5-(a-5/a+4)=a^2-7/a+4 5-(a-5/a+4)=a^2-7/a+4 @Mathematics

Question

anonymous · Answer

is the answer 12?

agreene · Answer

$$5-\frac{a-5}{a+4}=\frac{a^2-7}{a+4}$$

is that the question?

anonymous · Answer

yup thats it

agreene · Answer

the answer isnt 12 O.o

agreene · Answer

start by multiplying both sides by a+4 and you get to
$$5(a+4)-a+5=a^2-7$$
$$5a+20-a+5=a^2-7$$
$$4a+25=a^2-7$$
$$-a^2+4a+32=0$$
$$a^2-4a-32=0$$
$$a^2-4a=32$$
add 4 to both sides so we can factor
$$a^2-4a+4=32$$
$$(a-2)^2=32$$
sqrt both sides
$$|a-2|=6$$
now we need to eliminate the absolute value:
a-2=-6                or                        a-2=6
a=-4                    or                       a=8
plug back in and you will see -4 doesnt work so the only solution is
$$\large a=8$$

anonymous · Answer

i was way off lol

agreene · Answer

well, that one was a bit tricky--much harder than the 1st one I answered, lol

anonymous · Answer

yea i had like a whole sheet of paper with the problem worked out lol