Question

If $ 4x + 3y = 120 $ , find how many positive integer solutions are possible?

Answer 1

Hint: We know that a linear equation is the combination of constant and variable of the first order, the given equation is thus linear. The solution of a linear equation means the value of x and y which satisfies the linear equation. In this question, we have to find the number of positive integer solutions, so by applying the given conditions on the linear equation, we can find out the correct answer.

Complete step-by-step answer:
We are given that $ 4x + 3y = 120 $
This equation can be rewritten to express y in terms of x as follows –
 $
  4x + 3y = 120 \\
   \Rightarrow 3y = 120 - 4x \\
   \Rightarrow y = \dfrac{4}{3}(30 - x) \;
  $
Now, for y to be positive,
 $
  \dfrac{4}{3}(30 - x) > 0 \\
   \Rightarrow 30 - x > 0 \\
   \Rightarrow x < 30 \;
  $
 $ \dfrac{4}{3}(30 - x) $ can also be written as $ 40 - \dfrac{4}{3}x $ , from this simplified form, we see that for y to be an integer, x needs to be a multiple of 3 that is $ x = 3k $
As the solutions required are positive integers, so $ x > 0 $
So x lies between 0 and 30 and is a multiple of 3, thus x can take values $ 3,6,9,12....27 $ .
Or $ k = 1,2....9 $
 $ \Rightarrow y = \dfrac{4}{3}(30 - 3k) = 4(10 - k) = 40 - 4k $
Therefore, the solution set is $ (3k,40 - 4k) $ where $ k = 1,2,3....9 $
Thus, 9 positive integer solutions are possible.
So, the correct answer is “9”.

Note: Integers include both positive and negative numbers, the number that can be written without a fraction component or that are not in decimal form are called integers, so positive integers means the numbers greater than 0. For example, $ 1,2,3... $ are positive integers. We take $ x = 3k $ because by using this value, y is also obtained as an integer and not in the form of fraction or decimal.