Question

solve for a.
\frac{-5a}{5a} = \frac{1}{a - 7}
there may be 1 or 2 solutions.
a = \square or a = \square

Explanation:

Step1: Simplify the left - hand side

The left - hand side of the equation is \(\frac{- 5a}{5a}\). For \(a
eq0\) (because if \(a = 0\), the denominator \(5a=0\) and the fraction is undefined), we can simplify \(\frac{-5a}{5a}\) by canceling out the common factors \(5a\) in the numerator and the denominator. So \(\frac{-5a}{5a}=- 1\) (when \(a
eq0\)).
The equation becomes \(-1=\frac{1}{a - 7}\).

Step2: Cross - multiply

Cross - multiply the equation \(-1=\frac{1}{a - 7}\) to get \(-1\times(a - 7)=1\).

Step3: Expand and solve for \(a\)

Expand the left - hand side: \(-a + 7=1\).
Subtract 7 from both sides: \(-a=1 - 7=-6\).
Multiply both sides by \(- 1\): \(a = 6\).

We also need to check the domain restrictions. The original equation \(\frac{-5a}{5a}=\frac{1}{a - 7}\) has restrictions: \(a
eq0\) (because of the denominator \(5a\)) and \(a
eq7\) (because of the denominator \(a - 7\)). The solution \(a = 6\) does not violate these restrictions. Also, we note that when we simplified \(\frac{-5a}{5a}\), we assumed \(a
eq0\), and \(a = 0\) would make the left - hand side undefined (since \(\frac{-5\times0}{5\times0}\) is undefined), so \(a = 0\) is not a solution.

Answer:

\(a = 6\) (There is only one valid solution. The initial thought of 1 or 2 solutions, but \(a = 0\) is extraneous as it makes the original fraction \(\frac{-5a}{5a}\) undefined)