15/x-15/x+2=2
\[\frac{ 15 }{ x } - \frac{ 15 }{ x+2 } = 2\]
multiply each term so that they both have a common denominator.
the common denominator should be x(x+2)
multiple the numerator and the denominator of each term until every term (on the left) has that denominator.
so it is \[\frac{ 30x }{ x(x+2)} = 2\]
now multiply the left by x(x+2) and the right by the same.
that will get rid of the fraction.
so 30=2(x(x+2) @seitys
you're missing an x on the left. 30=2(x(x+2)) 30=2(x^2+2x) 30=2x^2+4x 0=2x^2+4x-30 0=2(x^2+2x-15) 0=2(x+5)(x-3)
ignore my first line, you're not missing an x on the left.
the answer in my book says -5, 3; i get how the -5 and 3 work but where does the 2 go because its not in the answer
@seitys
x can be -5 or 3.
if you plug in -5 or 3 for x, it satisfies the equation which is equal to 0.
2 is a constant, it doesn't have a role. if that 2 was 1,000,000 instead, your zeroes are still -5 and 3.
because only -5 OR 3 would make that equation equal to 0.
ah ok, thank you!
Join our real-time social learning platform and learn together with your friends!