Solve 9a+bsqrt[3]{4}+1=8

$$9 a + b \sqrt[ 3 ] { 4 } + 1 = 8$$

Solve for a

$$a=\frac{7-2^{\frac{2}{3}}b}{9}$$

Steps for Solving Linear Equation

Subtract b\sqrt[3]{4} from both sides.

$$9a+1=8-b\sqrt[3]{4}$$

Subtract 1 from both sides.

$$9a=8-b\sqrt[3]{4}-1$$

Subtract 1 from 8 to get 7.

$$9a=7-b\sqrt[3]{4}$$

The equation is in standard form.

$$9a=-\sqrt[3]{4}b+7$$

Divide both sides by 9.

$$\frac{9a}{9}=\frac{-\sqrt[3]{4}b+7}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$a=\frac{-\sqrt[3]{4}b+7}{9}$$

Divide 7-b\sqrt[3]{4} by 9.

$$a=\frac{7-2^{\frac{2}{3}}b}{9}$$

Solve for b

$$b=-\frac{4^{\frac{2}{3}}\left(9a-7\right)}{4}$$

Steps for Solving Linear Equation

Subtract 9a from both sides.

$$b\sqrt[3]{4}+1=8-9a$$

Subtract 1 from both sides.

$$b\sqrt[3]{4}=8-9a-1$$

Subtract 1 from 8 to get 7.

$$b\sqrt[3]{4}=7-9a$$

The equation is in standard form.

$$\sqrt[3]{4}b=7-9a$$

Divide both sides by \sqrt[3]{4}.

$$\frac{\sqrt[3]{4}b}{\sqrt[3]{4}}=\frac{7-9a}{\sqrt[3]{4}}$$

Dividing by \sqrt[3]{4} undoes the multiplication by \sqrt[3]{4}.

$$b=\frac{7-9a}{\sqrt[3]{4}}$$