How do you simplify #(x-5)/(4x-28)# and what are the restrictions?

1 Answer
May 17, 2018

#(x-5)/(4(x-7))#, with the restriction #x != 7#.

Explanation:

The numerator #(x-5)# can not be made simpler by factoring, but the denominator can.

In #4x-28#, both #4x# and #–28# are multiples of #4.# We can simplify #4x-28# by factoring out this multiple of 4:

#4x-28 = (4)x-(4)7#

#color(white)(4x-28) = 4(x-7)#

We now write our original fraction with this new denominator:

#color(white)= (x-5)/(4x-28)#

#=(x-5)/(4(x-7))#

At this point, we would check to see if there are any factors common to both the numerator and denominator. If there were, we could cancel the pairs off. By sight, we can see there are not, so this is as far as we can simplify.

Since this expression involves a denominator (i.e. division), we must be sure to remember that we are not allowed to divide by zero. (Division by 0 is undefined.) Because our denominator has a variable in it, we must not allow that variable to be anything that makes the denominator zero.

To find the restricted values, we set out to find what value(s) of #x# would make the denominator 0, so we can avoid those values.

#4(x-7) = 0#
#color(white)"4("x-7color(white)(")")=0#
#color(white)("4(")xcolor(white)(-0"   ")=7#

Thus, a value of #x=7# would cause the denominator to be 0, so this is the value of #x# we need to avoid.