Solve 10/3=a(8/3)^2+b*8/3+3

$$\frac { 10 } { 3 } = a ( \frac { 8 } { 3 } ) ^ { 2 } + b \times \frac { 8 } { 3 } + 3$$

Solve for a

$$a=-\frac{3b}{8}+\frac{3}{64}$$

Steps for Solving Linear Equation

Calculate \frac{8}{3} to the power of 2 and get \frac{64}{9}.

$$\frac{10}{3}=a\times \frac{64}{9}+b\times \frac{8}{3}+3$$

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

$$a\times \frac{64}{9}+b\times \frac{8}{3}+3=\frac{10}{3}$$

Subtract b\times \frac{8}{3} from both sides.

$$a\times \frac{64}{9}+3=\frac{10}{3}-b\times \frac{8}{3}$$

Subtract 3 from both sides.

$$a\times \frac{64}{9}=\frac{10}{3}-b\times \frac{8}{3}-3$$

Multiply -1 and \frac{8}{3} to get -\frac{8}{3}.

$$a\times \frac{64}{9}=\frac{10}{3}-\frac{8}{3}b-3$$

Subtract 3 from \frac{10}{3} to get \frac{1}{3}.

$$a\times \frac{64}{9}=\frac{1}{3}-\frac{8}{3}b$$

The equation is in standard form.

$$\frac{64}{9}a=\frac{1-8b}{3}$$

Divide both sides of the equation by \frac{64}{9}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{64}{9}a}{\frac{64}{9}}=\frac{1-8b}{3\times \frac{64}{9}}$$

Dividing by \frac{64}{9} undoes the multiplication by \frac{64}{9}.

$$a=\frac{1-8b}{3\times \frac{64}{9}}$$

Divide \frac{1-8b}{3} by \frac{64}{9} by multiplying \frac{1-8b}{3} by the reciprocal of \frac{64}{9}.

$$a=-\frac{3b}{8}+\frac{3}{64}$$

Solve for b

$$b=-\frac{8a}{3}+\frac{1}{8}$$

Steps for Solving Linear Equation

Calculate \frac{8}{3} to the power of 2 and get \frac{64}{9}.

$$\frac{10}{3}=a\times \frac{64}{9}+b\times \frac{8}{3}+3$$

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

$$a\times \frac{64}{9}+b\times \frac{8}{3}+3=\frac{10}{3}$$

Subtract a\times \frac{64}{9} from both sides.

$$b\times \frac{8}{3}+3=\frac{10}{3}-a\times \frac{64}{9}$$

Subtract 3 from both sides.

$$b\times \frac{8}{3}=\frac{10}{3}-a\times \frac{64}{9}-3$$

Multiply -1 and \frac{64}{9} to get -\frac{64}{9}.

$$b\times \frac{8}{3}=\frac{10}{3}-\frac{64}{9}a-3$$

Subtract 3 from \frac{10}{3} to get \frac{1}{3}.

$$b\times \frac{8}{3}=\frac{1}{3}-\frac{64}{9}a$$

The equation is in standard form.

$$\frac{8}{3}b=-\frac{64a}{9}+\frac{1}{3}$$

Divide both sides of the equation by \frac{8}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{8}{3}b}{\frac{8}{3}}=\frac{-\frac{64a}{9}+\frac{1}{3}}{\frac{8}{3}}$$

Dividing by \frac{8}{3} undoes the multiplication by \frac{8}{3}.

$$b=\frac{-\frac{64a}{9}+\frac{1}{3}}{\frac{8}{3}}$$

Divide \frac{1}{3}-\frac{64a}{9} by \frac{8}{3} by multiplying \frac{1}{3}-\frac{64a}{9} by the reciprocal of \frac{8}{3}.

$$b=-\frac{8a}{3}+\frac{1}{8}$$