Elements of Algebra

190 ELEMENTS SECT. IV. And in this, $x + 6a = 20 + 3a$, we have $x = 20 + 3a - 6a$, or $x = 20 - 3a$. 576. If the original equation have this form, $x - a + b = c$, we may begin by adding $a$ to both sides, which will give $x + b = c + a$; and then subtracting $b$ from both sides, we shall find $x = c + a - b$. But we might also add $+a - b$ at once to both sides; and thus obtain immediately $x = c + a - b$. So likewise in the following examples: If $x - 2a + 3b = 0$, we have $x = 2a - 3b$. If $x - 3a + 2b = 25 + a + 2b$, we have $x = 25 + 4a$. If $x - 9 + 6a = 25 + 2a$, we have $x = 34 - 4a$. 577. When the given equation has the form $ax = b$, we only divide the two sides by $a$, to obtain $x = \frac{b}{a}$. But if the equation has the form $ax + b - c = d$, we must first make the terms that accompany $ax$ vanish, by adding to both sides $-b + c$; and then dividing the new equation $ax = d - b + c$ by $a$, we shall have $x = \frac{d-b+c}{a}$. The same value of $x$ would have been found by subtracting $+b - c$ from the given equation; that is, we should have had, in the same form, $ax = d - b + c$, and $x = \frac{d-b+c}{a}$. Hence, If $2x + 5 = 17$, we have $2x = 12$, and $x = 6$. If $3x - 8 = 7$, we have $3x = 15$, and $x = 5$. If $4x - 5 - 3a = 15 + 9a$, we have $4x = 20 + 12a$, and consequently $x = 5 + 3a$. 578. When the first equation has the form $\frac{x}{a} = b$, we multiply both sides by $a$, in order to have $x = ab$. But if it is $\frac{x}{a} + b - c = d$, we must first make $\frac{x}{a} = d - b + c$; after which we find $x = (d - b + c)a = ad - ab + ac$. Let $\frac{1}{2}x - 3 = 4$, then $\frac{1}{2}x = 7$, and $x = 14$. Let $\frac{1}{3}x - 1 + 2a = 3 + a$, then $\frac{1}{3}x = 4 - a$, and $x = 12 - 3a$. Let $\frac{x}{a-1} - 1 = a$, then $\frac{x}{a-1} = a + 1$, and $x = a^2 - 1$. 579. When we have arrived at such an equation as