How do you solve $\frac { 2} { z + 1} = \frac { 1} { z - 8}$ ?

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Answer 1 · 2017-04-03T02:29:14

Multiply both sides by the least common denominator and solve for $z$ .

$z = 17$

To solve the problem it is helpful to remove the denominators.

To do this multiple both sides by the LCM of the denominators.

The LCM is $(z + 1)( z - 8 )$

$cancel(( z + 1))( z -8) xx 2 / cancel((z +1)) = ( z+ 1)(cancel( z -8)) xx 1 /cancel(( z-8))$

Multiplying this out removes the denominators. which results in:

$( z -8) xx 2 = (z + 1) xx 1 " "$ removing the parentheses results in

$2z - 16 = z + 1" "$ add 16 to both sides of the equation

$2z -16 + 16 = z + 1 +16$

$2z = z + 17" "$ Subtract $1z$ from both sides to isolate $z$

$2z- 1z = 1z - 1z + 17$ This gives

$z = 17$

How do you solve \frac { 2} { z + 1} = \frac { 1} { z - 8}\frac { 2} { z + 1} = \frac { 1} { z - 8}?