solving rectangle Diagonal of a rectangle is $13$ cm. If we extend the length of the rectangle for $4$ cm and width for $7$ cm, then diagonal will be longer for $7$ cm as well. Find sides (length and width) of the rectangle.
So $d^2=a^2+b^2 \implies b^2=d^2-a^2 \implies b^2=13^2-a^2 \implies b^2=169-a^2$ and $a^2+b^2=169$
\begin{align*}
d'^2 & =a'^2+b'^2\\
(d+7)^2 & =(a+4)^2+(b+7)^2\\
20^2 & = a^2+8a+16+b^2+14b+49\\
400 & = 8a+16+14b+49+169\\
\frac{166}{2} & = \frac{8a+14b}{2}\\
83 & =4a+7b
\end{align*}
... 
That's what I've done. What should I do next?

Let us rework the problem; this might make the answer easier to see. Let $a$ and $b$ be the side lengths of the rectangle, in which case $13 = \sqrt{a^{2} + b^{2}}$ is the diagonal. As you have already noted, we have $169 = a^{2} + b^{2}.$
When we enlarge the rectangle, we can use Pythagoras once again to get the following relation:
$$20 = \sqrt{(a + 4)^{2} + (b + 7)^{2}}.$$
This expands to
$$400 = (a + 4)^{2} + (b + 7)^{2},$$
which you have discovered to simplify into $4a + 7b = 83.$
Now, we begin the messy work. Rearranging the previous equation, $a = \frac{83 - 7b}{4}.$ We substitute this into $169 = a^{2} + b^{2}$ and solve as follows:
$$\left(\frac{83 - 7b}{4}\right)^{2} + b^{2} = 169$$
$$\frac{6889 - 1162b + 49b^{2}}{4} + b^{2} = 169,$$
which yields the quadratic form
$$\frac{53}{4}b^{2} - \frac{581}{2}b + \frac{6889}{4} = 0.$$
But the discriminant of the equation is less than $0,$ signaling no real solutions. Looking back at the problem, this makes sense. If you expand the width by $7$ units and the length by a non-zero amount, the diagonal should expand by more than $7$ units.
