Solve {l}{x-1/2-y-1/3=-13/30}{x-11/3+frac{9+1}{2}=-frac{2}{3}}{z=x-1-2y}{a=z}{text{Solvefor}btext{where}}{b=a}

$$\left. \begin{array} { l } { \frac{x - 1}{2} - \frac{y - 1}{3} = -\frac{13}{30} }\\ { \frac{x - 11}{3} + \frac{9 + 1}{2} = -\frac{2}{3} }\\ { z = x - 1 - 2 y }\\ { a = z }\\ { \text{Solve for } b \text{ where} } \\ { b = a } \end{array} \right.$$

Solve for x, y, z, a, b

$$b = \frac{47}{5} = 9\frac{2}{5} = 9.4$$

Solution Steps

Consider the second equation. Multiply both sides of the equation by 6, the least common multiple of 3,2.

$$2\left(x-11\right)+3\left(9+1\right)=-4$$

Use the distributive property to multiply 2 by x-11.

$$2x-22+3\left(9+1\right)=-4$$

Add 9 and 1 to get 10.

$$2x-22+3\times 10=-4$$

Multiply 3 and 10 to get 30.

$$2x-22+30=-4$$

Add -22 and 30 to get 8.

$$2x+8=-4$$

Subtract 8 from both sides.

$$2x=-4-8$$

Subtract 8 from -4 to get -12.

$$2x=-12$$

Divide both sides by 2.

$$x=\frac{-12}{2}$$

Divide -12 by 2 to get -6.

$$x=-6$$

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

$$\frac{-6-1}{2}-\frac{y-1}{3}=-\frac{13}{30}$$

Multiply both sides of the equation by 30, the least common multiple of 2,3,30.

$$15\left(-6-1\right)-10\left(y-1\right)=-13$$

Subtract 1 from -6 to get -7.

$$15\left(-7\right)-10\left(y-1\right)=-13$$

Multiply 15 and -7 to get -105.

$$-105-10\left(y-1\right)=-13$$

Use the distributive property to multiply -10 by y-1.

$$-105-10y+10=-13$$

Add -105 and 10 to get -95.

$$-95-10y=-13$$

Add 95 to both sides.

$$-10y=-13+95$$

Add -13 and 95 to get 82.

$$-10y=82$$

Divide both sides by -10.

$$y=\frac{82}{-10}$$

Reduce the fraction \frac{82}{-10} to lowest terms by extracting and canceling out 2.

$$y=-\frac{41}{5}$$

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

$$z=-6-1-2\left(-\frac{41}{5}\right)$$

Subtract 1 from -6 to get -7.

$$z=-7-2\left(-\frac{41}{5}\right)$$

Multiply -2 and -\frac{41}{5} to get \frac{82}{5}.

$$z=-7+\frac{82}{5}$$

Add -7 and \frac{82}{5} to get \frac{47}{5}.

$$z=\frac{47}{5}$$

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

$$a=\frac{47}{5}$$

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

$$b=\frac{47}{5}$$

The system is now solved.

$$x=-6 y=-\frac{41}{5} z=\frac{47}{5} a=\frac{47}{5} b=\frac{47}{5}$$