Solve 3a+5b=sqrt{3}a+sqrt{3}b+6quada*b=?

$$3 a + 5 b = \sqrt { 3 } a + \sqrt { 3 } b + 6 \quad a \cdot b = ?$$

Solve for a

$$a=-\frac{\left(5-\sqrt{3}\right)b}{-6b+3-\sqrt{3}}$$

$$b\neq -\frac{\sqrt{3}}{6}+\frac{1}{2}$$

Steps for Solving Linear Equation

Subtract \sqrt{3}a from both sides.

$$3a+5b-\sqrt{3}a=\sqrt{3}b+6ab$$

Subtract 6ab from both sides.

$$3a+5b-\sqrt{3}a-6ab=\sqrt{3}b$$

Subtract 5b from both sides.

$$3a-\sqrt{3}a-6ab=\sqrt{3}b-5b$$

Combine all terms containing a.

$$\left(3-\sqrt{3}-6b\right)a=\sqrt{3}b-5b$$

The equation is in standard form.

$$\left(-6b+3-\sqrt{3}\right)a=\sqrt{3}b-5b$$

Divide both sides by 3-\sqrt{3}-6b.

$$\frac{\left(-6b+3-\sqrt{3}\right)a}{-6b+3-\sqrt{3}}=\frac{\left(\sqrt{3}-5\right)b}{-6b+3-\sqrt{3}}$$

Dividing by 3-\sqrt{3}-6b undoes the multiplication by 3-\sqrt{3}-6b.

$$a=\frac{\left(\sqrt{3}-5\right)b}{-6b+3-\sqrt{3}}$$

Solve for b

$$b=-\frac{\sqrt{3}\left(\sqrt{3}-1\right)a}{-6a+5-\sqrt{3}}$$

$$a\neq \frac{5-\sqrt{3}}{6}$$

Steps for Solving Linear Equation

Subtract \sqrt{3}b from both sides.

$$3a+5b-\sqrt{3}b=\sqrt{3}a+6ab$$

Subtract 6ab from both sides.

$$3a+5b-\sqrt{3}b-6ab=\sqrt{3}a$$

Subtract 3a from both sides.

$$5b-\sqrt{3}b-6ab=\sqrt{3}a-3a$$

Combine all terms containing b.

$$\left(5-\sqrt{3}-6a\right)b=\sqrt{3}a-3a$$

The equation is in standard form.

$$\left(-6a+5-\sqrt{3}\right)b=\sqrt{3}a-3a$$

Divide both sides by 5-\sqrt{3}-6a.

$$\frac{\left(-6a+5-\sqrt{3}\right)b}{-6a+5-\sqrt{3}}=\frac{\left(\sqrt{3}-3\right)a}{-6a+5-\sqrt{3}}$$

Dividing by 5-\sqrt{3}-6a undoes the multiplication by 5-\sqrt{3}-6a.

$$b=\frac{\left(\sqrt{3}-3\right)a}{-6a+5-\sqrt{3}}$$