If x 0, how many integer values of (x, y) will satisfy the equation

Question

If x > 0, how many integer values of (x, y) will satisfy the equation 5x + 4|y| = 55?

(A) 3
(B) 6
(C) 5
(D) 4
(E) 2

GMATBusters · Accepted Answer

Please see the sketch for procedure to find the integral solution of a linear equation. IMG-20180417-WA0066.jpg

GMATBusters · Answer

The equation will be
5x +4y =55 , when y > or =0
Considering integer solution, (x,y) can be (3,10),(7,5),(11,0)
5x - 4y =55 , when y<0
Considering integer solutions, (x,y) can be (3,-10),(7,-5)
Hence there will be 5 integral solutions.
Answer C.

Jahfors · Answer

Greetings!
How did u calculate solutions?

push12345 · Answer

To get x as intger y need to be multiple of 5 
So possible values of y are 0,5,10,15....

But only after and with 15 x becomes -ve but x cannot be negative

So values of y are 0,5,10
Y is |y| so -ve values need to be considered

Total values of y are 0,5,-5,10,-10

With each of these y values x value also comes
So answer is 5

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robertops · Answer

Good one, I did it using the 'monkey' method, trying x,y numbers...

Doer01 · Answer

Hi,
 Given inequality 5x=4|y| = 55 could be written as
|y| = (55-5x)/4 = 5(11-x)/4
so, (11-x) has to be divisible by 4 for y to be an integer
and x>0
so it could be 8 when x = 3 or 4 when x = 7 or 0 when x = 11.
|y| = 10 when x =3 
|y| = 5 when x =7 
|y| = 0 when x =11
now, y=+-10, +-5,0
considering pairs (3,10)(3,-10),(7,5),(7,-5),(11,0)

Fdambro294 · Answer

We are given the linear indeterminate equation:

5X + 4[Y] = 55

Scenario 1: when X > 0 and Y < 0

Since 5 divides 55 evenly, we can start with:

X = 11 and Y = 0 ——- satisfy equation

Since Y < 0 ———> [Y] = -Y

So we have:

5X - 4Y = 55

Rule: to find the integer values that will satisfy the equation, The X value will decrease by the value of the Y coefficient (-4 in this case) and Y will similarly decrease by the X coefficient (-5 in this case)

X = 7 and Y = -5 ——— satisfies

X = 3 and Y = -10 ———- satisfies

We can not go any lower because X > 0 is a constraint and the next values that will satisfy the equation when Y < 0 is: X = -1. and Y = -15

So we have 2 cases in which Y is negative and 1 case in which Y = 0

3 cases so far

Scenario 2: when Y > 0

Absolute value of [Y] when Y > 0 ———-> [Y] = Y

5X + 4Y = 55

Again, we can start out with the case when Y = 0 as a starting point (though we already counted it

X = 11 — and — Y = 0

Rule: to find the integer values that satisfy, X will decrease by the coefficient of Y (-4 in this case) and Y will increase by the coefficient of X (+5 in this case)

X = +7 — and — Y = +5 ———- satisfies

X = +3 — and — Y = +10

The next case that will satisfy is X = -1 and Y = +15 , but we have the constraint where X > 0

That is 2 more cases in which Y is positive

In total, 5 integer values of (X , Y) satisfy the equation.

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If x > 0, how many integer values of (x, y) will satisfy the equation

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