If #p=3, q=-5# and #r=-1#, what is #(5p+q)/(q-8r) + (4q^2)/(a+5p)#?

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Answer 1 · 2017-05-03T08:07:39

#(10/3)+(100/(a+15))#

If you assign these values into the equation:

#(5*3-5)/(-5-(8*-1)) + (4*(-5^2))/(a+5*3)#

It will be:

#((15-5)/(-5+8)) + (100/(a+15))#

#10/3 + 100/(a+15)#

This is the answer of the problem. If you know a, you can get a better answer.

#10/3 + 100/(a+15)#

Answer 2 · 2017-05-03T08:11:29

#(5p+q)/(q-8r)+(4q^2)/(a+5p)=(10a+450)/(3a+15)#

As #p=3#, #q=-5# and #r=-1#

#(5p+q)/(q-8r)+(4q^2)/(a+5p)#

= #(5xx3+(-5))/((-5)-8(-1))+(4(-5)^2)/(a+5xx3)#

= #(15-5)/((-5)+8)+(4xx25)/(a+15)#

= #10/3+100/(a+15)#

= #(10(a+15)+100xx3)/(3(a+5))#

= #(10a+150+300)/(3(a+5))#

= #(10a+450)/(3a+15)#