Pre-Algebra Examples

$\frac{3 x}{5} - \frac{2 x^{2}}{5} = - 1$

Step 1

Multiply each term in $\frac{3 x}{5} - \frac{2 x^{2}}{5} = - 1$ by $5$ to eliminate the fractions.

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Step 1.1

Multiply each term in $\frac{3 x}{5} - \frac{2 x^{2}}{5} = - 1$ by $5$ .

$\frac{3 x}{5} \cdot 5 - \frac{2 x^{2}}{5} \cdot 5 = - 1 \cdot 5$

Step 1.2

Simplify the left side.

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Step 1.2.1

Simplify each term.

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Step 1.2.1.1

Cancel the common factor of $5$ .

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Step 1.2.1.1.1

Cancel the common factor.

$\frac{3 x}{5} \cdot 5 - \frac{2 x^{2}}{5} \cdot 5 = - 1 \cdot 5$

Step 1.2.1.1.2

Rewrite the expression.

$3 x - \frac{2 x^{2}}{5} \cdot 5 = - 1 \cdot 5$

Step 1.2.1.2

Cancel the common factor of $5$ .

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Step 1.2.1.2.1

Move the leading negative in $- \frac{2 x^{2}}{5}$ into the numerator.

$3 x + \frac{- 2 x^{2}}{5} \cdot 5 = - 1 \cdot 5$

Step 1.2.1.2.2

Cancel the common factor.

$3 x + \frac{- 2 x^{2}}{5} \cdot 5 = - 1 \cdot 5$

Step 1.2.1.2.3

Rewrite the expression.

$3 x - 2 x^{2} = - 1 \cdot 5$

Step 1.3

Simplify the right side.

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Step 1.3.1

Multiply $- 1$ by $5$ .

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

Step 2

Add $5$ to both sides of the equation.

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

Step 3

Factor the left side of the equation.

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Step 3.1

Factor $- 1$ out of $3 x - 2 x^{2} + 5$ .

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Step 3.1.1

Reorder $3 x$ and $- 2 x^{2}$ .

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

Step 3.1.2

Factor $- 1$ out of $- 2 x^{2}$ .

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

Step 3.1.3

Factor $- 1$ out of $3 x$ .

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

Step 3.1.4

Rewrite $5$ as $- 1 (- 5)$ .

$- (2 x^{2}) - (- 3 x) - 1 \cdot - 5 = 0$

Step 3.1.5

Factor $- 1$ out of $- (2 x^{2}) - (- 3 x)$ .

$- (2 x^{2} - 3 x) - 1 \cdot - 5 = 0$

Step 3.1.6

Factor $- 1$ out of $- (2 x^{2} - 3 x) - 1 (- 5)$ .

$- (2 x^{2} - 3 x - 5) = 0$

Step 3.2

Factor.

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Step 3.2.1

Factor by grouping.

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Step 3.2.1.1

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 $a \cdot c = 2 \cdot - 5 = - 10$ and whose sum is $b = - 3$ .

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Step 3.2.1.1.1

Factor $- 3$ out of $- 3 x$ .

$- (2 x^{2} - 3 x - 5) = 0$

Step 3.2.1.1.2

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

$- (2 x^{2} + (2 - 5) x - 5) = 0$

Step 3.2.1.1.3

Apply the distributive property.

$- (2 x^{2} + 2 x - 5 x - 5) = 0$

Step 3.2.1.2

Factor out the greatest common factor from each group.

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Step 3.2.1.2.1

Group the first two terms and the last two terms.

$- ((2 x^{2} + 2 x) - 5 x - 5) = 0$

Step 3.2.1.2.2

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

$- (2 x (x + 1) - 5 (x + 1)) = 0$

Step 3.2.1.3

Factor the polynomial by factoring out the greatest common factor, $x + 1$ .

$- ((x + 1) (2 x - 5)) = 0$

Step 3.2.2

Remove unnecessary parentheses.

$- (x + 1) (2 x - 5) = 0$

Step 4

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

$x + 1 = 0$

$2 x - 5 = 0$

Step 5

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

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Step 5.1

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

$x + 1 = 0$

Step 5.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 6

Set $2 x - 5$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $2 x - 5$ equal to $0$ .

$2 x - 5 = 0$

Step 6.2

Solve $2 x - 5 = 0$ for $x$ .

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Step 6.2.1

Add $5$ to both sides of the equation.

$2 x = 5$

Step 6.2.2

Divide each term in $2 x = 5$ by $2$ and simplify.

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Step 6.2.2.1

Divide each term in $2 x = 5$ by $2$ .

$\frac{2 x}{2} = \frac{5}{2}$

Step 6.2.2.2

Simplify the left side.

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Step 6.2.2.2.1

Cancel the common factor of $2$ .

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Step 6.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{5}{2}$

Step 6.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{5}{2}$

Step 7

The final solution is all the values that make $- (x + 1) (2 x - 5) = 0$ true.

$x = - 1, \frac{5}{2}$