Solve 3/(sqrt(3) - sqrt(2)) = a * sqrt(3) - b * sqrt(2) | 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

$$\frac{3}{\sqrt{3}-\sqrt{2}}=a\sqrt{3}-b\sqrt{2}$$

Solve for a

$a=\frac{\sqrt{6}b}{3}+\sqrt{6}+3$

Steps for Solving Linear Equation

Rationalize the denominator of $\frac{3}{\sqrt{3}-\sqrt{2}}$ by multiplying numerator and denominator by $\sqrt{3}+\sqrt{2}$.

$$\frac{3\left(\sqrt{3}+\sqrt{2}\right)}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}=a\sqrt{3}-b\sqrt{2}$$

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

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

Square $\sqrt{3}$. Square $\sqrt{2}$.

$$\frac{3\left(\sqrt{3}+\sqrt{2}\right)}{3-2}=a\sqrt{3}-b\sqrt{2}$$

Subtract $2$ from $3$ to get $1$.

$$\frac{3\left(\sqrt{3}+\sqrt{2}\right)}{1}=a\sqrt{3}-b\sqrt{2}$$

Anything divided by one gives itself.

$$3\left(\sqrt{3}+\sqrt{2}\right)=a\sqrt{3}-b\sqrt{2}$$

Use the distributive property to multiply $3$ by $\sqrt{3}+\sqrt{2}$.

$$3\sqrt{3}+3\sqrt{2}=a\sqrt{3}-b\sqrt{2}$$

Swap sides so that all variable terms are on the left hand side.

$$a\sqrt{3}-b\sqrt{2}=3\sqrt{3}+3\sqrt{2}$$

Add $b\sqrt{2}$ to both sides.

$$a\sqrt{3}=3\sqrt{3}+3\sqrt{2}+b\sqrt{2}$$

The equation is in standard form.

$$\sqrt{3}a=\sqrt{2}b+3\sqrt{2}+3\sqrt{3}$$

Divide both sides by $\sqrt{3}$.

$$\frac{\sqrt{3}a}{\sqrt{3}}=\frac{\sqrt{2}b+3\sqrt{2}+3\sqrt{3}}{\sqrt{3}}$$

Dividing by $\sqrt{3}$ undoes the multiplication by $\sqrt{3}$.

$$a=\frac{\sqrt{2}b+3\sqrt{2}+3\sqrt{3}}{\sqrt{3}}$$

Divide $3\sqrt{3}+3\sqrt{2}+b\sqrt{2}$ by $\sqrt{3}$.

$$a=\frac{\sqrt{3}\left(\sqrt{2}b+3\sqrt{2}+3\sqrt{3}\right)}{3}$$

Solve for b

$b=\frac{\sqrt{6}\left(a-3\right)-6}{2}$

Steps for Solving Linear Equation

Rationalize the denominator of $\frac{3}{\sqrt{3}-\sqrt{2}}$ by multiplying numerator and denominator by $\sqrt{3}+\sqrt{2}$.

$$\frac{3\left(\sqrt{3}+\sqrt{2}\right)}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}=a\sqrt{3}-b\sqrt{2}$$

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

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

Square $\sqrt{3}$. Square $\sqrt{2}$.

$$\frac{3\left(\sqrt{3}+\sqrt{2}\right)}{3-2}=a\sqrt{3}-b\sqrt{2}$$

Subtract $2$ from $3$ to get $1$.

$$\frac{3\left(\sqrt{3}+\sqrt{2}\right)}{1}=a\sqrt{3}-b\sqrt{2}$$

Anything divided by one gives itself.

$$3\left(\sqrt{3}+\sqrt{2}\right)=a\sqrt{3}-b\sqrt{2}$$

Use the distributive property to multiply $3$ by $\sqrt{3}+\sqrt{2}$.

$$3\sqrt{3}+3\sqrt{2}=a\sqrt{3}-b\sqrt{2}$$

Swap sides so that all variable terms are on the left hand side.

$$a\sqrt{3}-b\sqrt{2}=3\sqrt{3}+3\sqrt{2}$$

Subtract $a\sqrt{3}$ from both sides.

$$-b\sqrt{2}=3\sqrt{3}+3\sqrt{2}-a\sqrt{3}$$

The equation is in standard form.

$$\left(-\sqrt{2}\right)b=-\sqrt{3}a+3\sqrt{2}+3\sqrt{3}$$

Divide both sides by $-\sqrt{2}$.

$$\frac{\left(-\sqrt{2}\right)b}{-\sqrt{2}}=\frac{\sqrt{3}\left(-a+\sqrt{6}+3\right)}{-\sqrt{2}}$$

Dividing by $-\sqrt{2}$ undoes the multiplication by $-\sqrt{2}$.

$$b=\frac{\sqrt{3}\left(-a+\sqrt{6}+3\right)}{-\sqrt{2}}$$

Divide $\left(3+\sqrt{6}-a\right)\sqrt{3}$ by $-\sqrt{2}$.

$$b=-\frac{\sqrt{6}\left(-a+\sqrt{6}+3\right)}{2}$$