Solve for x (x^2)/5-x/3=-2/15

$\frac{x^{2}}{5} - \frac{x}{3} = - \frac{2}{15}$

Multiply each term in $\frac{x^{2}}{5} - \frac{x}{3} = - \frac{2}{15}$ by $15$ to eliminate the fractions.

Multiply each term in $\frac{x^{2}}{5} - \frac{x}{3} = - \frac{2}{15}$ by $15$ .

$\frac{x^{2}}{5} \cdot 15 - \frac{x}{3} \cdot 15 = - \frac{2}{15} \cdot 15$

Simplify the left side.

Simplify each term.

Cancel the common factor of $5$ .

Factor $5$ out of $15$ .

$\frac{x^{2}}{5} \cdot (5 (3)) - \frac{x}{3} \cdot 15 = - \frac{2}{15} \cdot 15$

Cancel the common factor.

$\frac{x^{2}}{5} \cdot (5 \cdot 3) - \frac{x}{3} \cdot 15 = - \frac{2}{15} \cdot 15$

Rewrite the expression.

$x^{2} \cdot 3 - \frac{x}{3} \cdot 15 = - \frac{2}{15} \cdot 15$

Move $3$ to the left of $x^{2}$ .

$3 \cdot x^{2} - \frac{x}{3} \cdot 15 = - \frac{2}{15} \cdot 15$

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{x}{3}$ into the numerator.

$3 x^{2} + \frac{- x}{3} \cdot 15 = - \frac{2}{15} \cdot 15$

Factor $3$ out of $15$ .

$3 x^{2} + \frac{- x}{3} \cdot (3 (5)) = - \frac{2}{15} \cdot 15$

Cancel the common factor.

$3 x^{2} + \frac{- x}{3} \cdot (3 \cdot 5) = - \frac{2}{15} \cdot 15$

Rewrite the expression.

$3 x^{2} - x \cdot 5 = - \frac{2}{15} \cdot 15$

Multiply $5$ by $- 1$ .

$3 x^{2} - 5 x = - \frac{2}{15} \cdot 15$

Simplify the right side.

Cancel the common factor of $15$ .

Move the leading negative in $- \frac{2}{15}$ into the numerator.

$3 x^{2} - 5 x = \frac{- 2}{15} \cdot 15$

Cancel the common factor.

$3 x^{2} - 5 x = \frac{- 2}{15} \cdot 15$

Rewrite the expression.

$3 x^{2} - 5 x = - 2$

Add $2$ to both sides of the equation.

$3 x^{2} - 5 x + 2 = 0$

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 $a \cdot c = 3 \cdot 2 = 6$ and whose sum is $b = - 5$ .

Factor $- 5$ out of $- 5 x$ .

$3 x^{2} - 5 x + 2 = 0$

Rewrite $- 5$ as $- 2$ plus $- 3$

$3 x^{2} + (- 2 - 3) x + 2 = 0$

Apply the distributive property.

$3 x^{2} - 2 x - 3 x + 2 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(3 x^{2} - 2 x) - 3 x + 2 = 0$

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

$x (3 x - 2) - (3 x - 2) = 0$

Factor the polynomial by factoring out the greatest common factor, $3 x - 2$ .

$(3 x - 2) (x - 1) = 0$

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

$3 x - 2 = 0$

$x - 1 = 0$

Set $3 x - 2$ equal to $0$ and solve for $x$ .

Set $3 x - 2$ equal to $0$ .

$3 x - 2 = 0$

Solve $3 x - 2 = 0$ for $x$ .

Add $2$ to both sides of the equation.

$3 x = 2$

Divide each term in $3 x = 2$ by $3$ and simplify.

Divide each term in $3 x = 2$ by $3$ .

$\frac{3 x}{3} = \frac{2}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{2}{3}$

Divide $x$ by $1$ .

$x = \frac{2}{3}$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

The final solution is all the values that make $(3 x - 2) (x - 1) = 0$ true.

$x = \frac{2}{3}, 1$

The result can be shown in multiple forms.

Exact Form:

$x = \frac{2}{3}, 1$

Decimal Form:

$x = 0.\overline{6}, 1$

Solve for x (x^2)/5-x/3=-2/15

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