Solve 2A-(3)/(4)B=((2)/((3)/(sqrt(2)))-(5)/((3)/(2)))+((1)/((4)/(sqrt(2)))-(3)/((5)/(-6))) | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$2A- \frac{ 3 }{ 4 } B=( \frac{ 2 }{ \frac{ 3 }{ \sqrt{ 2 } } } - \frac{ 5 }{ \frac{ 3 }{ 2 } } )+( \frac{ 1 }{ \frac{ 4 }{ \sqrt{ 2 } } } - \frac{ 3 }{ \frac{ 5 }{ -6 } } )$$

Solve for A

$A=\frac{3B}{8}+\frac{11\sqrt{2}}{24}+\frac{2}{15}$

Steps for Solving Linear Equation

Divide $2$ by $\frac{3}{\sqrt{2}}$ by multiplying $2$ by the reciprocal of $\frac{3}{\sqrt{2}}$.

$$2A-\frac{3}{4}B=\frac{2\sqrt{2}}{3}-\frac{5}{\frac{3}{2}}+\frac{1}{\frac{4}{\sqrt{2}}}-\frac{3}{\frac{5}{-6}}$$

Divide $5$ by $\frac{3}{2}$ by multiplying $5$ by the reciprocal of $\frac{3}{2}$.

$$2A-\frac{3}{4}B=\frac{2\sqrt{2}}{3}-5\times \frac{2}{3}+\frac{1}{\frac{4}{\sqrt{2}}}-\frac{3}{\frac{5}{-6}}$$

Multiply $5$ and $\frac{2}{3}$ to get $\frac{10}{3}$.

$$2A-\frac{3}{4}B=\frac{2\sqrt{2}}{3}-\frac{10}{3}+\frac{1}{\frac{4}{\sqrt{2}}}-\frac{3}{\frac{5}{-6}}$$

Divide $1$ by $\frac{4}{\sqrt{2}}$ by multiplying $1$ by the reciprocal of $\frac{4}{\sqrt{2}}$.

$$2A-\frac{3}{4}B=\frac{2\sqrt{2}}{3}-\frac{10}{3}+\frac{\sqrt{2}}{4}-\frac{3}{\frac{5}{-6}}$$

Combine $\frac{2\sqrt{2}}{3}$ and $\frac{\sqrt{2}}{4}$ to get $\frac{11}{12}\sqrt{2}$.

$$2A-\frac{3}{4}B=\frac{11}{12}\sqrt{2}-\frac{10}{3}-\frac{3}{\frac{5}{-6}}$$

Divide $3$ by $\frac{5}{-6}$ by multiplying $3$ by the reciprocal of $\frac{5}{-6}$.

$$2A-\frac{3}{4}B=\frac{11}{12}\sqrt{2}-\frac{10}{3}-\frac{3\left(-6\right)}{5}$$

Multiply $3$ and $-6$ to get $-18$.

$$2A-\frac{3}{4}B=\frac{11}{12}\sqrt{2}-\frac{10}{3}-\frac{-18}{5}$$

Fraction $\frac{-18}{5}$ can be rewritten as $-\frac{18}{5}$ by extracting the negative sign.

$$2A-\frac{3}{4}B=\frac{11}{12}\sqrt{2}-\frac{10}{3}-\left(-\frac{18}{5}\right)$$

The opposite of $-\frac{18}{5}$ is $\frac{18}{5}$.

$$2A-\frac{3}{4}B=\frac{11}{12}\sqrt{2}-\frac{10}{3}+\frac{18}{5}$$

Add $-\frac{10}{3}$ and $\frac{18}{5}$ to get $\frac{4}{15}$.

$$2A-\frac{3}{4}B=\frac{11}{12}\sqrt{2}+\frac{4}{15}$$

Add $\frac{3}{4}B$ to both sides.

$$2A=\frac{11}{12}\sqrt{2}+\frac{4}{15}+\frac{3}{4}B$$

The equation is in standard form.

$$2A=\frac{3B}{4}+\frac{11\sqrt{2}}{12}+\frac{4}{15}$$

Divide both sides by $2$.

$$\frac{2A}{2}=\frac{\frac{3B}{4}+\frac{11\sqrt{2}}{12}+\frac{4}{15}}{2}$$

Dividing by $2$ undoes the multiplication by $2$.

$$A=\frac{\frac{3B}{4}+\frac{11\sqrt{2}}{12}+\frac{4}{15}}{2}$$

Divide $\frac{11\sqrt{2}}{12}+\frac{4}{15}+\frac{3B}{4}$ by $2$.

$$A=\frac{3B}{8}+\frac{11\sqrt{2}}{24}+\frac{2}{15}$$

Solve for B

$B=\frac{8A}{3}-\frac{11\sqrt{2}}{9}-\frac{16}{45}$

Steps for Solving Linear Equation

Divide $2$ by $\frac{3}{\sqrt{2}}$ by multiplying $2$ by the reciprocal of $\frac{3}{\sqrt{2}}$.

$$2A-\frac{3}{4}B=\frac{2\sqrt{2}}{3}-\frac{5}{\frac{3}{2}}+\frac{1}{\frac{4}{\sqrt{2}}}-\frac{3}{\frac{5}{-6}}$$

Divide $5$ by $\frac{3}{2}$ by multiplying $5$ by the reciprocal of $\frac{3}{2}$.

$$2A-\frac{3}{4}B=\frac{2\sqrt{2}}{3}-5\times \frac{2}{3}+\frac{1}{\frac{4}{\sqrt{2}}}-\frac{3}{\frac{5}{-6}}$$

Multiply $5$ and $\frac{2}{3}$ to get $\frac{10}{3}$.

$$2A-\frac{3}{4}B=\frac{2\sqrt{2}}{3}-\frac{10}{3}+\frac{1}{\frac{4}{\sqrt{2}}}-\frac{3}{\frac{5}{-6}}$$

Divide $1$ by $\frac{4}{\sqrt{2}}$ by multiplying $1$ by the reciprocal of $\frac{4}{\sqrt{2}}$.

$$2A-\frac{3}{4}B=\frac{2\sqrt{2}}{3}-\frac{10}{3}+\frac{\sqrt{2}}{4}-\frac{3}{\frac{5}{-6}}$$

Combine $\frac{2\sqrt{2}}{3}$ and $\frac{\sqrt{2}}{4}$ to get $\frac{11}{12}\sqrt{2}$.

$$2A-\frac{3}{4}B=\frac{11}{12}\sqrt{2}-\frac{10}{3}-\frac{3}{\frac{5}{-6}}$$

Divide $3$ by $\frac{5}{-6}$ by multiplying $3$ by the reciprocal of $\frac{5}{-6}$.

$$2A-\frac{3}{4}B=\frac{11}{12}\sqrt{2}-\frac{10}{3}-\frac{3\left(-6\right)}{5}$$

Multiply $3$ and $-6$ to get $-18$.

$$2A-\frac{3}{4}B=\frac{11}{12}\sqrt{2}-\frac{10}{3}-\frac{-18}{5}$$

Fraction $\frac{-18}{5}$ can be rewritten as $-\frac{18}{5}$ by extracting the negative sign.

$$2A-\frac{3}{4}B=\frac{11}{12}\sqrt{2}-\frac{10}{3}-\left(-\frac{18}{5}\right)$$

The opposite of $-\frac{18}{5}$ is $\frac{18}{5}$.

$$2A-\frac{3}{4}B=\frac{11}{12}\sqrt{2}-\frac{10}{3}+\frac{18}{5}$$

Add $-\frac{10}{3}$ and $\frac{18}{5}$ to get $\frac{4}{15}$.

$$2A-\frac{3}{4}B=\frac{11}{12}\sqrt{2}+\frac{4}{15}$$

Subtract $2A$ from both sides.

$$-\frac{3}{4}B=\frac{11}{12}\sqrt{2}+\frac{4}{15}-2A$$

The equation is in standard form.

$$-\frac{3}{4}B=-2A+\frac{11\sqrt{2}}{12}+\frac{4}{15}$$

Divide both sides of the equation by $-\frac{3}{4}$, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{3}{4}B}{-\frac{3}{4}}=\frac{-2A+\frac{11\sqrt{2}}{12}+\frac{4}{15}}{-\frac{3}{4}}$$

Dividing by $-\frac{3}{4}$ undoes the multiplication by $-\frac{3}{4}$.

$$B=\frac{-2A+\frac{11\sqrt{2}}{12}+\frac{4}{15}}{-\frac{3}{4}}$$

Divide $\frac{11\sqrt{2}}{12}+\frac{4}{15}-2A$ by $-\frac{3}{4}$ by multiplying $\frac{11\sqrt{2}}{12}+\frac{4}{15}-2A$ by the reciprocal of $-\frac{3}{4}$.

$$B=\frac{8A}{3}-\frac{11\sqrt{2}}{9}-\frac{16}{45}$$