can i get help with some algebra 1?

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can i get help with some algebra 1?

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\[\frac{ x-7 }{ x }+2=\frac{ x-3 }{ }\]
  • phi
is there something missing on the right side? Is anything below x-3 ?
sorry x

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x−7 x−3 ---- +2= ---- x x
  • phi
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 \[\frac{ x(x-7) }{ x^2 } = (\frac{ x(x-3) }{ x^2 })\]
  • phi
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?
  • phi
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
x^2-3x over x?
  • phi
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: \[ \frac{\cancel{x}(x-3)}{\cancel{x}} = (x-3) \]
  • phi
\[ x \cdot \left(\frac{(x-7)}{x} +2\right)= x -3 \] 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.

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