How do you solve #3-((a-1)/(a+3)) = ((a^2-5)/(a+3))#?

1 Answer
Apr 17, 2018

Answer:

Multiply both sides by #(a+3)# and then solve the resulting quadratic to find #a=5#

Explanation:

The denominator is the same for both fractions in the equation. We can use this to our advantage by multiplying both sides by the denominator to 'clear' it from our equation:

#color(red)((a+3))(3-(a-1)/(a+3))=cancel(color(red)((a+3)))((a^2-5)/cancel(a+3))#

#color(red)((a+3))xx3-cancel(color(red)((a+3)))((a-1)/cancel(a+3))=a^2-5#

#color(red)((a+3))xx3-(a-1)=a^2-5#

Now that we've 'cleared' the denominators from the equation, we simplify all the terms:

#3a+9-(a-1)=a^2-5#

#3a+9-a+1=a^2-5#

#2a+10=a^2-5#

#-a^2+2a+10=-5#

#-a^2+2a+15=0#

Now, we have a quadratic equation with the following coefficients:

#c_1=-1#
#c_2=2#
#c_3=15#

I wrote the equation using #c_n# terms instead of #a# #b# and #c# to differentiate from the #a# we are trying to solve. Let's plug those into the quadratic formula:

#a=(-c_2+-sqrt(c_2^2-4c_1c_3))/(2c_1)#

#a=(-2+-sqrt(2^2-4(-1)(15)))/(2(-1))#

#a=(-2+-sqrt(4+60))/(-2)=(-2+-sqrt(64))/(-2)#

#a=(-2+-8)/(-2)#

#a={(-2+8)/(-2),(-2-8)/(-2)}#

#a={6/(-2),(-10)/(-2)}#

#a={-3,5}#

EDIT: Updating results

Finally, we'll plug the solutions for #a# back in to the original equation:

When #a=-3#:

#3-((-3)-1)/((-3)+3)=((-3)^2-5)/((-3)+3)#

#3-(-4)/(0)=(4)/(0)#

Since we have a solution where there's division by zero, this solution is considered extraneous, making it invalid for this question.

When #a=5#:

#3-((5)-1)/((5)+3)=((5)^2-5)/((5)+3)#

#3-(4)/(8)=(20)/(8)#

#3=20/8+4/8=(20+4)/8=24/8#

#3=3#

This means that the positive root is valid, and gives us our final answer:

#color(green)(a=5)#