Solve 3=aleft(2*1+1right)^2+b

$$3 = a { \left(2 \times 1+1 \right) }^{ 2 } +b$$

Solve for a

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

Steps for Solving Linear Equation

Multiply 2 and 1 to get 2.

$$3=a\left(2+1\right)^{2}+b$$

Add 2 and 1 to get 3.

$$3=a\times 3^{2}+b$$

Calculate 3 to the power of 2 and get 9.

$$3=a\times 9+b$$

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

$$a\times 9+b=3$$

Subtract b from both sides.

$$a\times 9=3-b$$

The equation is in standard form.

$$9a=3-b$$

Divide both sides by 9.

$$\frac{9a}{9}=\frac{3-b}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$a=\frac{3-b}{9}$$

Divide 3-b by 9.

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

Solve for b

$$b=3-9a$$

Solution Steps

Multiply 2 and 1 to get 2.

$$3=a\left(2+1\right)^{2}+b$$

Add 2 and 1 to get 3.

$$3=a\times 3^{2}+b$$

Calculate 3 to the power of 2 and get 9.

$$3=a\times 9+b$$

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

$$a\times 9+b=3$$

Subtract a\times 9 from both sides.

$$b=3-a\times 9$$

Multiply -1 and 9 to get -9.

$$b=3-9a$$