How do you solve 1/(x+3)+1/(x+5)=11x+3+1x+5=1?

1 Answer
Mar 21, 2016

Resolving this gives: "x^2+6x+7=0x2+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) 1=x+5x+5 you do not change the value but you do change the way it looks.

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

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

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

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

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

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

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

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