Solve sqrt{9}+sqrt{27}+sqrt{12}=a+bsqrt{3}

$$\sqrt { 9 } + \sqrt { 27 } + \sqrt { 12 } = a + b \sqrt { 3 }$$

Solve for b

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

Steps for Solving Linear Equation

Calculate the square root of 9 and get 3.

$$3+\sqrt{27}+\sqrt{12}=a+b\sqrt{3}$$

Factor 27=3^{2}\times 3. Rewrite the square root of the product \sqrt{3^{2}\times 3} as the product of square roots \sqrt{3^{2}}\sqrt{3}. Take the square root of 3^{2}.

$$3+3\sqrt{3}+\sqrt{12}=a+b\sqrt{3}$$

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

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

Combine 3\sqrt{3} and 2\sqrt{3} to get 5\sqrt{3}.

$$3+5\sqrt{3}=a+b\sqrt{3}$$

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

$$a+b\sqrt{3}=3+5\sqrt{3}$$

Subtract a from both sides.

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

The equation is in standard form.

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

Divide both sides by \sqrt{3}.

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

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

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

Divide 3+5\sqrt{3}-a by \sqrt{3}.

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

Solve for a

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