Question

Solve for x. Show all your work and check you answers.
1 over x^2-7x+10=x over x-5+1over x-2

The *overs* would be a fraction but I haven't quite figured out how to make them fractions yet so sorry about that

Answer 1

\[\frac{ 1 }{ x ^{2}-7x+10 }=\frac{ x }{ x-5 }+\frac{ 1 }{ x-2 }\]

Answer 2

1/(x^2-7x+10)=[x/*(x-5)]+[1/(x-2)]

Answer 3

\[\frac{ 1 }{ (x-5)(x-2) } = \frac{ x }{ x-5 } + \frac{ 1 }{ x-2 }\]

Answer 4

to just the right side of the equation
\[[\frac{ x }{ x-5 }\times\frac{ x-2 }{ x-2 }]+[\frac{ 1 }{ x-2 }\times\frac{ x-5 }{ x-5 }]+\]

Answer 5

\[= \frac{ x(x-2)+1(x-5) }{ (x-2)(x-5) }\]

Answer 6

\[=\frac{ (x^2-2x+x-5) }{ (x-2)(x-5 }\]

Answer 7

\[=\frac{ (x^2-x-5) }{ (x-2)(x-5) }\]

Answer 8

\[\frac{ 1 }{ (x-5)(x-2) }=\frac{ x^2-x-5 }{ (x-2)(x-5) }\]

Answer 9

cross multiply top left with bottom right = top right with bottom left

Answer 10

\[1\times(x-2)(x-5)=(x^2-x-5)(x-5)(x-2)\]

Answer 11

add them together set them on one side where other side is =0
and use b square all over two a stuff
and you get x=-2 and x=3

Answer 12

-b +/- square root of b squared-4(a)(c)divided by 2a

Answer 13

example:1x^2+3x+4=0
a=1
b=3
c=4

Answer 14

\[\small\begin{align}
\frac{ 1 }{ x ^{2}-7x+10 }=\frac{ x }{ x-5 }+\frac{ 1 }{ x-2 }
\ &: ~ \frac{ 1 }{ (x-5)(x-2) } = \frac{ x }{ x-5 } + \frac{ 1 }{ x-2 }
\\&: ~ \frac{1}{(x-5)(x-2)}=\frac{x(x-2)}{(x-5)(x-2)}+\frac{1(x-5)}{(x-5)(x-2)}
\\&: ~ (x-5)(x-2)\left[\frac{1}{(x-5)(x-2)}=\frac{x(x-2)}{(x-5)(x-2)}+ \\
\frac{1(x-5)}{(x-5)(x-2)}\right]\\
\\&: ~ 1=x(x-2)+1(x-5)
\\&: ~ 1 = x^2-2x +x-5
\\&: ~ 1=x^2 -x-5
\\&:~ 0=x^2 -x-6
\end{align}\]Find two numbers that multiply to give you -6 but add to give you -1, write it out in factored form, and you're done.