Question

can i get help with some algebra 1?

Answer 1

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

Answer 2

is there something missing on the right side? Is anything below x-3 ?

Answer 3

sorry x

Answer 4

x−7 x−3
---- +2= ----
x x

Answer 5

my first thought is to multiply both sides by x
like this
\[ x \cdot \left(\frac{(x-7)}{x} +2\right)= x \cdot \frac{(x-3)}{x} \]

Answer 6

ok \[\frac{ x(x-7) }{ x^2 } = (\frac{ x(x-3) }{ x^2 })\]

Answer 7

interesting... time to learn a few algebra rules.

First,
\[ c\cdot \frac{a}{b} \text{ is the same as } \frac{c}{1}\cdot \frac{a}{b}\]
you multiply top times top and bottom times bottom

so for the right hand side
\[ x \cdot \frac{(x-3)}{x} \]
you could write it as
\[ \frac{x(x-3)}{x}\]
but the x on top and the x on the bottom make
\[ \frac{x}{x} = 1\]
anything divided by itself is 1

Answer 8

but how would i multiply the equation?

Answer 9

start with
\[ x \cdot \left(\frac{(x-7)}{x} +2\right)= x \cdot \frac{(x-3)}{x} \]

what do you get for the right-hand side of the = ?
We'll get to the left after we get the right side done

Answer 10

x^2-3x over x?

Answer 11

@phi

Answer 12

yes, but you should always look for the same thing in the top and bottom, because they divide out.

so rather than doing x(x-3)/x --> (x^2 -3x)/x (which is correct)
you should say:
\[ \frac{\cancel{x}(x-3)}{\cancel{x}} = (x-3) \]

Answer 13

\[ x \cdot \left(\frac{(x-7)}{x} +2\right)= x -3 \]

on the left side, "distribute the x". that means multiply all the terms inside the parens by x
(don't forget the 2)

and notice that you will get x/x so you can cancel them.