equation system

srcc25anu wrote:if a and y are integers and 2x-y = 11, then 4x+y cannot be:

a. -5
b. 1
c. 13
d. 17
e. 55

I suppose it is "x and y are integers" not "a and "y".
Given that 2x-y=11
=> 2x= 11+y
=> 4x = 22 + 2y
then, 4x+y = 22 + 2y +y = 22 + 3y

Now if we try plugging in

for option (a) 22 + 3y = -5 , this leads to y = -9 ( an integer )
for option (b) 22 + 3y = 1 , this leads to y = -7 ( an integer )
for option (c) 22 + 3y = 13 , this leads to y = -3 ( an integer )
for option (d) 22 + 3y = 17 , this leads to y = -5/3 ( not an integer )
for option (e) 22 + 3y = 55 , this leads to y = 11 ( an integer )

Hence , we can mark "D" as answer.

Correct D should be the answer.. i think that's the easiest way.

srcc25anu wrote:if a and y are integers and 2x-y = 11, then 4x+y cannot be:

a. -5
b. 1
c. 13
d. 17
e. 55

2x-y = 11.
4x+y = z.

Adding the two equations, we get:
6x = 11+z.

Now we plug in the answer choices for z to see which turns x into a non-integer.

Answer choice D: z = 17
6x = 11+17
6x = 28
x = 28/6 = 14/3.
Not an integer.

The correct answer is D.

If 2x - y = 11, 4x = 22 + 2y and 4x + y = 22 + 3y

Thus, 4x + y is 1 greater than a multiple of 3. 17 is 2 greater than a multiple of 3, so 17 cannot be the value of 4x + y