Question

If x and y are $+ove$ integers, then find the solutions of the following equations separately :
(i) $7x+12y=220$
(ii) $14x-11y=29$

Answer 1

(i) $x=\frac{1}{7}(220-12y)$ or $y=\frac{1}{12}(220-7x)$
Now give $+ive$ integral values to y such that x also becomes $+ive$ integral. The first $+ive$ integral value of y is $2$ (as $y=1$ will not make x, $+ive$ and integral) so that
$x=\frac{1}{7}(220-122)=\frac{196}{7}=28$
$\therefore$ $x=28,y=2$
Now the values of x form an A.P. whose common difference is $-12$
$\because$ $T_n=pn+q$ forms an A.P. whose common difference is p and first term is $p+q$. Hence the other $+ive$ integral values of x will be $16,4,-8,-20$,.....
We choose only $x=16$ and $4$, which are $+ive$
$\therefore$ $12y=220-112=108$ $\therefore$ $y=9$.
$12y=220-28=192$ $\therefore$ $y=16$
Hence $(x,y)=(28,2),(16,9)(4,16)$ only.
Note : You observe that the values of y, i.e. $16,9,2$,..... form an A.P. of common difference $-7$.
(ii) Ans. $(x,y)=(6,5),(17,9),(28,33),(39,47)$,.....