Solve for y in \(\sqrt{75}\) - \(\sqrt{12}\) + \(\sqrt{27}\) = y\(\sqrt{3}\)

Answer Details

To solve for y in the equation

\[ \sqrt{75} - \sqrt{12} + \sqrt{27} = y \sqrt{3} \]

we need to express each square root on the left in terms of \(\sqrt{3}\).

• \(\sqrt{75}\) can be written as \(\sqrt{25 \times 3} = \sqrt{25} \times \sqrt{3} = 5\sqrt{3}\)
• \(\sqrt{12}\) can be written as \(\sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3}\)
• \(\sqrt{27}\) can be written as \(\sqrt{9 \times 3} = \sqrt{9} \times \sqrt{3} = 3\sqrt{3}\)

Now substitute these into the original equation:

\[ 5\sqrt{3} - 2\sqrt{3} + 3\sqrt{3} = y\sqrt{3} \]

Combine the coefficients of \(\sqrt{3}\):

\[ (5 - 2 + 3)\sqrt{3} = y\sqrt{3} \]

\[ 6\sqrt{3} = y\sqrt{3} \]

Since the terms on both sides have \(\sqrt{3}\), divide both sides by \(\sqrt{3}\):

\[ y = 6 \]

This means the correct value of \(y\) is 6, so the simplified expression is \(6\sqrt{3}\).

Underlying concept: The solution relies on simplifying square roots by factoring out perfect squares, and then collecting like terms. Any square root like \(\sqrt{ab}\) can be written as \(\sqrt{a} \cdot \sqrt{b}\), which helps combine expressions involving the same irrational factor.