simple t-test calculation question 
What values of $x$ satisfy the following equations?
(a) $P\left(-x \leq T_{22} \leq x\right)=0.98$
(b) $P\left(T_{13} \geq x\right)=0.85$
(c) $P\left(T_{26}<x\right)=0.95$

Hi guys,
So I am working on this problem here, and I am a bit confused because I have a solutions guide that does problem b confusingly.
It takes $P\left(T_{13} \geq x\right)=0.85$ and does $1 - P\left(T_{13} \leq x\right)=0.85$ and $P\left(T_{13} \geq -x\right)=0.5$, then using the t-table chart giving that $x = -1.079$
My question is why do we have to do those steps? Why can't we look up the t-table chart directly given that $P(T_{13} \geq x) = 0.85$? where $x = 1.079$
 A: The first thing to remember about t tables is that (in the usual format)
probabilities are along the top margin and $t$-values are in the body of
the table. This is the reverse of normal tables, where the probabilities
are in the body of the table and $z$ values in the margins.
The second thing to remember is that there are many different t distributions,
distinguished by their degrees of freedom (given along the left margin of the
table and denoted by 'DF' or $\nu).$ A typical t table has only about half
a dozen probabilities for each DF. By contrast, a normal table has a full
page (or two) of probability values.
Now to try to make sense of part (b): You seek $x$ such that $P(T_{13} \ge x) = .85.$ From what you say, I believe that the subscript $13$ means that
DF = 13. So the only relevant row of the table is the one marked 13. You want to cut probability $0.15$ from the lower tail of the distribution (so that
probability $0.85$ is above). The value $x$ has to be negative. (The distribution
is symmetrical about $0,$ so any $x$ that cuts less than 50% of the probability from the lower tail must be negative.) But there are no negative
values in the table, so we have to use symmetry. We find the positive $x$ that cuts probability $0.15$ from the upper tail, and put a negative sign in front of it. 
Look along the top margin for the column that cuts 15% of the area from
the upper tail, then follow that column down to the row for DF = 13, and
you should see $x = 1.079.$ Then make it negative, and you have the desired
answer $x = -1.0794.$ So that  $P(T_{13} \ge -1.079) = .85,$ as required.
The figure below shows the density function for $\mathsf{T}(DF = 13).$ 
The dotted red line at the right (at $1.079)$ cuts probability $0.15$ from the
upper tail of the distribution; it corresponds to the value from the table.
The solid red line at the left (at $-1.079)$ cuts probability $0.15$ from the 
lower tail.

Some kinds of statistical software and statistical calculators allow for a more direct approach. For example, in R statistical software qt is the inverse CDF function of a t distribution with specified DF:
 qt(.15, 13); q
 ## -1.079469

