Elements of Algebra

CHAP. II. OF ALGEBRA. 191 $\frac{ax}{b} = c$, we first multiply by $b$, in order to have $ax = bc$, and then dividing by $a$, we find $x = \frac{bc}{a}$. If $\frac{ax}{b} - c = d$, we begin by giving the equation this form $\frac{ax}{b} = d + c$; after which we obtain the value of $ax = bd + bc$, and then that of $x = \frac{bd+bc}{a}$. Let $\frac{2}{3}x - 4 = 1$, then $\frac{2}{3}x = 5$, and $2x = 15$; whence $x = \frac{15}{2}$, $= 7\frac{1}{2}$. If $\frac{1}{2}x + \frac{1}{3} = 5$, we have $\frac{1}{2}x = 5 - \frac{1}{3} = \frac{9}{2}$; whence $3x = 18$, and $x = 6$. 580. Let us now consider a case, which may frequently occur; that is, when two or more terms contain the letter $x$, either on one side of the equation, or on both. If those terms are all on the same side, as in the equation $x + \frac{1}{2}x + 5 = 11$, we have $x + \frac{1}{2}x = 6$; or $\frac{3}{2}x = 12$; and lastly, $x = 4$. Let $x + \frac{1}{2}x + \frac{1}{3}x = 44$, be an equation, in which the value of $x$ is required. If we first multiply by 3, we have $4x + \frac{3}{2}x = 132$; then multiplying by 2, we have $11x = 264$; wherefore $x = 24$. We might have proceeded in a more concise manner, by beginning with the reduction of the three terms which contain $x$ to the single term $\frac{11}{6}x$; and then dividing the equation $\frac{11}{6}x = 44$ by 11. This would have given $\frac{1}{6}x = 4$, and $x = 24$, as before. Let $\frac{2}{3}x - \frac{3}{4}x + \frac{1}{2}x = 1$. We shall have, by reduction, $\frac{5}{12}x = 1$, $5x = 12$, and $x = 2\frac{2}{5}$. And, generally, let $ax - bx + cx = d$; which is the same as $(a - b + c)x = d$, and, by division, we derive $x = \frac{d}{a-b+c}$. 581. When there are terms containing $x$ on both sides of the equation, we begin by making such terms vanish from that side from which it is most easily expunged; that is to say, in which there are the fewest terms so involved. If we have, for example, the equation $3x + 2 = x + 10$, we must first subtract $x$ from both sides, which gives $2x + 2 = 10$; wherefore $2x = 8$, and $x = 4$. Let $x + 4 = 20 - x$; here it is evident that $2x + 4 = 20$; and consequently $2x = 16$, and $x = 8$.