-x+1/4y+5/4=0;x=5

To solve the equation \(-x + \frac{1}{4}y + \frac{5}{4} = 0\) for \(y\) and express it in the desired format, let’s follow these steps:

1. First, isolate \(y\):
\[
\frac{1}{4}y = x - \frac{5}{4}
\]

2. Multiply both sides by 4 to eliminate the fraction:
\[
y = 4x - 5
\]

Now, you mentioned setting \(x = 5\) to find the corresponding value of \(y\):

3. Substitute \(x = 5\) into \(y = 4x - 5\):
\[
y = 4(5) - 5 = 20 - 5 = 15
\]

So, the equations obtained are:
\[
y = 4x - 5, \quad y = 15
\]

These satisfy your requirements:
- The variable \(y\) is on the left, before the constant.
- There are no mixed fractions or decimals.
- Each equation is expressed in proper descending order.

For how we are doing it in my class we first clear off the fractions in our equation by multiplying by our common denominator so what I have is 4 as the common denominator of 1/4 and 5/4 is that right?

Yes, you are correct! The common denominator for the fractions \( \frac{1}{4} \) and \( \frac{5}{4} \) is indeed \( 4 \).

To clear the fractions from the equation \(-x + \frac{1}{4}y + \frac{5}{4} = 0\), you can multiply the entire equation by \( 4 \) to eliminate the fractions:

\[
4(-x) + 4\left(\frac{1}{4}y\right) + 4\left(\frac{5}{4}\right) = 0
\]

This simplifies to:

\[
-4x + y + 5 = 0
\]

Now, isolate \(y\):

\[
y = 4x - 5
\]

This is the same result as before, and it keeps the variable before the constant. Now you can substitute \( x = 5 \):

\[
y = 4(5) - 5 = 20 - 5 = 15
\]

So to summarize:
- The cleared equation is: \(-4x + y + 5 = 0\) or, rearranged, \(y = 4x - 5\).
- When \(x = 5\), you find \(y = 15\).

Your understanding of clearing fractions with a common denominator is spot on!