$ according to $LM$
relation. And, as $i$ increases, $y$ decreases according to $IS$
relation. As both these relations have to be satisfied, some of the
initial increase is moderated. My question is: what ensures that $y$
in fact doesn't decrease? If the $LM$ curve determines interest rates
that are super sensitive to increases in $y,$ can it not be that
$i$ increases sufficiently that $y$ in equilibrium decreases? Graphically,
this is not possible- in the extreme case, the $LM$ curve is vertical,
$y$ does not change. But, what condition ensures that $y$ does not decrease?
Write $i=I(y,a)$ for the IS function (expressed as some function of $y$). Likewise, let $i=L(y)$ be the LM function. Assume the following properties:


$\frac{\partial I(y,a)}{\partial a}&gt;0$ : $a$ is a parameter that causes the IS curve to shift up.
$\frac{\partial I(y,a)}{\partial y}&lt;0$: the IS function slopes downwards
$\frac{\partial L(y)}{\partial y}&gt;0$ : the LM function slopes upwards


For any $a$, we find the equilibrium $y^*(a)$ at the intersection of the two functions:
$$I(y^*(a),a)- L(y^*(a))=0.$$
Applying the implicit function theorem:
$$\frac{\partial I}{\partial a}+\frac{\partial y^*}{\partial a}\left(\frac{\partial I}{\partial y^*}-\frac{\partial L}{\partial y^*}\right)=0,$$
which implies
$$\frac{\partial y^*}{\partial a}=-\frac{\frac{\partial I}{\partial a}}{\frac{\partial I}{\partial y^*}-\frac{\partial L}{\partial y^*}}.$$

The numerator is positive by hypothesis. Thus, for $y^*$ to increase with $a$ we require that the denominator be negative: $\frac{\partial I}{\partial y^*}&lt;\frac{\partial L}{\partial y^*}$. This is clearly true because $I$ slopes downward and $L$ slopes upward. But $y^*$ would also increase even if $I$ sloped upward—provided itdoes so more gradually than $L$.