Elements of Algebra

192 ELEMENTS SECT. IV. Let $x + 8 = 32 - 3x$, this gives us $4x + 8 = 32$; or $4x = 24$, whence $x = 6$. Let $15 - x = 20 - 2x$, here we shall have $15 + x = 20$, and $x = 5$. Let $1 + x = 5 - \frac{1}{2}x$; this becomes $1 + \frac{3}{2}x = 5$, or $\frac{3}{2}x = 4$; therefore $3x = 8$; and lastly, $x = \frac{8}{3} = 2\frac{2}{3}$. If $\frac{1}{2} - \frac{1}{3}x = \frac{1}{4} - \frac{1}{6}x$, we must add $\frac{1}{3}x$, which gives $\frac{1}{2} = \frac{1}{4} + \frac{1}{6}x$; subtracting $\frac{1}{4}$, and transposing the terms, there remains $\frac{1}{6}x = \frac{1}{4}$; then multiplying by 12, we obtain $x=2$. If $1\frac{1}{2} - \frac{2}{3}x = \frac{1}{4} + \frac{1}{2}x$, we add $\frac{2}{3}x$, which gives $1\frac{1}{2} = \frac{1}{4} + \frac{7}{6}x$; then subtracting $\frac{1}{4}$, and transposing, we have $\frac{7}{6}x = 1\frac{1}{4}$, whence we deduce $x = 1\frac{1}{4} = \frac{15}{14}$ by multiplying by 6 and dividing by 7. 582. If we have an equation in which the unknown number $x$ is a denominator, we must make the fraction vanish by multiplying the whole equation by that denominator. Suppose that we have found $\frac{100}{x} - 8 = 12$, then, adding 8, we have $\frac{100}{x} = 20$; and multiplying by $x$, it becomes $100 = 20x$; lastly, dividing by 20, we find $x = 5$. Let now $\frac{5x+3}{x-1} = 7$; here multiplying by $x-1$, we have $5x + 3 = 7x - 7$; and subtracting $5x$, there remains $3 = 2x - 7$; then adding 7, we have $2x = 10$; whence $x = 5$. 583. Sometimes, also, radical signs are found in equations of the first degree. For example: A number $x$, below 100, is required, such, that the square root of $100 - x$ may be equal to 8; or $\sqrt{(100 - x)} = 8$. The square of both sides will give $100 - x = 64$, and adding $x$, we have $100 = 64 + x$; whence we obtain $x = 100 - 64 = 36$. Or, since $100 - x = 64$, we might have subtracted 100 from both sides; which would have given $-x = -36$; or, multiplying by $-1$, $x = 36$. 584. Lastly, the unknown number $x$ is sometimes found as an exponent, of which we have already seen some examples; and, in this case, we must have recourse to logarithms. Thus, when we have $2^x = 512$, we take the logarithms of both sides; whence we obtain $x \log. 2 = \log. 512$; and dividing by $\log. 2$, we find $x = \frac{\log. 512}{\log. 2}$. The Tables then