Solve for j square root of j+ square root of j+14=3 square root of j+10

$\sqrt{j} + \sqrt{j} + 14 = 3 \sqrt{j} + 10$

Add $\sqrt{j}$ and $\sqrt{j}$ .

$2 \sqrt{j} + 14 = 3 \sqrt{j} + 10$

Move all terms containing $\sqrt{j}$ to the left side of the equation.

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Subtract $3 \sqrt{j}$ from both sides of the equation.

$2 \sqrt{j} + 14 - 3 \sqrt{j} = 10$

Subtract $3 \sqrt{j}$ from $2 \sqrt{j}$ .

$- \sqrt{j} + 14 = 10$

Move all terms not containing $\sqrt{j}$ to the right side of the equation.

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Subtract $14$ from both sides of the equation.

$- \sqrt{j} = 10 - 14$

Subtract $14$ from $10$ .

$- \sqrt{j} = - 4$

Multiply each term in $- \sqrt{j} = - 4$ by $- 1$

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Multiply each term in $- \sqrt{j} = - 4$ by $- 1$ .

$(- \sqrt{j}) \cdot - 1 = (- 4) \cdot - 1$

Multiply $- \sqrt{j} \cdot - 1$ .

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Multiply $- 1$ by $- 1$ .

$1 \sqrt{j} = (- 4) \cdot - 1$

Multiply $\sqrt{j}$ by $1$ .

$\sqrt{j} = (- 4) \cdot - 1$

Multiply $- 4$ by $- 1$ .

$\sqrt{j} = 4$

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

${(\sqrt{j})}^{2} = {(4)}^{2}$

Simplify each side of the equation.

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Simplify the left side of the equation.

$j = {(4)}^{2}$

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

$j = 16$

Solve for j square root of j+ square root of j+14=3 square root of j+10

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