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Spartan_Of_Ares

  • 3 years ago

can i get help with some algebra 1?

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  1. Spartan_Of_Ares
    • 3 years ago
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    \[\frac{ x-7 }{ x }+2=\frac{ x-3 }{ }\]

  2. phi
    • 3 years ago
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    is there something missing on the right side? Is anything below x-3 ?

  3. Spartan_Of_Ares
    • 3 years ago
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    sorry x

  4. Spartan_Of_Ares
    • 3 years ago
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    x−7 x−3 ---- +2= ---- x x

  5. phi
    • 3 years ago
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    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
    • 3 years ago
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    ok \[\frac{ x(x-7) }{ x^2 } = (\frac{ x(x-3) }{ x^2 })\]

  7. phi
    • 3 years ago
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    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
    • 3 years ago
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    but how would i multiply the equation?

  9. phi
    • 3 years ago
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    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
    • 3 years ago
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    x^2-3x over x?

  11. Spartan_Of_Ares
    • 3 years ago
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    @phi

  12. phi
    • 3 years ago
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    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
    • 3 years ago
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    \[ x \cdot \left(\frac{(x-7)}{x} +2\right)= x -3 \] 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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