Solve (x+0,1)(2x-1/3)

$$( x + 0,1 ) ( 2 x - \frac { 1 } { 3 } )$$

Evaluate

$$\frac{\left(6x-1\right)\left(x+0,1\right)}{3}$$

Solution Steps

Apply the distributive property by multiplying each term of x+0,1 by each term of 2x-\frac{1}{3}.

$$2x^{2}+x\left(-\frac{1}{3}\right)+0,2x+0,1\left(-\frac{1}{3}\right)$$

Combine x\left(-\frac{1}{3}\right) and 0,2x to get -\frac{2}{15}x.

$$2x^{2}-\frac{2}{15}x+0,1\left(-\frac{1}{3}\right)$$

Convert decimal number 0,1 to fraction \frac{1}{10}.

$$2x^{2}-\frac{2}{15}x+\frac{1}{10}\left(-\frac{1}{3}\right)$$

Multiply \frac{1}{10} times -\frac{1}{3} by multiplying numerator times numerator and denominator times denominator.

$$2x^{2}-\frac{2}{15}x+\frac{1\left(-1\right)}{10\times 3}$$

Do the multiplications in the fraction \frac{1\left(-1\right)}{10\times 3}.

$$2x^{2}-\frac{2}{15}x+\frac{-1}{30}$$

Fraction \frac{-1}{30} can be rewritten as -\frac{1}{30} by extracting the negative sign.

$$2x^{2}-\frac{2}{15}x-\frac{1}{30}$$

Expand

$$2x^{2}-\frac{2x}{15}-\frac{1}{30}$$

Solution Steps

Apply the distributive property by multiplying each term of x+0,1 by each term of 2x-\frac{1}{3}.

$$2x^{2}+x\left(-\frac{1}{3}\right)+0,2x+0,1\left(-\frac{1}{3}\right)$$

Combine x\left(-\frac{1}{3}\right) and 0,2x to get -\frac{2}{15}x.

$$2x^{2}-\frac{2}{15}x+0,1\left(-\frac{1}{3}\right)$$

Convert decimal number 0,1 to fraction \frac{1}{10}.

$$2x^{2}-\frac{2}{15}x+\frac{1}{10}\left(-\frac{1}{3}\right)$$

Multiply \frac{1}{10} times -\frac{1}{3} by multiplying numerator times numerator and denominator times denominator.

$$2x^{2}-\frac{2}{15}x+\frac{1\left(-1\right)}{10\times 3}$$

Do the multiplications in the fraction \frac{1\left(-1\right)}{10\times 3}.

$$2x^{2}-\frac{2}{15}x+\frac{-1}{30}$$

Fraction \frac{-1}{30} can be rewritten as -\frac{1}{30} by extracting the negative sign.

$$2x^{2}-\frac{2}{15}x-\frac{1}{30}$$