How do you simplify and find the restrictions for #(y-3)/(y+5)#?

1 Answer
May 4, 2016

Answer:

restriction: #y!=-5#

Explanation:

The given expression is already simplified and cannot be simplified any further.

To find the restriction, recall that in any fraction, the denominator must not equal to #0#. In the given fraction,

#(y-3)/(y+5)#

the denominator is expressed as a variable plus a constant. When setting the denominator to equal to #0#, you are solving for the value of #y# that will produce a denominator or #0#. The #y# value would be your restriction.

Thus,

#y+5=0#

#y+5color(white)(i)color(red)(-5)=0color(white)(i)color(red)(-5)#

#y=-5#

restriction: #color(green)(|bar(ul(color(white)(a/a)color(black)(y!=-5)color(white)(a/a)|)))#

To check your answer, you can plug in #y=-5# into the given expression to check if the denominator equals #0#. If it does, then you know the restriction is correct.

Plugging in #y=-5#,

#(y-3)/(y+5)#

#=(-5-3)/(-5+5)#

#=-8/0#

undefined

#:.#, the restriction is correct.