Solve for d square root of 1/2y-1=3/4y+d

$\sqrt{\frac{1}{2} y - 1} = \frac{3}{4} y + d$

Step 1

Rewrite the equation as $\frac{3}{4} y + d = \sqrt{\frac{1}{2} y - 1}$ .

$\frac{3}{4} y + d = \sqrt{\frac{1}{2} y - 1}$

Step 2

Combine $\frac{3}{4}$ and $y$ .

$\frac{3 y}{4} + d = \sqrt{\frac{1}{2} y - 1}$

Step 3

Simplify $\sqrt{\frac{1}{2} y - 1}$ .

Combine $\frac{1}{2}$ and $y$ .

$\frac{3 y}{4} + d = \sqrt{\frac{y}{2} - 1}$

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{3 y}{4} + d = \sqrt{\frac{y}{2} - 1 \cdot \frac{2}{2}}$

Combine $- 1$ and $\frac{2}{2}$ .

$\frac{3 y}{4} + d = \sqrt{\frac{y}{2} + \frac{- 1 \cdot 2}{2}}$

Combine the numerators over the common denominator.

$\frac{3 y}{4} + d = \sqrt{\frac{y - 1 \cdot 2}{2}}$

Multiply $- 1$ by $2$ .

$\frac{3 y}{4} + d = \sqrt{\frac{y - 2}{2}}$

Rewrite $\sqrt{\frac{y - 2}{2}}$ as $\frac{\sqrt{y - 2}}{\sqrt{2}}$ .

$\frac{3 y}{4} + d = \frac{\sqrt{y - 2}}{\sqrt{2}}$

Multiply $\frac{\sqrt{y - 2}}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$\frac{3 y}{4} + d = \frac{\sqrt{y - 2}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}$

Combine and simplify the denominator.

Multiply $\frac{\sqrt{y - 2}}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$\frac{3 y}{4} + d = \frac{\sqrt{y - 2} \sqrt{2}}{\sqrt{2} \sqrt{2}}$

Raise $\sqrt{2}$ to the power of $1$ .

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

Raise $\sqrt{2}$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $1$ and $1$ .

$\frac{3 y}{4} + d = \frac{\sqrt{y - 2} \sqrt{2}}{{\sqrt{2}}^{2}}$

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

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

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

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

$\frac{3 y}{4} + d = \frac{\sqrt{y - 2} \sqrt{2}}{2^{\frac{1}{2} \cdot 2}}$

Combine $\frac{1}{2}$ and $2$ .

$\frac{3 y}{4} + d = \frac{\sqrt{y - 2} \sqrt{2}}{2^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{3 y}{4} + d = \frac{\sqrt{y - 2} \sqrt{2}}{2^{\frac{2}{2}}}$

Rewrite the expression.

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

Evaluate the exponent.

$\frac{3 y}{4} + d = \frac{\sqrt{y - 2} \sqrt{2}}{2}$

Combine using the product rule for radicals.

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

Reorder factors in $\frac{\sqrt{(y - 2) \cdot 2}}{2}$ .

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

Step 4

Subtract $\frac{3 y}{4}$ from both sides of the equation.

$d = \frac{\sqrt{2 (y - 2)}}{2} - \frac{3 y}{4}$

Solve for d square root of 1/2y-1=3/4y+d

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