We first analyze the given equation \(-3x + 20 = 5x - 4\). To find the value of \(x\), we rearrange the terms:
\[20 + 4 = 5x + 3x\]
\[24 = 8x \implies x = 3\]
Since the solution is \(x = 3\), any equation that simplifies to \(x = 3\) is equivalent. Three such equations are:
Next, we examine the equations \(7r + 9 = 2r - 3 + 4r\) and \(5r + 10 = 4r - 2\).
Simplifying the first equation:
\[7r + 9 = 6r - 3 \implies r = -12\]
Simplifying the second equation:
\[5r - 4r = -2 - 10 \implies r = -12\]
Since both equations yield the same solution \(r = -12\), they are equivalent.
Given the system \(w = 3y\) and \(8w = 48\), we first solve for \(w\):
\[w = \frac{48}{8} = 6\]
Then, substitute \(w = 6\) into the first equation:
\[6 = 3y \implies y = 2\]
We are given \(9(x - y) = 30\). We want to find \(6(x - y)\).
First, solve for \((x - y)\):
\[x - y = \frac{30}{9} = \frac{10}{3}\]
Now, multiply by 6:
\[6(x - y) = 6 \times \frac{10}{3} = 2 \times 10 = 20\]
The prompt provides variables \(m, f, s\) representing the number of teachers but does not define any relationships, equations, or tasks related to them. Therefore, this part of the problem is incomplete and cannot be solved without further information.
Finally, we solve the two given equations for \(k\):
Both equations result in \(k = 2\).
These algebraic manipulation techniques are used daily in financial planning and resource management. For example, if you are determining how much money to save (\(x\)) based on your income and expenses, setting up and solving these linear equations ensures your budget balances perfectly. Similarly, identifying equivalent equations helps in simplifying complex billing models into manageable formulas, ensuring accurate accounting in business operations.