Solve 9x^2/4-4y^2/9=

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

Evaluate

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

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4 and 9 is 36. Multiply \frac{9x^{2}}{4} times \frac{9}{9}. Multiply \frac{4y^{2}}{9} times \frac{4}{4}.

$$\frac{9\times 9x^{2}}{36}-\frac{4\times 4y^{2}}{36}$$

Since \frac{9\times 9x^{2}}{36} and \frac{4\times 4y^{2}}{36} have the same denominator, subtract them by subtracting their numerators.

$$\frac{9\times 9x^{2}-4\times 4y^{2}}{36}$$

Do the multiplications in 9\times 9x^{2}-4\times 4y^{2}.

$$\frac{81x^{2}-16y^{2}}{36}$$

Factor

$$\frac{\left(9x-4y\right)\left(9x+4y\right)}{36}$$

Solution Steps

Factor out \frac{1}{36}.

$$\frac{81x^{2}-16y^{2}}{36}$$

Consider 81x^{2}-16y^{2}. Rewrite 81x^{2}-16y^{2} as \left(9x\right)^{2}-\left(4y\right)^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(9x-4y\right)\left(9x+4y\right)$$

Rewrite the complete factored expression.

$$\frac{\left(9x-4y\right)\left(9x+4y\right)}{36}$$