How to solve (-2/5-5)-(3/10-5)

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve (-2/5-5)-(3/10-5) | Equatix

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Question

$$(-2/5-5)-(3/10-5)$$

Evaluate

$-\frac{7}{10}=-0.7$

Solution Steps

Fraction $\frac{-2}{5}$ can be rewritten as $-\frac{2}{5}$ by extracting the negative sign.

$$-\frac{2}{5}-5-\left(\frac{3}{10}-5\right)$$

Convert $5$ to fraction $\frac{25}{5}$.

$$-\frac{2}{5}-\frac{25}{5}-\left(\frac{3}{10}-5\right)$$

Since $-\frac{2}{5}$ and $\frac{25}{5}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{-2-25}{5}-\left(\frac{3}{10}-5\right)$$

Subtract $25$ from $-2$ to get $-27$.

$$-\frac{27}{5}-\left(\frac{3}{10}-5\right)$$

Convert $5$ to fraction $\frac{50}{10}$.

$$-\frac{27}{5}-\left(\frac{3}{10}-\frac{50}{10}\right)$$

Since $\frac{3}{10}$ and $\frac{50}{10}$ have the same denominator, subtract them by subtracting their numerators.

$$-\frac{27}{5}-\frac{3-50}{10}$$

Subtract $50$ from $3$ to get $-47$.

$$-\frac{27}{5}-\left(-\frac{47}{10}\right)$$

The opposite of $-\frac{47}{10}$ is $\frac{47}{10}$.

$$-\frac{27}{5}+\frac{47}{10}$$

Least common multiple of $5$ and $10$ is $10$. Convert $-\frac{27}{5}$ and $\frac{47}{10}$ to fractions with denominator $10$.

$$-\frac{54}{10}+\frac{47}{10}$$

Since $-\frac{54}{10}$ and $\frac{47}{10}$ have the same denominator, add them by adding their numerators.

$$\frac{-54+47}{10}$$

Add $-54$ and $47$ to get $-7$.

$$-\frac{7}{10}$$

Show Solution

Factor

$-\frac{7}{10} = -0.7$