If x and y are nonnegative integers and 5x + 2y = 32, how many values of x are there?

A. 1
B. 2
C. 3
D. 4
E. 6
_________________

we are looking foe the number of x may take . Trail and Error is the best in my sense.

we know that both x and y are integers and they are non-negative in nature.

even - even = even, which is divisible by 2. Plug even values for x only.

5x + 2y = 32.

1. x = 0

5*0 + 2y = 32 = y = 16.

2. x=2

5*2 + 2y = 32

y = 11

3. x = 4

5*4 + 2y = 32

y = 6

4. x = 6

5*6 + 2y = 32

y = 1.


Therefore the equation is true when we put these values of x. one may plug the value of y and look for the integer value of x instead of y.

The best answer is D.

I am not able to understand any of the solutions for the given problem. Can someone explain it?

Solution:

Simplifying, we have:

5x = 32 - 2y

5x = 2(16 - y)

x = 2(16 - y)/5

The expression (16 - y) must be a multiple of 5, in order for the entire expression on the right side of the equation to be a non-negative integer. Thus, we see that y can be 1, 6, 11, or 16. Since there are 4 possible values for y, there are also 4 possible values for x.

Alternate Solution:

Notice x and y are nonnegative integers, we see that in order for 5x + 2y = 32 to hold, x can’t be greater than 6 (since 5 * 7 = 35 already). Therefore, x can only be any integers from 0 to 6, inclusive. Furthermore, if x is odd, then 5x is odd and 2y = 32 - 5x will be odd. However, since 2y is always even regardless of what y is, we see that x can’t be odd. On the other hand, if x is even, then 5x is even and 2y = 32 - 5x will be even. Therefore, x can only be 0, 2, 4 or 6,

Answer: D
_________________