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RenB
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If x is an integer and y is a non-negative integer such that 4x-3y = 3 18 Sep 2023, 21:04
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85% (hard)

Question Stats:

59% (02:44) correct 41% (02:59) wrong based on 564 sessions

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If x is an integer and y is a non-negative integer such that 4x-3y = 36 and |y - 3| < 16, then how many ordered pairs (x, y) satisfy the given conditions?

A. 2
B. 3
C. 4
D. 5
E. 6
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D
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If x is an integer and y is a non-negative integer such that 4x-3y = 3 18 Sep 2023, 21:54
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RenB
If x is an integer and y is a non-negative integer such that 4x-3y = 36 and ly - 3| < 16, then how many ordered pairs (x, y) satisfy the given conditions?
A. 2
B. 3
C. 4
D. 5
E. 6
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y is non negative and |y-3|<16
Two cases:
1) y-3<16 or y<19
2) 3-y<16 or y>-13
Thus, \(0\leq y<19\)

Now, \(4x-3y=36\)
\(x=\frac{36+3y}{4}=9+\frac{3y}{4}\)
So, for y as multiple of 4 will give an integer value for x.
Thus, y can be 0,4,8,12 and 16.

Answer: 5 values.

(9,0), (12,4), (15,8), (18,12), (21,15)

D
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If x is an integer and y is a non-negative integer such that 4x-3y = 3 18 Sep 2023, 23:06
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RenB
If x is an integer and y is a non-negative integer such that 4x-3y = 36 and ly - 3| < 16, then how many ordered pairs (x, y) satisfy the given conditions?
A. 2
B. 3
C. 4
D. 5
E. 6
Show more

Given

\(|y - 3| < 16\)

\(-16 < y - 3 < 16\)

Adding three on both sides of the equation

\(-16 + 3 < y - 3 + 3 < 16 + 3\)

\(-13 < y < 19\)

\(4x-3y = 36\)

\(3y = -36 + 3x + x\)

Divide by 3 on both sides of the equality

\(y = -12 + x + \frac{x}{3}\)

The question premise mentions that y is an integer. Hence,

x = 3 ➜ y = -12 + 3 + 1 = -8

Further values can be found as shown below -

Attachment:
Screenshot 2023-09-19 112748.jpg
Screenshot 2023-09-19 112748.jpg [ 74.87 KiB | Viewed 5464 times ]

Option D
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If x is an integer and y is a non-negative integer such that 4x-3y = 3 10 Jan 2025, 06:44
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|y - 3| < 16
=> -16 < y - 3 < 16

Adding 3 on all the sides
=> -16 + 3 < y - 3 + 3 < 16 + 3
=> -13 < y < 19

But y is non-negative
=> 0 ≤ y < 19

4x-3y = 36
=> 4x = 3y + 36
=> x = \(\frac{3y + 36}{4}\) = \(\frac{3y}{4} + 9\)

Since, x is an integer so only those values of y which will make [m]\frac{3y}{4} an integer are possible
=> y will be a multiple of 4

=> Multiplies of 4 between 0 and 19, which are 0, 4, 8, 12, 16
=> 5 values

So, Answer will be D
Hope it helps!

Watch the following video to MASTER Absolute Values

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If x is an integer and y is a non-negative integer such that 4x-3y = 3 25 Mar 2026, 15:06
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