1Cademy - Solving $$ \sqrt{3y + 5} + 2 = 5 $$

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How to Solve a Radical Equation

Example

Solving $\sqrt{3y + 5} + 2 = 5$

Solve the radical equation $\sqrt{3y + 5} + 2 = 5$ by applying the four-step procedure.

Step 1 — Isolate the radical. Subtract $2$ from both sides to get the square root by itself: $\sqrt{3y + 5} + 2 - 2 = 5 - 2$ $\sqrt{3y + 5} = 3$

Step 2 — Square both sides. Apply the Squaring Property: $(\sqrt{3y + 5})^2 = 3^2$ $3y + 5 = 9$

Step 3 — Solve the new equation. Subtract $5$ from both sides to get $3y = 4$ . Divide both sides by $3$ to obtain: $y = \frac{4}{3}$

Step 4 — Check the answer. Substitute $y = \frac{4}{3}$ into the original equation: $\sqrt{3 \cdot \frac{4}{3} + 5} + 2 = \sqrt{4 + 5} + 2 = \sqrt{9} + 2 = 3 + 2 = 5$

Since $5 = 5$ is a true statement, $y = \frac{4}{3}$ is confirmed as the valid solution. This example demonstrates that radical equations can produce fractional solutions. In the check step, substituting the fraction $\frac{4}{3}$ into the expression $3 \cdot \frac{4}{3}$ conveniently simplifies to the integer $4$ , making the verification straightforward.

Updated 2026-06-30

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OpenStax Elementary Algebra

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