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Spartan_Of_Ares
can i get help with some algebra 1?
\[\frac{ x-7 }{ x }+2=\frac{ x-3 }{ }\]
is there something missing on the right side? Is anything below x-3 ?
x−7 x−3 ---- +2= ---- x x
my first thought is to multiply both sides by x like this \[ x \cdot \left(\frac{(x-7)}{x} +2\right)= x \cdot \frac{(x-3)}{x} \]
ok \[\frac{ x(x-7) }{ x^2 } = (\frac{ x(x-3) }{ x^2 })\]
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
but how would i multiply the equation?
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
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: \[ \frac{\cancel{x}(x-3)}{\cancel{x}} = (x-3) \]
\[ x \cdot \left(\frac{(x-7)}{x} +2\right)= x -3 \] 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.