How can adding a 2nd IV make the 1st IV significant? I have what is probably a simple question, but it is baffling me right now, so I am hoping you can help me out.
I have a least squares regression model, with one independent variable and one dependent variable. The relationship is not significant. Now I add a second independent variable. Now the relationship between the first independent variable and the dependent variable becomes significant.
How does this work? This is probably demonstrating some issue with my understanding, but to me, but I do not see how adding this second independent variable can make the first significant.
 A: It feels like the OP's question can be interpreted in two different ways:


*

*Mathematically, how does OLS work, such that adding an independent variable can change results in an unexpected way?

*How can modifying my model by adding one variable change the effect of another, independent variable in the model?
There are several good answers already for question #1. And question #2 may be so obvious to the experts that they assume the OP must be asking question #1 instead. But I think question #2 deserves an answer, which would be something like:
Let's start with an example. Say that you had the heights, age, gender, etc, of a number of children, and you wanted to do a regression to predict their height.
You start with a naive model that uses gender as the independent variable. And it's not statistically significant. (How could it be, you're mixing 3-year-olds and teen-agers.)
Then you add in age and suddenly not only is age significant, but so is gender. How could that be?
Of course, in my example, you can clearly see that age is an important factor in the height of a child/teen. Probably the most important factor that you have data on. Gender can matter, too, especially for older children and adults, but gender alone is a poor model of how tall a child is.
Age plus gender is a reasonable (though, of course simplified) model that's adequate for the task. If you add other data -- interaction of age and gender, diet, height of parents, etc -- you could make an even better model, which would of course still be simplified compared to the host of factors that actually determine a child's height, but then again all models are simplified versions of reality. (A map of the world that's 1:1 scale isn't too useful for a traveler.)
Your original model (gender only) is too simplified -- so simplified that it's essentially broken. But that doesn't mean that gender is not useful in a better model.
EDIT: added gung's suggestion re: the interaction term of age and gender.
A: This thread has already three excellent answers (+1 to each). My answer is an extended comment and illustration to the point made by @gung (which took me some time to understand):

There are two basic possibilities: First, the other IV may absorb some of the residual variability and thus increase the power of the statistical test of the initial IV. The second possibility is that you have a suppressor variable.

For me, the clearest conceptual way to think about multiple regression is geometric. Consider two IVs $x_1$ and $x_2$, and a DV $y$. Let them be centered, so that we do not need to care about intercept. Then if we have $n$ data points in the dataset, all three variables can be imagined as vectors in $\mathbb R^n$; the length of each vector corresponds to the variance and the angle between any two of them corresponds to the correlation. Crucially, performing multiple OLS regression is nothing else than projecting dependent variable $\mathbf y$ onto the plane spanned by $\mathbf x_1$ and $\mathbf x_2$ (with the "hat matrix" simply being a projector). Readers unfamiliar with this approach can look e.g. in The Elements of Statistical Learning, Section 3.2, or in many other books.
"Enhancement"
The following Figure shows both possibilities listed by @gung. Consider only the blue part at first (i.e. ignore all the red lines):

Here $\mathbf x_1$ and $\mathbf x_2$ are orthogonal predictors spanning a plane (called "plane $X$"). Dependent variable $\mathbf y$ is projected onto this plane, and its projection OD is what is usually called $\hat y$. Then OD is decomposed into OF (contribution of IV1) and OE (contribution of IV2). Note that OE is much longer than OF.
Now imagine that there is no second predictor $\mathbf x_2$. Regressing $\mathbf y$ onto $\mathbf x_1$ would result in projecting it onto OF as well. But the angle AOC ($\alpha$) is close to $90^\circ$; an appropriate statistical test would conclude that there is almost no association between $y$ and $x_1$ and that $x_1$ is hence not significant. 
When $x_2$ is added, the projection OF does not change (because $\mathbf x_1$ and $\mathbf x_2$ are orthogonal). However, to test whether $x_1$ is significant, we now need to look at what is left unexplained after $x_2$. The second predictor $x_2$ explains a large portion of $y$, OE, with only a smaller part EC remaining unexplained. For clarity, I copied this vector to the origin and called it OG: notice that the angle GOF ($\beta$) is much smaller than $\alpha$. It can easily be small enough for the test to conclude that it is "significantly smaller than $90^\circ$", i.e. that $x_1$ is now a significant predictor.
Another way to put it is that the test is now comparing the length of OF to OG, and not to OC as before; OF is tiny and "insignificant" compared to OC, but big enough to be "significant" compared to OG.
This is exactly the situation presented by @whuber, @gung, and @Wayne in their answers. I don't know if this effect has a standard name in the regression literature, so I will call it "enhancement".
Suppression
Notice that in the above, if $\alpha=90^\circ$ then $\beta=90^\circ$ as well; in other words, "enhancement" can only enhance the power to detect significant predictor, but if the effect of $x_1$ alone was exactly zero, it will stay exactly zero.
Not so in suppression. 
Imagine that we add $x_3$ to $x_1$ (instead of $x_2$) -- please consider the red part of the drawing. The vector $\mathbf x_3$ lies in the same plane $X$, but is not orthogonal to $\mathbf x_1$ (meaning that $x_3$ is correlated with $x_1$). Since the plane $X$ is the same as before, projection OD of $\mathbf y$ also stays the same. However, the decomposition of OD into contributions of both predictors changes drastically: now OD is decomposed into OF' and OE'. 
Notice how OF' is much longer than OF used to be. A statistical test would compare the length of OF' to E'C and conclude that the contribution of $x_1$ is significant. This means that a predictor $x_1$ that has exactly zero correlation with $y$ turns out to be a significant predictor. This situation is (very confusingly, in my opinion!) known as "suppression"; see here as to why: Suppression effect in regression: definition and visual explanation/depiction -- @ttnphns illustrates his great answer with a lot of figures similar to mine here (only better done).
A: I don't think any of the answers have explicitly mentioned the mathematical intuition for the orthogonal/uncorrelated case, so I will show this here, but I don't believe this answer will be 100% complete.
Suppose that $x_1$ and $x_2$ are uncorrelated, which implies that their centered versions are orthogonal, i.e., $(x_1 - \bar{x}_1 ) \perp (x_2 - \bar{x}_2)$.
Now consider the estimators:
$$
\hat{\beta} = (X^TX)^{-1}X^Ty 
$$
Without loss of generality (constant shifts doesn't affect $\hat{\beta}$), suppose that $X$ here consist of the centered versions of $x_1, x_2$. We can also assume that $X$ here does not include the intercept, which is fine since it's centered and we can simply compute the intercept as $\hat{\beta}_0 = \bar{y}$, so $X \in \mathbb{R}^{n \times 2}$ and $(X^TX)^{-1}$ is diagonal, which will result in the estimators from multiple linear regression being the same as that of separate regression $y$ on $x_1$ and $y$ on $x_2$.
Now consider the t score, which is used to compute p values and measure significance. We have
$$
t = \frac{\hat{\beta}_j - \beta_j}{SE(\hat{\beta}_j)}
$$
We want to test for $\beta_j \neq 0$, so our null hypothesis is $\beta_j = 0$, and we have
$$
t = \frac{\hat{\beta}_j}{SE(\hat{\beta}_j)}
$$
As we saw earlier, $\hat{\beta}_j$ doesn't change when a predictor that is orthogonal to $x_j$ is added. So for this situation, the only thing that would affect the t score/significance is $SE(\hat{\beta})$, which we know to be
$$
SE(\beta) = \operatorname{var}\left(\hat{\beta} \right) = \sigma^2 (X^TX)^{-1}
$$
Again note that $(X^TX)^{-1}$ is a diagonal matrix, so this component remains the same for the simple linear regression and multiple linear regression case, so the only thing that could change the t score is $\sigma^2$. If we know the population variance, then the t score wouldn't change, but we typically estimate the population variance with
$$
\hat{\sigma}^2 = \frac{1}{n - p - 1}\sum_{i=1}^n (y_i - \hat{y}_i)^2
$$
The estimator of the population variance is non-decreasing when additional predictors are added (orthogonal or not) -- this is equivalent to that the plain vanilla $R^2$ cannot decrease when additional predictors are added, because the sum of squares of residuals can only remain the same (which occurs when the added predictor can be written as a linear combination of the current predictors) or increase. An intuitive way to think about this is if having $p+1$ predictors with non-zero coefficients would get you a worse fit than a p-subset of these $p+1$ predictors, then least squares would just return the smaller model with p non-zero predictors and 1 zeroed predictor
So we see that $SE(\hat{\beta})$ is non-decreasing, which means that the t score increases monotonically, which would contribute to a decrease in the p-value; however, the degrees of freedom of the t-distribution decreases with increasing predictors, and this results in an increase in p-value. So there are competing effects going on here.
