## Spartan_Of_Ares 2 years ago can i get help with some algebra 1?

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1. Spartan_Of_Ares

$\frac{ x-7 }{ x }+2=\frac{ x-3 }{ }$

2. phi

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

3. Spartan_Of_Ares

sorry x

4. Spartan_Of_Ares

x−7 x−3 ---- +2= ---- x x

5. phi

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

6. Spartan_Of_Ares

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

7. 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

8. Spartan_Of_Ares

but how would i multiply the equation?

9. 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

10. Spartan_Of_Ares

x^2-3x over x?

11. Spartan_Of_Ares

@phi

12. 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: $\frac{\cancel{x}(x-3)}{\cancel{x}} = (x-3)$

13. 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.