Solve {l}{3x-1=100}{y+x-1=97}{z-x*y=1}{a=z-x+y}{text{Solvefor}btext{where}}{b=a}

$$\left. \begin{array} { l } { 3 x - 1 = 100 }\\ { y + x - 1 = 97 }\\ { z - x \cdot y = 1 }\\ { a = z - x + y }\\ { \text{Solve for } b \text{ where} } \\ { b = a } \end{array} \right.$$

Solve for x, y, z, a, b

$$b = \frac{19778}{9} = 2197\frac{5}{9} \approx 2197.555555556$$

Solution Steps

Consider the first equation. Add 1 to both sides.

$$3x=100+1$$

Add 100 and 1 to get 101.

$$3x=101$$

Divide both sides by 3.

$$x=\frac{101}{3}$$

Consider the second equation. Insert the known values of variables into the equation.

$$y+\frac{101}{3}-1=97$$

Subtract 1 from \frac{101}{3} to get \frac{98}{3}.

$$y+\frac{98}{3}=97$$

Subtract \frac{98}{3} from both sides.

$$y=97-\frac{98}{3}$$

Subtract \frac{98}{3} from 97 to get \frac{193}{3}.

$$y=\frac{193}{3}$$

Consider the third equation. Insert the known values of variables into the equation.

$$z-\frac{101}{3}\times \frac{193}{3}=1$$

Multiply -\frac{101}{3} and \frac{193}{3} to get -\frac{19493}{9}.

$$z-\frac{19493}{9}=1$$

Add \frac{19493}{9} to both sides.

$$z=1+\frac{19493}{9}$$

Add 1 and \frac{19493}{9} to get \frac{19502}{9}.

$$z=\frac{19502}{9}$$

Consider the fourth equation. Insert the known values of variables into the equation.

$$a=\frac{19502}{9}-\frac{101}{3}+\frac{193}{3}$$

Subtract \frac{101}{3} from \frac{19502}{9} to get \frac{19199}{9}.

$$a=\frac{19199}{9}+\frac{193}{3}$$

Add \frac{19199}{9} and \frac{193}{3} to get \frac{19778}{9}.

$$a=\frac{19778}{9}$$

Consider the fifth equation. Insert the known values of variables into the equation.

$$b=\frac{19778}{9}$$

The system is now solved.

$$x=\frac{101}{3} y=\frac{193}{3} z=\frac{19502}{9} a=\frac{19778}{9} b=\frac{19778}{9}$$