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

1 Answer
Sep 7, 2016

Answer:

#x = 3.064 " or " x = 1.436#

Explanation:

If you have an algebraic expression which has fractions, you have to work with them and perform whatever operation is involved.

However, if you have an equation which has fractions, you can get rid of the denominators by the process of:

Multiply through by the LCD and cancel the denominators.

In this case the LCD = #color(blue)(2(x-3)(2-x)#

#(color(blue)(2(x-3)(2-x))xx1)/(2(x-3))+(color(blue)(2(x-3)(2-x))xx3)/(2-x)= (color(blue)(2(x-3)(2-x))xx5#

cancel the denominators:

#(cancel2cancel(x-3)(2-x)xx1)/(cancel2cancel(x-3))+(color(blue)(2(x-3)cancel(2-x))xx3)/cancel(2-x)= color(blue)(2(x-3)(2-x))xx5#

#(2-x) + 6(x-3) = 10(x-3)(2-x)#

#2-x +6x-18= 10(5x-x^2-6)#

#5x-16 = 50x-10x^2-60 " "larr # make it = 0

#10x^2-45x +44= 0#

Use the formula

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#x = (-(-45)+-sqrt((-45)^2-4(10)(44)))/(2(10))#

#x = (45+-sqrt(2025-1760))/(20)#

#x = (45+sqrt265)/20" or " x = (45-sqrt265)/20#

#x = 3.064 " or " x = 1.436#