Solve (2/3a-b/3)^2+4/9ab=4/9a^2+frac{b^2}{9}

$$( \frac { 2 } { 3 } a - \frac { b } { 3 } ) ^ { 2 } + \frac { 4 } { 9 } a b = \frac { 4 } { 9 } a ^ { 2 } + \frac { b ^ { 2 } } { 9 }$$

Solve for a

$$a\in \mathrm{R}$$

Short Solution Steps

Multiply both sides of the equation by 9.

$$9\left(\frac{2}{3}a-\frac{b}{3}\right)^{2}+4ab=4a^{2}+b^{2}$$

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(\frac{2}{3}a-\frac{b}{3}\right)^{2}.

$$9\left(\frac{4}{9}a^{2}+\frac{4}{3}a\left(-\frac{b}{3}\right)+\left(-\frac{b}{3}\right)^{2}\right)+4ab=4a^{2}+b^{2}$$

Multiply \frac{4}{3} times -\frac{b}{3} by multiplying numerator times numerator and denominator times denominator.

$$9\left(\frac{4}{9}a^{2}+\frac{-4b}{3\times 3}a+\left(-\frac{b}{3}\right)^{2}\right)+4ab=4a^{2}+b^{2}$$

Calculate -\frac{b}{3} to the power of 2 and get \left(\frac{b}{3}\right)^{2}.

$$9\left(\frac{4}{9}a^{2}+\frac{-4b}{3\times 3}a+\left(\frac{b}{3}\right)^{2}\right)+4ab=4a^{2}+b^{2}$$

Use the distributive property to multiply 9 by \frac{4}{9}a^{2}+\frac{-4b}{3\times 3}a+\left(\frac{b}{3}\right)^{2}.

$$4a^{2}+9\times \frac{-4b}{3\times 3}a+9\times \left(\frac{b}{3}\right)^{2}+4ab=4a^{2}+b^{2}$$

Multiply 3 and 3 to get 9.

$$4a^{2}+9\times \frac{-4b}{9}a+9\times \left(\frac{b}{3}\right)^{2}+4ab=4a^{2}+b^{2}$$

Express 9\times \frac{-4b}{9} as a single fraction.

$$4a^{2}+\frac{9\left(-4\right)b}{9}a+9\times \left(\frac{b}{3}\right)^{2}+4ab=4a^{2}+b^{2}$$

Cancel out 9 and 9.

$$4a^{2}-4ba+9\times \left(\frac{b}{3}\right)^{2}+4ab=4a^{2}+b^{2}$$

To raise \frac{b}{3} to a power, raise both numerator and denominator to the power and then divide.

$$4a^{2}-4ba+9\times \frac{b^{2}}{3^{2}}+4ab=4a^{2}+b^{2}$$

Express 9\times \frac{b^{2}}{3^{2}} as a single fraction.

$$4a^{2}-4ba+\frac{9b^{2}}{3^{2}}+4ab=4a^{2}+b^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 4a^{2}-4ba times \frac{3^{2}}{3^{2}}.

$$\frac{\left(4a^{2}-4ba\right)\times 3^{2}}{3^{2}}+\frac{9b^{2}}{3^{2}}+4ab=4a^{2}+b^{2}$$

Since \frac{\left(4a^{2}-4ba\right)\times 3^{2}}{3^{2}} and \frac{9b^{2}}{3^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(4a^{2}-4ba\right)\times 3^{2}+9b^{2}}{3^{2}}+4ab=4a^{2}+b^{2}$$

Do the multiplications in \left(4a^{2}-4ba\right)\times 3^{2}+9b^{2}.

$$\frac{36a^{2}-36ba+9b^{2}}{3^{2}}+4ab=4a^{2}+b^{2}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{36a^{2}-36ba+9b^{2}}{9}+4ab=4a^{2}+b^{2}$$

Divide each term of 36a^{2}-36ba+9b^{2} by 9 to get b^{2}-4ba+4a^{2}.

$$b^{2}-4ba+4a^{2}+4ab=4a^{2}+b^{2}$$

Combine -4ba and 4ab to get 0.

$$b^{2}+4a^{2}=4a^{2}+b^{2}$$

Subtract 4a^{2} from both sides.

$$b^{2}+4a^{2}-4a^{2}=b^{2}$$

Combine 4a^{2} and -4a^{2} to get 0.

$$b^{2}=b^{2}$$

Reorder the terms.

$$\text{true}$$

This is true for any a.

$$a\in \mathrm{R}$$

Solve for b

$$b\in \mathrm{R}$$

Short Solution Steps

Multiply both sides of the equation by 9.

$$9\left(\frac{2}{3}a-\frac{b}{3}\right)^{2}+4ab=4a^{2}+b^{2}$$

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(\frac{2}{3}a-\frac{b}{3}\right)^{2}.

$$9\left(\frac{4}{9}a^{2}+\frac{4}{3}a\left(-\frac{b}{3}\right)+\left(-\frac{b}{3}\right)^{2}\right)+4ab=4a^{2}+b^{2}$$

Multiply \frac{4}{3} times -\frac{b}{3} by multiplying numerator times numerator and denominator times denominator.

$$9\left(\frac{4}{9}a^{2}+\frac{-4b}{3\times 3}a+\left(-\frac{b}{3}\right)^{2}\right)+4ab=4a^{2}+b^{2}$$

Calculate -\frac{b}{3} to the power of 2 and get \left(\frac{b}{3}\right)^{2}.

$$9\left(\frac{4}{9}a^{2}+\frac{-4b}{3\times 3}a+\left(\frac{b}{3}\right)^{2}\right)+4ab=4a^{2}+b^{2}$$

Use the distributive property to multiply 9 by \frac{4}{9}a^{2}+\frac{-4b}{3\times 3}a+\left(\frac{b}{3}\right)^{2}.

$$4a^{2}+9\times \frac{-4b}{3\times 3}a+9\times \left(\frac{b}{3}\right)^{2}+4ab=4a^{2}+b^{2}$$

Multiply 3 and 3 to get 9.

$$4a^{2}+9\times \frac{-4b}{9}a+9\times \left(\frac{b}{3}\right)^{2}+4ab=4a^{2}+b^{2}$$

Express 9\times \frac{-4b}{9} as a single fraction.

$$4a^{2}+\frac{9\left(-4\right)b}{9}a+9\times \left(\frac{b}{3}\right)^{2}+4ab=4a^{2}+b^{2}$$

Cancel out 9 and 9.

$$4a^{2}-4ba+9\times \left(\frac{b}{3}\right)^{2}+4ab=4a^{2}+b^{2}$$

To raise \frac{b}{3} to a power, raise both numerator and denominator to the power and then divide.

$$4a^{2}-4ba+9\times \frac{b^{2}}{3^{2}}+4ab=4a^{2}+b^{2}$$

Express 9\times \frac{b^{2}}{3^{2}} as a single fraction.

$$4a^{2}-4ba+\frac{9b^{2}}{3^{2}}+4ab=4a^{2}+b^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 4a^{2}-4ba times \frac{3^{2}}{3^{2}}.

$$\frac{\left(4a^{2}-4ba\right)\times 3^{2}}{3^{2}}+\frac{9b^{2}}{3^{2}}+4ab=4a^{2}+b^{2}$$

Since \frac{\left(4a^{2}-4ba\right)\times 3^{2}}{3^{2}} and \frac{9b^{2}}{3^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(4a^{2}-4ba\right)\times 3^{2}+9b^{2}}{3^{2}}+4ab=4a^{2}+b^{2}$$

Do the multiplications in \left(4a^{2}-4ba\right)\times 3^{2}+9b^{2}.

$$\frac{36a^{2}-36ba+9b^{2}}{3^{2}}+4ab=4a^{2}+b^{2}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{36a^{2}-36ba+9b^{2}}{9}+4ab=4a^{2}+b^{2}$$

Divide each term of 36a^{2}-36ba+9b^{2} by 9 to get 4a^{2}-4ab+b^{2}.

$$4a^{2}-4ab+b^{2}+4ab=4a^{2}+b^{2}$$

Combine -4ab and 4ab to get 0.

$$4a^{2}+b^{2}=4a^{2}+b^{2}$$

Subtract b^{2} from both sides.

$$4a^{2}+b^{2}-b^{2}=4a^{2}$$

Combine b^{2} and -b^{2} to get 0.

$$4a^{2}=4a^{2}$$

Cancel out 4 on both sides.

$$a^{2}=a^{2}$$

Reorder the terms.

$$\text{true}$$

This is true for any b.

$$b\in \mathrm{R}$$