Question

Solve the equation.

Answer 1

whats the equation?

Answer 2

\[\frac{ 7x+1 }{ x-2}+ \frac{ 3 }{ x } = \frac{ -6 }{ x^2-2x }\]

Answer 3

find the common denominator for \[\frac{7x+1}{x-2}and\frac{3}{x}\]

Answer 4

-4/7

Answer 5

x=-4/7

Answer 6

\[\frac{(7x+1)x}{(x-2)x}+\frac{3(x-2)}{x(x-2)}=\frac{-6}{x^2-2x}\]
\[\frac{7x^2+x}{x^2-2x}+\frac{3x-6}{x^2-2x}=\frac{-6}{x^2-2x}\]
you can cancel out the denominator, remove it, so you get just
\[7x^2+x+3x-6=-6\]add 6 to both sides,\[7x^2+x+3x-6+6=-6+6 ---> 7x^2+x+3x=0\]
add like terms, or add 3x+x\[7x^2+x+3x=0-->7x^2+4x=0\]
Now that you have \[7x^2+4x=0\]
can you solve it on your own?

Answer 7

factor out of x, so you get \[7x^2+4x=0--->x(7x+4)=0\]
so either
x=0 or 7x+4=0 ----> 7x=-4 ---> x=-4/7

Answer 8

Note that the problem is undefined for x = 0 and so you will have to discard that solution.

Answer 9

Yes! , true!