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

1 Answer
Mar 21, 2016

Answer:

Resolving this gives: #"x^2+6x+7=0#

I will let you finish that off (you need to use the formula).

Explanation:

The bottom number/expression (denominator) need to be the same to enable direct addition.

Multiply by 1 and you do not change the value. Multiply by 1 but in the form of #1=(x+5)/(x+5) # you do not change the value but you do change the way it looks.

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Multiply #1/(x+3)# by 1 but in the form of #1=(x+5)/(x+5)#

Giving #color(brown)(1/(x+3)xx(x+5)/(x+5) = (x+5)/((x+3)(x+5)))#

Multiply #1/(x+5)# by 1 but in the form of #1=(x+3)/(x+3)#

Giving #color(brown)(1/(x+5)xx(x+3)/(x+3) = (x+3)/((x+3)(x+5)))#

#color(blue)("Putting it all together")#

#" "(x+5)/((x+3)(x+5)) + (x+3)/((x+3)(x+5))=1#

# " "(2x+8)/((x+3)(x+5))=1#

Multiply both sides by #(x+3)(x+5)# giving

# " "(2x+8)xx(cancel((x+3)(x+5)))/(cancel((x+3)(x+5)))=1xx(x+3)(x+5)#

#" "2x+8=x^2+8x+15#

#" "x^2+6x+7=0#

Use the formula to solve for #x#

I will let you do that

#y=ax^2+bx+c" "# where #" "x=(-b+-sqrt(b^2-4ac))/(2a)#
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Tony B