Solve {l}{x+1/y=14/3}{y+1/z=13/21}{2+frac{1}{x}=frac{27}{7}}{a=21xyz}{text{Solvefor}btext{where}}{b=a}

$$\left. \begin{array} { l } { x + \frac{1}{y} = \frac{14}{3} }\\ { y + \frac{1}{z} = \frac{13}{21} }\\ { 2 + \frac{1}{x} = \frac{27}{7} }\\ { a = 21 x y z }\\ { \text{Solve for } b \text{ where} } \\ { b = a } \end{array} \right.$$

Solve for x, y, z, a, b

$$b = \frac{189}{26} = 7\frac{7}{26} \approx 7.269230769$$

Solution Steps

Consider the third equation. Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 7x, the least common multiple of x,7.

$$7x\times 2+7=27x$$

Multiply 7 and 2 to get 14.

$$14x+7=27x$$

Subtract 27x from both sides.

$$14x+7-27x=0$$

Combine 14x and -27x to get -13x.

$$-13x+7=0$$

Subtract 7 from both sides. Anything subtracted from zero gives its negation.

$$-13x=-7$$

Divide both sides by -13.

$$x=\frac{-7}{-13}$$

Fraction \frac{-7}{-13} can be simplified to \frac{7}{13} by removing the negative sign from both the numerator and the denominator.

$$x=\frac{7}{13}$$

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

$$\frac{7}{13}+\frac{1}{y}=\frac{14}{3}$$

Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 39y, the least common multiple of 13,y,3.

$$39y\times \frac{7}{13}+39=182y$$

Multiply 39 and \frac{7}{13} to get 21.

$$21y+39=182y$$

Subtract 182y from both sides.

$$21y+39-182y=0$$

Combine 21y and -182y to get -161y.

$$-161y+39=0$$

Subtract 39 from both sides. Anything subtracted from zero gives its negation.

$$-161y=-39$$

Divide both sides by -161.

$$y=\frac{-39}{-161}$$

Fraction \frac{-39}{-161} can be simplified to \frac{39}{161} by removing the negative sign from both the numerator and the denominator.

$$y=\frac{39}{161}$$

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

$$\frac{39}{161}+\frac{1}{z}=\frac{13}{21}$$

Variable z cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 483z, the least common multiple of 161,z,21.

$$483z\times \frac{39}{161}+483=299z$$

Multiply 483 and \frac{39}{161} to get 117.

$$117z+483=299z$$

Subtract 299z from both sides.

$$117z+483-299z=0$$

Combine 117z and -299z to get -182z.

$$-182z+483=0$$

Subtract 483 from both sides. Anything subtracted from zero gives its negation.

$$-182z=-483$$

Divide both sides by -182.

$$z=\frac{-483}{-182}$$

Reduce the fraction \frac{-483}{-182} to lowest terms by extracting and canceling out -7.

$$z=\frac{69}{26}$$

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

$$a=21\times \frac{7}{13}\times \frac{39}{161}\times \frac{69}{26}$$

Multiply 21 and \frac{7}{13} to get \frac{147}{13}.

$$a=\frac{147}{13}\times \frac{39}{161}\times \frac{69}{26}$$

Multiply \frac{147}{13} and \frac{39}{161} to get \frac{63}{23}.

$$a=\frac{63}{23}\times \frac{69}{26}$$

Multiply \frac{63}{23} and \frac{69}{26} to get \frac{189}{26}.

$$a=\frac{189}{26}$$

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

$$b=\frac{189}{26}$$

The system is now solved.

$$x=\frac{7}{13} y=\frac{39}{161} z=\frac{69}{26} a=\frac{189}{26} b=\frac{189}{26}$$