Question

Find the value of x and y in positive integers, \[14x-11y=29\] .
(A) (5,0)
(B) (5,6)
(C) (6,5)
(D) (2,4)

Answer 1

Hint: In the given equation, we have the equation \[14x-11y=29\] . In this equation, the LHS is equal to \[14x-11y\] while the RHS is equal to 29. We have four options, A, B, C, and D. Put the values of x and y from all the options given. After putting the values of x and y in the LHS, if we have the value got its value equal to the RHS, i.e, 29. Then, those values of x and y satisfy the equation.

Complete step-by-step solution -
According to the question, we have the equation
\[14x-11y=29\] …………………….(1)
In the given equation, we have the LHS equal to \[14x-11y\] and RHS is equal to 29.
In LHS, we have
\[14x-11y\] ………………….(2)
The RHS of equation (1) is equal to 29 …………………………..(3)
We have four options given, A, B, C, and D.
In option (A), we have (5,0).
Now, putting \[x=5\] and \[y=0\] in equation (2), we get
\[\begin{align}
  & 14x-11y \\
 & =14\times 5-11\times 0 \\
 & =70-0 \\
\end{align}\]
\[=70\] ……………………..(4)
From equation (4), we have got the value of the LHS of the given equation after putting \[x=5\] and \[y=0\] .
Since the value of the LHS of the equation \[\left( 14x-11y \right)\] is given as 29 but from equation (4), we have got the value of \[\left( 14x-11y \right)\] equal to 70. So, we can say that the values \[x=5\] and \[y=0\] don’t satisfy the equation \[\left( 14x-11y \right)\] . Therefore, option (A) is not the correct option.
In option (B), we have (5,6).
Now, putting \[x=5\] and \[y=6\] in equation (2), we get
\[\begin{align}
  & 14x-11y \\
 & =14\times 5-11\times 6 \\
 & =70-66 \\
\end{align}\]
\[=4\] ……………………..(5)
From equation (5), we have got the value of the LHS of the given equation after putting \[x=5\] and \[y=6\] .
Since the value of the LHS of the equation \[\left( 14x-11y \right)\] is given as 29 but from equation (5), we have got the value of \[\left( 14x-11y \right)\] equal to 4. So, we can say that the values \[x=5\] and \[y=6\] don’t satisfy the equation \[\left( 14x-11y \right)\] . Therefore, option (B) is not the correct option.
In option (C), we have (6,5).
Now, putting \[x=6\] and \[y=5\] in equation (2), we get
\[\begin{align}
  & 14x-11y \\
 & =14\times 6-11\times 5 \\
 & =84-55 \\
\end{align}\]
\[=29\] ……………………..(6)
From equation (6), we have got the value of the LHS of the given equation after putting \[x=6\] and \[y=5\] .
Since the value of the LHS of the equation \[\left( 14x-11y \right)\] is given as 29 but from equation (6), we also have the value of \[\left( 14x-11y \right)\] equal to 29. So, we can say that the values \[x=5\] and \[y=6\] satisfy the equation \[\left( 14x-11y \right)\] . Therefore, option (C) is the correct option.
Hence, option (C) is the correct one.

Note: We can also solve this question in another way. First of all, divide the LHS and RHS of the equation by 11 and then make the equation as \[x-y+\dfrac{3x-7}{11}=2\] . As, x and y is an integer, so \[\dfrac{3x-7}{11}\] must be an integer. Multiply by 4 in \[\dfrac{3x-7}{11}\]and make this equation as \[x-2+\dfrac{x-6}{11}\] . As \[\dfrac{3x-7}{11}\] is an integer, so \[x-2+\dfrac{x-6}{11}\] must also be an integer. Since the expression \[x-2+\dfrac{x-6}{11}\] is an integer so, \[\dfrac{x-6}{11}\] should also be an integer. Consider \[\dfrac{x-6}{11}=p\] , then find x and y in terms of \[p\] and solve further by putting values of p and p should be greater than 0.