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The Process
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**"We need to turn this rational equation into a non-rational one to be able to solve it. The easiest way to do that, is to perform operations that give us a quotient (fraction) on each side of the equation. The steps are:
#1)# add #4# and #3/(x-2)# together by taking a common denominator and turning them into one fraction. The common denominator is #x-2# and #4# is actually #4/1#.
#4/1+3/(x-2)=(4(x-2)+3)/(x-2)#
What we did was divide the common denominator (#x-2#) by the denominator of the first term, i.e. #4#, and got #(x-2)#. Then we multiplied it by the numerator, i.e. #4# and put the result on top.
Then we divided the common denominator by the denominator of the second term and got #1#, i.e. #(x-2)/(x-2)#. Then we multiplied it by the numerator of the second term, i.e. #3# and put the result on top. This is how we ended up with the fraction on the right hand side.
#2)# After simplifying the new fraction, i.e. combining like terms and removing parentheses, we set it equal to the fraction we had on the left hand side.
#3)# Then we cross-multiplied the fractions to get a regular equation which, after simplifying, gave us a quadratic equation. This means an equation with the #x# having the power of #2#.
#4)# Then we used the quadratic formula to solve for #x#. If you have an equation in the form of:
#ax^2+bx+c=0#
The formula for #x# is:
#x=(-b+-sqrt(b^2-4ac))/(2a)#
In this equation:
#a=2#, #b=3#, and #c=-5#
We plugged them into the formula and got two answers for #x#.
I hope this helps."**
I am going to assume you meant the following:
#5/(x+4)=4+3/(x-2)#
#5/(x+4)=(4(x-2)+3)/(x-2)#
#5/(x+4)=(4x-8+3)/(x-2)#
#5/(x+4)=(4x-5)/(x-2)#
Cross-multiplying, we get:
#(x+4)(4x-5)=5(x-2)#
#4x^2-5x+16x-20=5x-10#
#4x^2+11x-20-5x+10=0#
#4x^2+6x-10=0#
#2(2x^2+3x-5)=0#
#2x^2+3x-5=0#
Using the quadratic formula:
#x=(-3+-sqrt(9-4(2)(-5)))/4=(-3+-sqrt49)/4=(-3+-7)/4#
#x=1#
#x=-5/2#
