Solve 4.9=a(1^2)+b(1)d

$$4.9 = a ( 1 ^ { 2 } ) + b ( 1 ) d$$

Solve for b (complex solution)

$$\left\{\begin{matrix}b=-\frac{a-4.9}{d}\text{, }&d\neq 0\\b\in \mathrm{C}\text{, }&a=\frac{49}{10}\text{ and }d=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Calculate 1 to the power of 2 and get 1.

$$4.9=a\times 1+b\times 1d$$

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

$$a\times 1+b\times 1d=4.9$$

Subtract a\times 1 from both sides.

$$b\times 1d=4.9-a\times 1$$

Reorder the terms.

$$bd=4.9-a$$

The equation is in standard form.

$$db=4.9-a$$

Divide both sides by d.

$$\frac{db}{d}=\frac{4.9-a}{d}$$

Dividing by d undoes the multiplication by d.

$$b=\frac{4.9-a}{d}$$

Solve for a

$$a=4.9-bd$$

Solution Steps

Calculate 1 to the power of 2 and get 1.

$$4.9=a\times 1+b\times 1d$$

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

$$a\times 1+b\times 1d=4.9$$

Subtract b\times 1d from both sides.

$$a\times 1=4.9-b\times 1d$$

Reorder the terms.

$$a=4.9-bd$$

Reorder the terms.

$$a=-bd+4.9$$

Solve for b

$$\left\{\begin{matrix}b=-\frac{a-4.9}{d}\text{, }&d\neq 0\\b\in \mathrm{R}\text{, }&a=\frac{49}{10}\text{ and }d=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Calculate 1 to the power of 2 and get 1.

$$4.9=a\times 1+b\times 1d$$

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

$$a\times 1+b\times 1d=4.9$$

Subtract a\times 1 from both sides.

$$b\times 1d=4.9-a\times 1$$

Reorder the terms.

$$bd=4.9-a$$

The equation is in standard form.

$$db=4.9-a$$

Divide both sides by d.

$$\frac{db}{d}=\frac{4.9-a}{d}$$

Dividing by d undoes the multiplication by d.

$$b=\frac{4.9-a}{d}$$