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2a+b=12 and |b| less than or equal to 12
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29 Feb 2024, 06:16
29 Feb 2024, 06:16
Expert Reply
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Question Stats:
47% (02:26) correct
52% (01:45) wrong based on 34 sessions
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\( 2a+b=12\) and \(|b| \leq 12\)
How many ordered pairs ab are solutions of the above system such that a and b both are integers?
(A) 9
(B) 10
(C) 11
(D) 12
(E) 13
How many ordered pairs ab are solutions of the above system such that a and b both are integers?
(A) 9
(B) 10
(C) 11
(D) 12
(E) 13
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Re: 2a+b=12 and |b| less than or equal to 12
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30 Mar 2024, 11:19
30 Mar 2024, 11:19
1
If |b| is less than or equal to 12, then b is more than or equal to -12 or less than or equal to 12.
Also, b can only be an even number because 2a is always even.
Now, if I understand it correctly, the ordered pairs are
0,12 (0+12)
1,10 (2+10)
2,8 (4+8)
3,6 (6+6)
4,4 (8+4)
5,2 (10+2)
6,0 (12+0)
7, -2 (14-2)
8, -4 (16-4)
9, -6 (18-6)
10, -8 (20-8)
11, -10 (22-10)
12, -12 (24-12)
Is there a way to solve it algebraically?
Also, b can only be an even number because 2a is always even.
Now, if I understand it correctly, the ordered pairs are
0,12 (0+12)
1,10 (2+10)
2,8 (4+8)
3,6 (6+6)
4,4 (8+4)
5,2 (10+2)
6,0 (12+0)
7, -2 (14-2)
8, -4 (16-4)
9, -6 (18-6)
10, -8 (20-8)
11, -10 (22-10)
12, -12 (24-12)
Is there a way to solve it algebraically?
Re: 2a+b=12 and |b| less than or equal to 12
[#permalink]
30 Mar 2024, 11:26
30 Mar 2024, 11:26
1
Expert Reply
We see that the value of b ranges from −12 to +12; this follows that b can have 25 number of integer values.
Now let us see how many integer values a can have.
\(2a+b=12 \)
\(a=12-\frac{b}{2}\)
\(a=6-\frac{b}{2}\)
We see that for a to be an integer, b/2 must be an integer; this follows that b must be an even number.
Out of 25 possible values of b, 13 are even (including 0); so for a to be an integer, the set of arrangement can only have 13 ordered pairs.
Now let us see how many integer values a can have.
\(2a+b=12 \)
\(a=12-\frac{b}{2}\)
\(a=6-\frac{b}{2}\)
We see that for a to be an integer, b/2 must be an integer; this follows that b must be an even number.
Out of 25 possible values of b, 13 are even (including 0); so for a to be an integer, the set of arrangement can only have 13 ordered pairs.
