Elements of Algebra

CHAP. IV. OF ALGEBRA. 207 only to substitute its value in one of the quantities that express $x$. 608. According to this rule, we obtain from the first equation, $x = \frac{c-by}{a}$, and from the second, $x = \frac{h-gy}{f}$: then putting these values equal to each other, we have this new equation: $\frac{c-by}{a} = \frac{h-gy}{f}$; multiplying by $a$, the product is $c-by = \frac{ah-agy}{f}$; and then by $f$, the product is $fc-fby=ah-agy$; adding $agy$, we have $fc-fby+agy = ah$; subtracting $fc$, gives $-fby+agy = ah-fc$; or $(ag-bf)y = ah-fc$; lastly, dividing by $ag-bf$, we have $y = \frac{ah-fc}{ag-bf}$. In order now to substitute this value of $y$ in one of the two values which we have found of $x$, as in the first, where $x = \frac{c-by}{a}$, we shall first have $-by = -\frac{abh-bcf}{ag-bf}$; whence $c-by = c - \frac{abh-bcf}{ag-bf} = \frac{acg-bcf-abh+bcf}{ag-bf} = \frac{acg-abh}{ag-bf}$; and, dividing by $a$, $x = \frac{c-by}{a} = \frac{cg-bh}{ag-bf}$. 609. Question 1. To illustrate this method by examples, let it be proposed to find two numbers, whose sum may be 15, and difference 7. Let us call the greater number $x$, and the less $y$: then we shall have $x + y = 15$, and $x - y = 7$. The first equation gives $x = 15-y$, and the second, $x = 7+y$; whence results this equation, $15-y=7+y$. So that $15=7+2y$; $2y=8$, and $y=4$; by which means we find $x=11$. So that the less number is 4, and the greater is 11. 610. Question 2. We may also generalise the preceding