Question

HELP! What is the largest number k for which 2x+3y=20 and 3x+4y=k have the same number of positive integer solutions (x,y)?

Answer 1

Given:

$2x+3y=20$

$3x\;+\;4y\;=\;k$

To find:

The largest number $k$ for the equations have the same number of positive integer solutions $(x,\;y)$

Step-by-step explanation:

$2x\;+\;3y\;=\;20\\\3y\;=\;-2x\;+\;20\\y\;=\;\frac{-2}3x\;+\;\frac{20}3$ ------------------- ($1$)

$3x\;+\;4y\;=\;k\\4y\;=\;-3x\;+\;k\\y\;=\;\frac{-3}4x\;+\;\frac k4$------------------- ($2$)

Equating ($1$) and ($2$),

$\\\frac{-2}3x\;+\;\frac{20}3\;=\;\frac{-3}4x\;+\;\frac k4$

⇒ $\frac{-8x\;+\;80}{12}\;=\;\frac{-9x\;+\;3k}{12}$

⇒ $-8x\;+\;80\;=\;-9x\;+\;3k\;$

⇒ $x\;=\;3k\;-\;80$

These are positive integer solutions and lie in the first quadrant.

From equation ($1$) we know that it is in the first quadrant for

$0\;<\;x\;<\;10$

$0\;<\;y\;<\;6.667$

Therefore, the possibilities of $(x,\;y)$ are

$x\;=\;\left\{1,2,3,4,5,6,7,8,9\right\}\\y\;=\;\left\{1,2,3,4,5,6\right\}$

Any pair of $(x,\;y)$ satisfies the positive integer solutions.

$x\;=\;3k\;-\;80\\9\;=\;3k\;-\;80\\k\;=\;\frac{89}3$

For $1\;\leq\;x\;\leq\;9$

$27\;\leq\;k\;\leq\;\frac{89}3$

$k\;=\;27,\;28,\;29$

Therefore, for $k\;=\;29$

$(x, y) \;=\; (7,2)$

Answer:

The largest number $k$ for the equations that have the same number of positive integer solutions $(x,\;y)$ is $29$

Answer 2

Largest number k = 48 for  which  3x+4y=k have the same number of positive integer solutions (x,y) as 2x+3y=20

Given:

• 3x+4y=k have the same number of positive integer solutions (x,y) as 2x+3y=20

To Find:

Largest possible value of k

Step 1:

Find number of  possible positive integer solutions for 2x+3y=20

3y = 20 - 2x

3y = 2(10 - x)

y must be a multiple of 2

y = 0 not possible as as only positive integers are the solutions

y = 2  => x = 7

y = 4  => x = 4

y = 6 => x  = 1

These are only possible positive integer solutions

as y = 8 will result in negative value of x.

Hence 3 possible solution

( 1 , 6) , (4 , 4) , ( 7 , 2)

Step 2:

Find maximum value of k

3x + 4y = k

Three possible solutions

( a , b) , (a + 4 , b - 3) , (a + 8 , b - 6)

where a can be 1 , 2 , 3 or  4

b - 6 can be 1 , 2 , 3  =>  b can be 7 , 8 , 9

Taking maximum value of a and b  for max value of k

(4 , 9) , ( 8 , 6) and ( 12 , 3) are the solution

3(4) + 4(9) = 48 = k

and 3x + 4y = 48  is the equation

Hence largest number k = 48