OK, let's start over:[EQN_1]: A * B + C = 18[EQN_2]: B * C = 16[EQN_3]: A + B = 7Hmmm. 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*18A*B^2 + B*C = B*18But, EQN_3 says B*C = 16, so:[EQN_9]: A*B^2 + 16 = B*18This 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*18Uh oh…7*B^2 - B^3 + 16 = 18*B-B^3 + 7*B^2 - 18*B + 16 = 0B^3 - 7*B^2 + 18*B - 16 = 0Wow… 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?
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.
Plugging A = 5, B = 2, C = 8 back into the original equations...[EQN_1]: A * B + C = 18(5) * (2) + (8) = 1810 + 8 = 1818 = 18 Check![EQN_2]: B * C = 16(2) * (8) = 1616 = 16 Check![EQN_3]: A + B = 7(5) + (2) = 77 = 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.phpand 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.phpand the solutions are:Answer:Solutions for x:x1 = 2x2 = 2.5 + i * 1.32288x3 = 2.5 - i * 1.32288Those 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.
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=16B=2, C=8B=4, C=4If we try B=1, C=16 and plug into A*B+C = 18, you get:A*1+16 = 18So, 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 = 184A = 14A = 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 = 182A = 18 - 8 = 10A = 5.But what about A+B = 7?(5)+(2) = 7 = 7 so that's it.
That's a cubic equation. You are allowed to just plug an equation like that into a calculator without actually solving it?
@abc3643 hi, yes here we can do that, fortunately
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