Solve 4x/3-3/4x/4-frac{3{4}}

$$\frac { \frac { 4 x } { 3 } - \frac { 3 } { 4 x } } { 4 - \frac { 3 } { 4 } }$$

Evaluate

$$\frac{16x}{39}-\frac{3}{13x}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3 and 4x is 12x. Multiply \frac{4x}{3} times \frac{4x}{4x}. Multiply \frac{3}{4x} times \frac{3}{3}.

$$\frac{\frac{4x\times 4x}{12x}-\frac{3\times 3}{12x}}{4-\frac{3}{4}}$$

Since \frac{4x\times 4x}{12x} and \frac{3\times 3}{12x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{4x\times 4x-3\times 3}{12x}}{4-\frac{3}{4}}$$

Do the multiplications in 4x\times 4x-3\times 3.

$$\frac{\frac{16x^{2}-9}{12x}}{4-\frac{3}{4}}$$

Convert 4 to fraction \frac{16}{4}.

$$\frac{\frac{16x^{2}-9}{12x}}{\frac{16}{4}-\frac{3}{4}}$$

Since \frac{16}{4} and \frac{3}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{16x^{2}-9}{12x}}{\frac{16-3}{4}}$$

Subtract 3 from 16 to get 13.

$$\frac{\frac{16x^{2}-9}{12x}}{\frac{13}{4}}$$

Divide \frac{16x^{2}-9}{12x} by \frac{13}{4} by multiplying \frac{16x^{2}-9}{12x} by the reciprocal of \frac{13}{4}.

$$\frac{\left(16x^{2}-9\right)\times 4}{12x\times 13}$$

Cancel out 4 in both numerator and denominator.

$$\frac{16x^{2}-9}{3\times 13x}$$

Multiply 3 and 13 to get 39.

$$\frac{16x^{2}-9}{39x}$$

Expand

$$\frac{16x}{39}-\frac{3}{13x}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3 and 4x is 12x. Multiply \frac{4x}{3} times \frac{4x}{4x}. Multiply \frac{3}{4x} times \frac{3}{3}.

$$\frac{\frac{4x\times 4x}{12x}-\frac{3\times 3}{12x}}{4-\frac{3}{4}}$$

Since \frac{4x\times 4x}{12x} and \frac{3\times 3}{12x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{4x\times 4x-3\times 3}{12x}}{4-\frac{3}{4}}$$

Do the multiplications in 4x\times 4x-3\times 3.

$$\frac{\frac{16x^{2}-9}{12x}}{4-\frac{3}{4}}$$

Convert 4 to fraction \frac{16}{4}.

$$\frac{\frac{16x^{2}-9}{12x}}{\frac{16}{4}-\frac{3}{4}}$$

Since \frac{16}{4} and \frac{3}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{16x^{2}-9}{12x}}{\frac{16-3}{4}}$$

Subtract 3 from 16 to get 13.

$$\frac{\frac{16x^{2}-9}{12x}}{\frac{13}{4}}$$

Divide \frac{16x^{2}-9}{12x} by \frac{13}{4} by multiplying \frac{16x^{2}-9}{12x} by the reciprocal of \frac{13}{4}.

$$\frac{\left(16x^{2}-9\right)\times 4}{12x\times 13}$$

Cancel out 4 in both numerator and denominator.

$$\frac{16x^{2}-9}{3\times 13x}$$

Multiply 3 and 13 to get 39.

$$\frac{16x^{2}-9}{39x}$$