Elements of Algebra

214 ELEMENTS SECT. IV. 620. As, in this example, each equation contains only two unknown quantities, we may obtain the solution required in an easier way. The first equation gives $y = 200 - 2x$, so that $y$ is determined by $x$; and if we substitute this value in the second equation, we have $200 - 2x + \frac{1}{3}z = 100$; therefore $\frac{1}{3}z = 2x - 100$, and $z = 6x - 300$. So that $z$ is also determined by $x$; and if we introduce this value into the third equation, we obtain $6x - 300 + \frac{1}{4}x = 100$, in which $x$ stands alone, and which, when reduced to $25x - 1600 = 0$, gives $x = 64$. Consequently, $y = 200 - 128 = 72$, and $z = 384 - 300 = 84$. 621. We may follow the same method, when we have a greater number of equations. Suppose, for example, that we have in general; 1. $u + \frac{x}{a} = n$, 2. $x + \frac{y}{b} = n$, 3. $y + \frac{z}{c} = n$, 4. $z + \frac{u}{d} = n$; or, destroying the fractions, these equations become, 1. $au + x = an$, 2. $bx + y = bn$, 3. $cy + z = cn$, 4. $dz + u = dn$. Here, the first gives immediately $x = an - au$, and, this value being substituted in the second, we have $abn - abu + y = bn$; so that $y = bn - abn + abu$; and the substitution of this value, in the third equation, gives $bcn - abcn + abcu + z = cn$; therefore $z = cn - bcn + abcn - abcu$. Substituting this in the fourth equation, we have $cdn - bcdn + abcdn - abcdu + u = dn$. So that $dn - cdn + bcdn - abcdn = abcdu - u$, or $(abcd - 1).u = abcdn - bcdn + cdn - dn$; whence we have $u = \frac{abcdn - bcdn + cdn - dn}{abcd - 1} = \frac{n.(abcd - bcd + cd - d)}{abcd - 1}$ And, consequently, by substituting this value of $u$ in the equation, $x = an - au$, we have $x = \frac{abcdn - acdn + adn - an}{abcd - 1} = \frac{n.(abcd - acd + ad - a)}{abcd - 1}$