Algebra Examples

$\sqrt{\frac{1}{2} (20) - 1} = \frac{3}{4} (20) + d$

Step 1

Rewrite the equation as $\frac{3}{4} \cdot 20 + d = \sqrt{\frac{1}{2} \cdot (20) - 1}$ .

$\frac{3}{4} \cdot 20 + d = \sqrt{\frac{1}{2} \cdot (20) - 1}$

Step 2

Simplify each term.

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Step 2.1

Cancel the common factor of $4$ .

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Step 2.1.1

Factor $4$ out of $20$ .

$\frac{3}{4} \cdot (4 (5)) + d = \sqrt{\frac{1}{2} \cdot (20) - 1}$

Step 2.1.2

Cancel the common factor.

$\frac{3}{4} \cdot (4 \cdot 5) + d = \sqrt{\frac{1}{2} \cdot (20) - 1}$

Step 2.1.3

Rewrite the expression.

$3 \cdot 5 + d = \sqrt{\frac{1}{2} \cdot (20) - 1}$

Step 2.2

Multiply $3$ by $5$ .

$15 + d = \sqrt{\frac{1}{2} \cdot (20) - 1}$

Step 3

Simplify $\sqrt{\frac{1}{2} \cdot (20) - 1}$ .

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Step 3.1

Cancel the common factor of $2$ .

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Step 3.1.1

Factor $2$ out of $20$ .

$15 + d = \sqrt{\frac{1}{2} \cdot (2 (10)) - 1}$

Step 3.1.2

Cancel the common factor.

$15 + d = \sqrt{\frac{1}{2} \cdot (2 \cdot 10) - 1}$

Step 3.1.3

Rewrite the expression.

$15 + d = \sqrt{10 - 1}$

Step 3.2

Simplify the expression.

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Step 3.2.1

Subtract $1$ from $10$ .

$15 + d = \sqrt{9}$

Step 3.2.2

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

$15 + d = \sqrt{3^{2}}$

Step 3.3

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

$15 + d = 3$

Step 4

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

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Step 4.1

Subtract $15$ from both sides of the equation.

$d = 3 - 15$

Step 4.2

Subtract $15$ from $3$ .

$d = - 12$