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

Question

Savannah Anderson · Accepted Answer

restriction: #y!=-5#

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.

Julia Adams · Answer

undefined To simplify the expression (y-3)/(y+5), you can divide both the numerator and denominator by their greatest common factor, which is 1 in this case. This simplifies the expression to (y-3)/(y+5).

To find the restrictions, we need to identify any values of y that would make the denominator equal to zero, as division by zero is undefined. In this case, the denominator is y+5. Therefore, the restriction is that y cannot be equal to -5.