MATHEMATICS HIGH SCHOOL

AB+C=50
BC+A=41
What is A, B, C

Answers

Answer 1
Answer:

For the given system of equations, when B = 2, the values are A = 133 and C = -216. Values will vary with different choices of B.

To solve for the values of A, B, and C in the system of equations:

AB + C = 50

BC + A = 41

We can use a systematic approach. Let's first isolate one variable in one equation and then substitute it into the other equation.

From the first equation (AB + C = 50), we can isolate C:

C = 50 - AB

Now, substitute this expression for C into the second equation:

B(50 - AB) + A = 41

Expand and simplify:

50B - AB^2 + A = 41

Rearrange terms:

AB^2 - 50B + A = 41

Now, let's consider this as a quadratic equation in terms of A and solve for A:

A = 41 - AB^2 + 50B

Now that we have expressions for A and C in terms of B, we can choose a value for B, and then calculate the corresponding values of A and C. For instance, let's say B = 2:

A = 41 - (2)(2^2) + 50(2) = 41 - 8 + 100 = 133

C = 50 - (2)(133) = 50 - 266 = -216

So, for B = 2, we have A = 133 and C = -216. You can similarly calculate values for different values of B.

For more such questions on equations

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Complete question below:

What are the values of A, B, and C in the system of equations:

AB + C = 50

BC + A = 41?
Answer 2
Answer: A=4
B=3
C=7

43+7=50
37+4=41

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What I did:
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Answers

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