Solve for y 2y = square root of 6-5y

$2 y = \sqrt{6 - 5 y}$

Step 1

Since the radical is on the right side of the equation, switch the sides so it is on the left side of the equation.

$\sqrt{6 - 5 y} = 2 y$

Step 2

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

${\sqrt{6 - 5 y}}^{2} = {(2 y)}^{2}$

Step 3

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{6 - 5 y}$ as ${(6 - 5 y)}^{\frac{1}{2}}$ .

${({(6 - 5 y)}^{\frac{1}{2}})}^{2} = {(2 y)}^{2}$

Simplify the left side.

Simplify ${({(6 - 5 y)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(6 - 5 y)}^{\frac{1}{2}})}^{2}$ .

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

${(6 - 5 y)}^{\frac{1}{2} \cdot 2} = {(2 y)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(6 - 5 y)}^{\frac{1}{2} \cdot 2} = {(2 y)}^{2}$

Rewrite the expression.

${(6 - 5 y)}^{1} = {(2 y)}^{2}$

Simplify.

$6 - 5 y = {(2 y)}^{2}$

Simplify the right side.

Simplify ${(2 y)}^{2}$ .

Apply the product rule to $2 y$ .

$6 - 5 y = 2^{2} y^{2}$

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

$6 - 5 y = 4 y^{2}$

Step 4

Solve for $y$ .

Subtract $4 y^{2}$ from both sides of the equation.

$6 - 5 y - 4 y^{2} = 0$

Factor the left side of the equation.

Factor $- 1$ out of $6 - 5 y - 4 y^{2}$ .

Reorder the expression.

Move $6$ .

$- 5 y - 4 y^{2} + 6 = 0$

Reorder $- 5 y$ and $- 4 y^{2}$ .

$- 4 y^{2} - 5 y + 6 = 0$

Factor $- 1$ out of $- 4 y^{2}$ .

$- (4 y^{2}) - 5 y + 6 = 0$

Factor $- 1$ out of $- 5 y$ .

$- (4 y^{2}) - (5 y) + 6 = 0$

Rewrite $6$ as $- 1 (- 6)$ .

$- (4 y^{2}) - (5 y) - 1 \cdot - 6 = 0$

Factor $- 1$ out of $- (4 y^{2}) - (5 y)$ .

$- (4 y^{2} + 5 y) - 1 \cdot - 6 = 0$

Factor $- 1$ out of $- (4 y^{2} + 5 y) - 1 (- 6)$ .

$- (4 y^{2} + 5 y - 6) = 0$

Factor.

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 4 \cdot - 6 = - 24$ and whose sum is $b = 5$ .

Factor $5$ out of $5 y$ .

$- (4 y^{2} + 5 (y) - 6) = 0$

Rewrite $5$ as $- 3$ plus $8$

$- (4 y^{2} + (- 3 + 8) y - 6) = 0$

Apply the distributive property.

$- (4 y^{2} - 3 y + 8 y - 6) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$- ((4 y^{2} - 3 y) + 8 y - 6) = 0$

Factor out the greatest common factor (GCF) from each group.

$- (y (4 y - 3) + 2 (4 y - 3)) = 0$

Factor the polynomial by factoring out the greatest common factor, $4 y - 3$ .

$- ((4 y - 3) (y + 2)) = 0$

Remove unnecessary parentheses.

$- (4 y - 3) (y + 2) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$4 y - 3 = 0$

$y + 2 = 0$

Set $4 y - 3$ equal to $0$ and solve for $y$ .

Set $4 y - 3$ equal to $0$ .

$4 y - 3 = 0$

Solve $4 y - 3 = 0$ for $y$ .

Add $3$ to both sides of the equation.

$4 y = 3$

Divide each term in $4 y = 3$ by $4$ and simplify.

Divide each term in $4 y = 3$ by $4$ .

$\frac{4 y}{4} = \frac{3}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 y}{4} = \frac{3}{4}$

Divide $y$ by $1$ .

$y = \frac{3}{4}$

Set $y + 2$ equal to $0$ and solve for $y$ .

Set $y + 2$ equal to $0$ .

$y + 2 = 0$

Subtract $2$ from both sides of the equation.

$y = - 2$

The final solution is all the values that make $- (4 y - 3) (y + 2) = 0$ true.

$y = \frac{3}{4}, - 2$

Step 5

Exclude the solutions that do not make $2 y = \sqrt{6 - 5 y}$ true.

$y = \frac{3}{4}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$y = \frac{3}{4}$

Decimal Form:

$y = 0.75$

Solve for y 2y = square root of 6-5y

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