Solve for y square root of 3y+1+ square root of 9=7

$\sqrt{3 y + 1} + \sqrt{9} = 7$

Step 1

Solve for $\sqrt{3 y + 1}$ .

Simplify each term.

Rewrite $9$ as $3^{2}$ .

$\sqrt{3 y + 1} + \sqrt{3^{2}} = 7$

Pull terms out from under the radical, assuming positive real numbers.

$\sqrt{3 y + 1} + 3 = 7$

Move all terms not containing $\sqrt{3 y + 1}$ to the right side of the equation.

Subtract $3$ from both sides of the equation.

$\sqrt{3 y + 1} = 7 - 3$

Subtract $3$ from $7$ .

$\sqrt{3 y + 1} = 4$

Step 2

To remove the radical on the left side of the equation, square both sides of the equation.

${\sqrt{3 y + 1}}^{2} = 4^{2}$

Step 3

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3 y + 1}$ as ${(3 y + 1)}^{\frac{1}{2}}$ .

${({(3 y + 1)}^{\frac{1}{2}})}^{2} = 4^{2}$

Simplify the left side.

Simplify ${({(3 y + 1)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(3 y + 1)}^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(3 y + 1)}^{\frac{1}{2} \cdot 2} = 4^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(3 y + 1)}^{\frac{1}{2} \cdot 2} = 4^{2}$

Rewrite the expression.

${(3 y + 1)}^{1} = 4^{2}$

Simplify.

$3 y + 1 = 4^{2}$

Simplify the right side.

Raise $4$ to the power of $2$ .

$3 y + 1 = 16$

Step 4

Solve for $y$ .

Move all terms not containing $y$ to the right side of the equation.

Subtract $1$ from both sides of the equation.

$3 y = 16 - 1$

Subtract $1$ from $16$ .

$3 y = 15$

Divide each term in $3 y = 15$ by $3$ and simplify.

Divide each term in $3 y = 15$ by $3$ .

$\frac{3 y}{3} = \frac{15}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 y}{3} = \frac{15}{3}$

Divide $y$ by $1$ .

$y = \frac{15}{3}$

Simplify the right side.

Divide $15$ by $3$ .

$y = 5$

Solve for y square root of 3y+1+ square root of 9=7

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