Help me with math?

[GuyProblems24_7]

A*b+c=18
b*c=16
a+b=7

[abc3643]

OK, let me re-write what you wrote first, but make all the letters in CAPS, add some spacing and start naming the equations.

[EQN_1 which means Equation 1]: A * B + C = 18

[EQN_2]: B * C = 16

[EQN_3]: A + B = 7

OK, so you have 3 nonlinear equations in 3 unknown variables.

Now, before I continue, I solved it in my head and the answer is:

A = 5, B = 2, C = 8

OK, let's now prove it. This next paragraph is going to lay-out what we're going to do...

Generally, that can suck, but we are in luck. EQN_2 and EQN_3 are equations that have only 2 of the 3 unknown variables and they are both linear. EQN_1 is nonlinear, but, more importantly, it has all 3 unknown variables. What's really good is that B is the common variable in EQN_2 and EQN_3, so, what we are going to do is solve EQN_2 and EQN_3 in the other variable in terms of B; that is, in EQN_2, we're going to solve for C in terms of B and, in EQN_3, we're going to solve for A in terms of B. Once we have A in terms of B and C in terms of B, we just plug those into EQN_1 which will just become an equation in just B. After that, you solve that equation for B and, because you know B, you can figure out A and C.

Let's get started. First, we're going to take EQN_2 and solve for C. Rewriting EQN_2:

[EQN_2]: B * C = 16

Solving for C, divide each side by B.

[EQN_4]: C = 16 / B

We're going to need EQN_4 later because we'll use that to replace C in EQN_1 and, at the end after we know the value of B, we'll need this to get the value of C.

Now for EQN_3 and solve for A. Rewriting EQN_3:

[EQN_3]: A + B = 7

Solving for A, subtract B from each side.

[EQN_5]: A = 7 - B

Again, we're going to need EQN_5 later because we'll use that to replace A in EQN_1 and, at the end after we know the value of B, we'll need this to get the value of A.

OK, so now we are onto the next phase: Plugging in EQN_4 and EQN_5 into EQN_1 to get a single equation in B. Rewriting EQN_1:

[EQN_1]: A * B + C = 18

Now we use EQN_4 to get rid of C.

[EQN_4]: C = 16 / B

So, substituting EQN_4 into EQN_1:

A * B + C = 18

[EQN_6]: A * B + (16 / B) = 18

Great! So, C is gone and we don't care about it until near the end after we know the value of B.

OK, so, think of EQN_6 is the new EQN_1 because we are going to use that now instead of EQN_1.

Now we use EQN_5 to get rid of A.

[EQN_5]: A = 7 - B

So, substituting EQN_5 into EQN_6 (and NOT EQN_1):

A * B + (16 / B) = 18

[EQN_7]: (7 - B) * B + (16 / B) = 18

EQN_7 is our single equation in B. It's ugly.

Let's see what we can do…

Multiply across by B.

B * [(7 - B) * B + (16 / B)] = B * 18

(7 * B - B * B) * B + 16 * B / B = 18 * B

7 * B * B - B * B * B + 16 = 18 * B

Doing some simplifying and cleaning up…

[EQN_8]: B^3 - 7B^2 + 18B - 16 = 0

Wow, that's a cubic equation and beyond your skills, so we need to try something a little smarter…

First, let's see if I was right…

I said B was 2. That's plug that into EQN_8:

B^3 - 7*B^2 + 18*B - 16 =?= 0

(2)^3 - 7*(2)^2 + 18*(2) - 16 =?= 0

8 - 7*4 + 36 - 16 =?= 0

8 - 28 + 36 - 16 =?= 0

Yes. So __one__ answer for B = 2. (A cubic equation will have 3 answers.)

(more)

[abc3643]

OK, let's start over:

[EQN_1]: A * B + C = 18

[EQN_2]: B * C = 16

[EQN_3]: A + B = 7

Hmmm. Looking at EQN_1, if we multiplied everything by B, we'd get a B*C term and we can use EQN_2 to plug that in.

B*(A * B + C) = B*18

A*B^2 + B*C = B*18

But, EQN_3 says B*C = 16, so:

[EQN_9]: A*B^2 + 16 = B*18

This is looking good! We only have a quadratic equation in B and we can solve that. However, we still have A. So, let's use EQN_5 and substitute into EQN_9:

A*B^2 + 16 = B*18

(7-B)*B^2 + 16 = B*18

Uh oh…

7*B^2 - B^3 + 16 = 18*B

-B^3 + 7*B^2 - 18*B + 16 = 0

B^3 - 7*B^2 + 18*B - 16 = 0

Wow… That's EQN_8 all over again.

At this point, if you can use a calculator and just plug in the cubic and have it solve, then I guess that's OK. When I was in school, you had to solve it. This is a cubic equation and there are methods to solve it like a quadratic equation, but that's still not easy and I doubt that you are responsible for that.

What is this class that you are taking?
Are you allowed to use a calculator to solve cubics?

[abc3643]

Again, though, once we know B, which is 2, then we get A and C...
EQN_4 says C=16/B, so C = 8.
EQN_5 says A = 7-B, so A = 5.

[abc3643]

Plugging A = 5, B = 2, C = 8 back into the original equations...

[EQN_1]: A * B + C = 18
(5) * (2) + (8) = 18
10 + 8 = 18
18 = 18 Check!

[EQN_2]: B * C = 16
(2) * (8) = 16
16 = 16 Check!

[EQN_3]: A + B = 7
(5) + (2) = 7
7 = 7 Check!

So, we know that A=5, B=2, and C=8 is an answer for that system of equations.
(Technically, there may be two other answers because we know that the system is actually a cubic in B so B will have 3 values that will solve that cubic.)

Using the cubic equation calculator here:
www.calculatorsoup.com/.../cubicequation.php
and plugging in the coefficients of 1, -7, 18, -16, it gave these as the solutions of the cubic; that is, the values of B that solve it. That's exact equation is here:
www.calculatorsoup.com/.../cubicequation.php
and the solutions are:
Answer:
Solutions for x:
x1 = 2
x2 = 2.5 + i * 1.32288
x3 = 2.5 - i * 1.32288

Those last two are complex numbers and, specifically, "complex conjugates". They are not real numbers because they have an imaginary part (the number to the right of "i" which is the square root of -1). Using these numbers would make A and C also be complex numbers.

So, the only value that you will likely care about in class is B = 2 which, again, results in A = 5 and C = 8.

[slatyb]

Whole numbers? There aren’t many choices for b given b*c=16. Just try them all.

[abc3643]

Yeah, this is a good way to go if you are told beforehand that A, B, C are whole numbers.

The factors of 16 are:
1,2,4,8,16. That's it.
So, either:
B=1, C=16
B=2, C=8
B=4, C=4

If we try B=1, C=16 and plug into A*B+C = 18, you get:
A*1+16 = 18
So, A would have to be 2, but what about A+B = 7?
(2)+(1) = 3 <> 7 so B=1 is out...

If we try B=4, C=4 and plug into A*B+C = 18, you get:
4A + 4 = 18
4A = 14
A = 14/4=3.5... that's not whole, so already this may be out.
But what about A+B = 7?
(14/4)+(4) = 7.5 <> 7 so B=4 is out...

That leaves B=2, C=8.
Plug into A*B+C = 18, you get:
2A + 8 = 18
2A = 18 - 8 = 10
A = 5.
But what about A+B = 7?
(5)+(2) = 7 = 7 so that's it.

[Monfrere]

A=7-b
(7-b)*b+c=18
7b-b^2+c=18
C=16/b
7b-b^2+16/b=18
7b^2-b^3+16-18b=0
Now use the calculator

[abc3643]

That's a cubic equation. You are allowed to just plug an equation like that into a calculator without actually solving it?

[Monfrere]

@abc3643 hi, yes here we can do that, fortunately

[andreasderjuengere]

a = 3
A = 3.5
b = 4
c = 4
:)

[AzzaBlue]

A = 5
B = 2
C = 8

[monkeynutts]

A= 3 maybe B and C are 4. Algebra is fun right.

[Leavesbound]

What's the question?

[Sam_The_Savage]

😂😂😂

[prettyboy57886]

A5 b2 c8