Elements of Algebra

194 ELEMENTS SECT. IV. 19. If $\sqrt{x} + \sqrt{(a + x)} = \frac{2a}{\sqrt{(a+x)}}$, then will $x = \frac{a}{3}$. 20. If $\sqrt{(a^2 + x^2)} = \sqrt[4]{(b^4 + x^4)}$, then will $x = \sqrt{\frac{b^4-a^4}{2a^2}}$. 21. If $y = \sqrt{(a^2 + \sqrt{(b^2 + x^2)})} - a$, then will $x = \frac{b^2}{4a} - a$. 22. If $\frac{128}{3x-4} = \frac{216}{5x-6}$, then will $x = 12$. 23. If $\frac{42x}{x-2} = \frac{35x}{x-3}$, then will $x = 8$. 24. If $\frac{45}{2x+3} = \frac{57}{4x-5}$, then will $x = 6$. 25. If $\frac{x^2-12}{3} = \frac{x^2-4}{4}$, then will $x = 6$. 26. If $615x - 7x^3 = 48x$, then will $x = 9$. *** CHAP. III. *Of the Solution of Questions relating to the preceding Chapter.* 585. Question 1. To divide 7 into two such parts that the greater may exceed the less by 3. Let the greater part be $x$, then the less will be $7-x$; so that $x = 7 - x + 3$, or $x = 10 - x$. Adding $x$, we have $2x = 10$; and dividing by 2, $x = 5$. The two parts therefore are 5 and 2. Question 2. It is required to divide $a$ into two parts, so that the greater may exceed the less by $b$. Let the greater part be $x$, then the other will be $a-x$; so that $x = a - x + b$. Adding $x$, we have $2x = a + b$; and dividing by 2, $x = \frac{a+b}{2}$. Another method of solution. Let the greater part $= x$; which as it exceeds the less by $b$, it is evident that this is less than the other by $b$, and therefore must be $= x - b$. Now,