Elements of Algebra

316 ELEMENTS PART II. whence it follows, that $19t + 3q$ must be less than $27$, and that, provided this condition be observed, we may give any value to $q$ and $t$. We have therefore to consider the following cases: 1. If $t = 0$ we have $p = 1$ $q = q$ $r = 27 - 3q$ $s = 72 + 2q$. 2. If $t = 1$ $p = 4$ $q = q$ $r = 8 - 3q$ $s = 88 + 2q$. We cannot make $t = 2$, because $r$ would then become negative. Now, in the first case, $q$ cannot exceed $9$; and, in the second, it cannot exceed $2$; so that these two cases give the following solutions, the first giving the following ten answers: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. $p = 1 \ 1 \ 1 \ 1 \ 1 \ 1 \ 1 \ 1 \ 1 \ 1$ $q = 0 \ 1 \ 2 \ 3 \ 4 \ 5 \ 6 \ 7 \ 8 \ 9$ $r = 27 \ 24 \ 21 \ 18 \ 15 \ 12 \ 9 \ 6 \ 3 \ 0$ $s = 72 \ 74 \ 76 \ 78 \ 80 \ 82 \ 84 \ 86 \ 88 \ 90$. And the second furnishes the three following answers: 1. 2. 3. $p = 4 \ 4 \ 4$ $q = 0 \ 1 \ 2$ $r = 8 \ 5 \ 2$ $s = 88 \ 90 \ 92$ There are, therefore, in all, thirteen answers, which are reduced to ten if we exclude those that contain zero, or $0$. 30. The method would still be the same, even if the letters in the first equation were multiplied by given numbers, as will be seen from the following example. Question 5. To find three such integer numbers, that if the first be multiplied by $3$, the second by $5$, and the third by $7$, the sum of the products may be $560$; and if we multiply the first by $9$, the second by $25$, and the third by $49$, the sum of the products may be $2920$. If the first number be $x$, the second $y$, and the third $z$, we shall have the two equations, 1. $3x + 5y + 7z = 560$ 2. $9x + 25y + 49z = 2920$. And here, if we subtract three times the first, or $9x+15y+21z=1680$, from the second, there remains $10y + 28z = 1240$; dividing by $2$, we have $5y + 14z = 620$; whence