One the first set, #1 figure, notice we have two angle tickmarks, so those angles are equal, however, notice at segment AB, is at the border of both triangles, so AB is just the same length for ∡ABC as it's for ∡ADB.
so, we have an Angle, then a Side, and then an Angle, so they're congruent by ASA.
now, on the 2nd set, #2 figure, we have an angle tickmark and a side tickmark, hmmmmm but notice angles 1 and 2, they're just across from each other at that junction, therefore, angles 1 and 2 are vertical angles adn therefore are also equal.
so, we have a Side, then an Angle and then another Side, the triangles are congruent by SAS.
Let us assume the first number to be = x Let us assume the second number to be = y Then x^2 - y^2 = 5 x^2 = y^2 + 5 And 3x^2 - 2y^2 = 19 Multiplying the first equation by -2, we get - 2x^2 + 2y^2 = - 10 Now subtracting the two equations, we get x^2 = 9 x = 3 Putting the value of "x" in the first equation, we get (3)^2 - y^2 = 5 9 - y^2 = 5 y^2 = 4 y = 2
I hope that the procedure is clear enough for you to understand and this is the answer that you were looking for.