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The equation you provided is:

\[
\tan(45^\circ) = \frac{\frac{9 - y}{5} - \frac{6}{11}}{1 + \left(\frac{9 - y}{5}\right)\left(\frac{6}{11}\right)}
\]

Since \(\tan(45^\circ) = 1\), we can set the left side equal to 1:

\[
1 = \frac{\frac{9 - y}{5} - \frac{6}{11}}{1 + \left(\frac{9 - y}{5}\right)\left(\frac{6}{11}\right)}
\]

Next, let's solve for \(y\).

### Step 1: Multiply both sides of the equation by the denominator on the right-hand side to clear the fraction:
\[
1 \times \left(1 + \left(\frac{9 - y}{5}\right)\left(\frac{6}{11}\right)\right) = \frac{9 - y}{5} - \frac{6}{11}
\]

This simplifies to:
\[
1 + \left(\frac{9 - y}{5}\right)\left(\frac{6}{11}\right) = \frac{9 - y}{5} - \frac{6}{11}
\]

### Step 2: Eliminate the fractions by multiplying through by the least common denominator (LCD) of 55 (since 5 and 11 are the denominators):
\[
55 \times \left(1 + \left(\frac{9 - y}{5} \times \frac{6}{11}\right)\right) = 55 \times \left(\frac{9 - y}{5} - \frac{6}{11}\right)
\]

Simplifying the fractions:
\[
55 + \left(55 \times \frac{9 - y}{5} \times \frac{6}{11}\right) = 11(9 - y) - 5(6)
\]

The term on the left simplifies as:
\[
55 + 6(9 - y) = 

Solve the equation: tan(45) = ( (9 - y) / 5 - 6 / 11 ) / ( 1 + ( (9 - y) / 5 ) * (6 / 11) ).

This problem involves solving the trigonometric equation tan(45) = ( (9 - y) / 5 - 6 / 11 ) / ( 1 + ( (9 - y) / 5 ) * (6 / 11) ) using algebraic manipulation and trigonometric identities. Suitable for students in Grades 10-12.

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