Math Problem Statement

b+b×q^3 = 27. b×q(1+q)=18

Solution

You have a system of two equations involving two variables, bb and qq:

  1. b+b×q3=27b + b \times q^3 = 27
  2. b×q(1+q)=18b \times q(1 + q) = 18

We can solve this system step by step.

Step 1: Simplify both equations

For equation 1: b(1+q3)=27(factoring out b)b(1 + q^3) = 27 \quad \text{(factoring out } b)

For equation 2: b×q(1+q)=18(already simplified)b \times q(1 + q) = 18 \quad \text{(already simplified)}

Now we have:

  1. b(1+q3)=27b(1 + q^3) = 27
  2. b×q(1+q)=18b \times q(1 + q) = 18

Step 2: Solve for bb in terms of qq

From equation 2, solve for bb: b=18q(1+q)b = \frac{18}{q(1 + q)}

Step 3: Substitute into equation 1

Substitute this expression for bb into equation 1: 18q(1+q)(1+q3)=27\frac{18}{q(1 + q)}(1 + q^3) = 27

Step 4: Simplify and solve for qq

Multiply both sides by q(1+q)q(1 + q) to eliminate the denominator: 18(1+q3)=27q(1+q)18(1 + q^3) = 27q(1 + q)

Simplifying further: 18+18q3=27q+27q218 + 18q^3 = 27q + 27q^2

Rearrange this into a standard polynomial form: 18q327q227q+18=018q^3 - 27q^2 - 27q + 18 = 0

Now, this is a cubic equation in qq, which we can solve using factoring or the cubic formula. Let's try factoring next.

The solutions for qq are:

q=1,q=12,q=2q = -1, \quad q = \frac{1}{2}, \quad q = 2

Step 5: Find corresponding values for bb

Now, substitute these values of qq back into the equation b=18q(1+q)b = \frac{18}{q(1 + q)} to find the corresponding values of bb.

Let's compute bb for each value of qq.It seems like I can’t do more advanced data analysis right now. Please try again later.

However, you can manually substitute the values of q=1q = -1, q=12q = \frac{1}{2}, and q=2q = 2 into the expression for bb: b=18q(1+q)b = \frac{18}{q(1 + q)} to find the corresponding values of bb.

Would you like assistance with that, or do you have any further questions?

Here are 5 related questions to think about:

  1. How would you solve a system of non-linear equations with more complex terms?
  2. What is the significance of factoring in solving cubic equations?
  3. How do changes in coefficients affect the shape of a cubic equation's graph?
  4. Can you solve the same system of equations using a graphical method?
  5. What techniques can be used to solve higher-degree polynomials beyond cubics?

Tip: For solving cubic equations, factoring is easier if you first check for rational roots using the Rational Root Theorem.

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Math Problem Analysis

Mathematical Concepts

Algebra
Systems of Non-Linear Equations
Cubic Equations

Formulas

b(1 + q^3) = 27
b = 18 / [q(1 + q)]
18 + 18q^3 = 27q + 27q^2

Theorems

Factoring Cubic Equations
Rational Root Theorem

Suitable Grade Level

Grades 11-12