Let's analyze the given system of equations. From the layout, it seems like we're dealing with a matrix equation or a set of related equalities. However, the information provided is a bit ambiguous in format. I'll assume that each fraction represents an equation comparing two ratios.

Here’s what the problem looks like:

$\frac{7}{d} = \frac{8}{-4}$

$\frac{-4}{-c} = \frac{-4}{5}$

$\frac{-8}{a} = \frac{5}{-3}$

$\frac{a}{b} = \frac{T-5}{T-5}$

The above equations can be solved for $a$, $b$, $c$, and $d$:

Equation 1: $\frac{7}{d} = \frac{8}{-4}$

$\frac{7}{d} = -2 \Rightarrow d = \frac{7}{-2} = -\frac{7}{2}$ So, $d = -\frac{7}{2}$.

Equation 2: $\frac{-4}{-c} = \frac{-4}{5}$

$\frac{-4}{-c} = \frac{-4}{5} \Rightarrow \frac{4}{c} = \frac{-4}{5} \Rightarrow 4c = -20 \Rightarrow c = -5$ So, $c = -5$.

Equation 3: $\frac{-8}{a} = \frac{5}{-3}$

$\frac{-8}{a} = -\frac{5}{3} \Rightarrow -8 \times -3 = 5a \Rightarrow 24 = 5a \Rightarrow a = \frac{24}{5}$ So, $a = \frac{24}{5}$.

Equation 4: $\frac{a}{b} = \frac{T-5}{T-5}$

Since $\frac{T-5}{T-5} = 1$, we have: $\frac{a}{b} = 1 \Rightarrow a = b$ So, $b = \frac{24}{5}$.

Thus, the solutions are:

• $a = \frac{24}{5}$
• $b = \frac{24}{5}$
• $c = -5$
• $d = -\frac{7}{2}$

Would you like further details on any of the steps? Here are five related questions for further exploration:

1. How do you solve a system of linear equations using matrix methods?
2. What are the conditions for two ratios to be equal?
3. How can you check the consistency of a solution in a system of equations?
4. What methods can be used to solve non-linear systems of equations?
5. How does the determinant of a matrix affect the solution to a system of equations?

Tip: When solving equations involving fractions, cross-multiply to eliminate the fractions and simplify the process.