Elements of Algebra

CHAP. I. OF ALGEBRA. 301 by 2. Let us, therefore, make $y-1=2z$; and we shall have $y=2z+1$, so that $x = 12-2z-1-z = 11-3z$. And, since $y$ cannot be greater than 8, we must not substitute any numbers for $z$ which would render $2z+1$ greater than 8; consequently, $z$ must be less than 4, that is to say, $z$ cannot be taken greater than 3, for which reasons we have the following answers: If we make $z=0 | z=1 | z=2 | z=3$, we have $y=1 | y=3 | y=5 | y=7$, and $x=11 | x=8 | x=5 | x=2$. Hence, the two parts of 25 sought, are $(2x+3y) = 22+3$; $16+9$; $10+15$; or $4+21$. 5. Question 2. To divide 100 into two such parts, that the one may be divisible by 7, and the other by 11. Let $7x$ be the first part, and $11y$ the second. Then we must have $7x+11y=100$; and, consequently, $x = \frac{100-11y}{7} = \frac{98+2-7y-4y}{7} = 14-y+\frac{2-4y}{7}$; wherefore $2-4y$, or $4y-2$, must be divisible by 7. Now, if we can divide $4y-2$ by 7, we may also divide its half, $2y-1$, by 7*. Let us therefore make $2y-1=7z$, or $2y=7z+1$, and we shall have $x=14-y-2z$; but, since $2y=7z+1=6z+z+1$, we shall have $y = 3z + \frac{z+1}{2}$. Let us therefore make $z+1=2u$, or $z=2u-1$; which supposition gives $y=3z+u$; and, consequently, we may substitute for $u$ every integer number that does not make $x$ or $y$ negative. Now, as $y$ becomes $=7u-3$, and $x=19-11u$, the first of these expressions shews that $7u$ must exceed 3; and, according to the second, $11u$ must be less than 19, or $u$ less than $\frac{19}{11}$; so that $u$ cannot be 2; and since it is impossible for this number to be 0, we must have $u=1$: which is the only value that this letter can have. Hence, we obtain $x=8$, and $y=4$; and the two parts of 100 which were required, are 56, and 44. * For $\frac{2-4y}{7}$, or $\frac{4y-2}{7}$, being a whole number, and 4 and 2 not being divisible by 7, the numerator, $4y-2$, and its half, $2y-1$, must necessarily be either 7, or some multiple of 7: and it may be observed, that, if any number divides the whole of another number, and also a part of it, it will likewise divide the remaining part.