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If x and y are positive integers, what are all the solutions (x comma y) of the equation (3 times x) + (2 times y) = 11

 Apr 25, 2014

Best Answer 

 #2
avatar+128566 
+5

Re: If x and y are positive integers, what are all the solutions (x,y) of the equation (3*x) + (2*y) = 11

-------------------------------------------------------------------------------------------------------------------------- What we have here is a line with the following equation.

y = (-3/2)x + 11/2  

It has  a y-intercept of (0, 11/2) and an x intercept of (11/3,0) ≈ (3.66, 0) Since you specified that x  and y must be positive integers, we're looking for any x values that lie between 1 and 3, inclusively.

So we have:

x = 1,   y = 4  

x = 2,   y = 5/2

x = 3,   y = 1

So, the only things that "work" are (1,4) and (3,1)  

 Apr 25, 2014
 #1
avatar+33616 
+5

Rewrite this as x = (11 -2y)/3

Now 2y is an even number, so 11 - 2y must be an odd number.  So we just look at odd numbers below 11 that are exactly divisible by 3.  We quickly see that there are only two, namely 3 and 9.

When 11 - 2y = 3 we must have x = 1, so y = 4

When 11 - 2y = 9 we must have x = 3, so y = 1

 Apr 25, 2014
 #2
avatar+128566 
+5
Best Answer

Re: If x and y are positive integers, what are all the solutions (x,y) of the equation (3*x) + (2*y) = 11

-------------------------------------------------------------------------------------------------------------------------- What we have here is a line with the following equation.

y = (-3/2)x + 11/2  

It has  a y-intercept of (0, 11/2) and an x intercept of (11/3,0) ≈ (3.66, 0) Since you specified that x  and y must be positive integers, we're looking for any x values that lie between 1 and 3, inclusively.

So we have:

x = 1,   y = 4  

x = 2,   y = 5/2

x = 3,   y = 1

So, the only things that "work" are (1,4) and (3,1)  

CPhill Apr 25, 2014

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