# Thread: Solving the System

1. ## Solving the System

• A + C = 3 (1)
• 11 = A e^(B(4)) + C (2)
• 22 = (Ae^(4B)-A)/B + 4C (3)

I pretty much only know how you would approach this by substitution on.

i keep getting massive answers with ln everywhere. Im not getting anywhere.

Is there a more logical method or do i just need to practice with my simplifying in MAD MATPS

2. ## Re: Solving the System

Originally Posted by asaver
• A + C = 3 (1)
• 11 = A e^(B(4)) + C (2)
• 22 = (Ae^(4B)-A)/B + 4C (3)

I pretty much only know how you would approach this by substitution on.

i keep getting massive answers with ln everywhere. Im not getting anywhere.

Is there a more logical method or do i just need to practice with my simplifying in MAD MATPS
Equation 1: $\displaystyle A+C=3 \implies C=3-A$.

Substituting into Equation 2:

\displaystyle \begin{align*} 11 &= Ae^{4B} + C \\ 11 &= Ae^{4B} + 3 - A \\ 8 &= Ae^{4B} - A \\ 8 &= A\left(e^{4B} - 1\right) \\ \frac{8}{e^{4B}-1} &= A \end{align*}

Substituting into Equation 3:

\displaystyle \begin{align*} 22 &= \frac{Ae^{4B} - A}{B} + 4C \\ 22 &= \frac{A\left(e^{4B}-1\right)}{B} + 4C \\ 22 &= \frac{\frac{8}{e^{4B} - 1}\left(e^{4B} - 1\right)}{B} + 4C \\ 22 &= \frac{8}{B} + 4C \\ 22 &= \frac{8}{B} + 4\left(3-A\right) \\ 22 &= \frac{8}{B} + 12 - 4A \\ 10 &= \frac{8}{B} - 4A \\ 10 &= \frac{8}{B} - 4\left(\frac{8}{e^{4B}-1}\right) \\ 10 &= \frac{8}{B} - \frac{32}{e^{4B}-1} \end{align*}

From here, get a common denominator and then use an iterative method to solve for $\displaystyle B$.