Math
posted by Andrea .
Can someone please help me with an algebra problem?
48b=12
I really get confused with the absolute value symbols and a negative sign outside of them.
Don't worry about it; it does tend to be a bit confusing.
I'll explain the difference between inside and outside negative signs first.
I'm assuming you know that the absolute value of a number is nothing more than its distance from zero on a number line.
Anyway, inside:
4
The negative sign is inside the absolute value sign. The absolute value of negative four is 4.
Outside:
4
First, we would want to find the absolute value of four, which is, of course, four. Now, the negative sign is not inside the absolute value sign, so it remains, resulting in negative four.
Now on to your problem.
48b=12
We first need to lose the absolute value signs, which means first getting rid of that four. Since it's a positive four within the confines of the absolute value sign, subtract four from both sides.
8b=8
Now we can take out that abs. val. sign.
The absolute value of 8b is 8b, since 8b is eight away from zero.
BUT there's that negative sign on the outside, which is unaffected by the abs. val. sign, and which is attached to the 8b, resulting in:
8b=8
Divide both sides by 8:
b=1
Sorry, but no. The absolute value is always a positive number. There is no solution to this problem.
We have 48b=12
If we multiply both sides by 1 we have
48b=12
No solution exists.
Okay, thankyou.
I'll just remark too that if the problem were
48b=12 then there would be two solutions, For the first you'd set
48b=12 and for the second you'd set
1(48b)=12
You should get b=1 to satisfy the first and b=2 for the second.
This would be a completely different problem though.
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