# A/b = 8 and ac - 5 = 11, then bc=?

Question:I can't figure this out? Any ideas?

bc=2 cuz:

a=8b

ac = 16

sub in a.

8b(c) = 16

bc=2
Part 1
if a/b = 8, then a = 8b

Part 2
ac - 5 = 11 is ac=16
plug a = 8b in
8bc = 16
c = 2/b

Part 3
bc = ??
b(2/b) = 2
errrr. A could be 16, C could be 1, and B could be 2. Then BC would be 2.
a/b=8, so a=8b
and
ac-5=11

By substitution,
8bc-5=11
8bc=16
bc=2
a/b = 8 and ac-5 = 11.

So ac=16.

Since we know that, we can say that ac/2=8.

Therefore, a/b=ac/2=8.

Looks to me like a=8, b=1 and c=2, so bc=2.
1) solve a/b=8 for b. b=a/8

2) solve ac-5=11 for c c=16/a

3) therefore bc would be (a/8)(16/a) a's cancel out and 16/8=2. So b times c = 2
First equation is:

a/b = 8

Second equation is:

ac - 5 = 11

To find what bc is, we need to find what is the value of a and c

So to find 'a' and 'c', we solve it like with solve simultaneous equations,

Therefore:

a/b = 8

Bring over the b to the other side, so from divide it becomes times

a = 8b --------> first equation

ac - 5 = 11
ac = 11 + 5
ac = 16 ---------> second equation

Put the First equation into the Second equation.

ac = 16

becomes

(8b)c = 16
8bc = 16
bc = 16/8

To only get bc like how you wanted it, you have to move the eight to the other side. So since 8bc is actually 8 times 'bc'. When you move it to the other side, you would reverse it and it becomes divide

Thus bc = 2

I hope you understand what I'm trying to explain though. If you need anymore explanation, I'll be obliged to help.

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