Question

If $x$ and $y$ are $+$ve integers, then find the solution of the following equation:

$14x-11y=29.$

• A. $(x,y)=(6,5),(17,19),(28,33),(39,47),\cdots$
• B. $(x,y)=(3,5),(17,16),(28,33),(39,47),\cdots$
• C. $(x,y)=(2,5),(27,19),(28,33),(39,37),\cdots$
• D. $(x,y)=(1,9),(17,79),(28,33),(69,47),\cdots$

Answer 1

The correct option is A $(x,y)=(6,5),(17,19),(28,33),(39,47),\cdots$

Write the equation in the form of $y$,

$14x-11y=29$
$\frac{14x-29}{11}=y$
For positive integral values of $y$, $(14x-29)$ should be a multiple of 11 and $14x>29$.
$x>2$...(i)

At $x=6$, we get .................(let's consider option A, as only it satisfies the required condition)
$y=\frac{70-29}{11}$
$y=\frac{55}{11}$
$y=5$
Similarly at $x=17$
$y=\frac{238-29}{11}$
$y=\frac{209}{19}$
$y=19$. ...(ii)
Hence x takes the values $6,17,28,\mathrm{39...}$

This is an A.P with the initial term as 6 and a common difference as 11.

Similarly, $y$ takes the value
$5,19,33,\mathrm{47...}$

This is also an A.P with initial term 5 and common difference 14.

Hence the ordered pairs of x and y are
$(x,y)=(6,5),(6+11,5+14),(6+22,5+28)...$
$\Rightarrow (x,y)=(6,5),(17,19),(28,33)...$