Solve (2x/3+1)*(2x/3-1)=

$$( \frac{ 2x }{ 3 } +1) \times ( \frac{ 2x }{ 3 } -1)=$$

Evaluate

$$\frac{4x^{2}}{9}-1$$

Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{3}{3}.

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

Since \frac{2x}{3} and \frac{3}{3} have the same denominator, add them by adding their numerators.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{3}{3}.

$$\frac{2x+3}{3}\left(\frac{2x}{3}-\frac{3}{3}\right)$$

Since \frac{2x}{3} and \frac{3}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2x+3}{3}\times \frac{2x-3}{3}$$

Multiply \frac{2x+3}{3} times \frac{2x-3}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(2x+3\right)\left(2x-3\right)}{3\times 3}$$

Multiply 3 and 3 to get 9.

$$\frac{\left(2x+3\right)\left(2x-3\right)}{9}$$

Consider \left(2x+3\right)\left(2x-3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(2x\right)^{2}-3^{2}}{9}$$

Expand \left(2x\right)^{2}.

$$\frac{2^{2}x^{2}-3^{2}}{9}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{4x^{2}-3^{2}}{9}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{4x^{2}-9}{9}$$

Expand

$$\frac{4x^{2}}{9}-1$$

Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{3}{3}.

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

Since \frac{2x}{3} and \frac{3}{3} have the same denominator, add them by adding their numerators.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{3}{3}.

$$\frac{2x+3}{3}\left(\frac{2x}{3}-\frac{3}{3}\right)$$

Since \frac{2x}{3} and \frac{3}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2x+3}{3}\times \frac{2x-3}{3}$$

Multiply \frac{2x+3}{3} times \frac{2x-3}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(2x+3\right)\left(2x-3\right)}{3\times 3}$$

Multiply 3 and 3 to get 9.

$$\frac{\left(2x+3\right)\left(2x-3\right)}{9}$$

Consider \left(2x+3\right)\left(2x-3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(2x\right)^{2}-3^{2}}{9}$$

Expand \left(2x\right)^{2}.

$$\frac{2^{2}x^{2}-3^{2}}{9}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{4x^{2}-3^{2}}{9}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{4x^{2}-9}{9}$$