Tuesday, Nov. 22nd

pdf icon ws-equationsolving-fractions-2.pdf
doc icon ws-mathcounts-20161122-OPLET.docx
In class today students worked on equation solving and mathcounts problems. I notice that generally they are comfortable solving multi-step equations like (equation 1)

72 + 4(5 - 3x) - (x + 3) - (4 - x) - 2(3 - 5x)

But many students were not sure how to solve (equation 2):

dfrac{3x - 4}{5} = dfrac{2x - 5}{2}

So we did this example in class. I showed students that division can be distributed just like multiplication so we can turn the equation into

dfrac{3}{5}x - dfrac{4}{5} = x - dfrac{5}{2}

and then solve it. Or, we can notice that if two fractions are equal, then their cross products are equal. For example, dfrac{color{red} 3}{color{blue}4} = dfrac{color{blue} 9}{color{red}12} it is also the case that {color{red} 3 cdot 12} = {color{blue} 4 cdot 9}. This is true for all equivalent fractions and so another way to approach solving equation 2 is to find the cross products and set them equal

2(3x - 4) = 5(2x - 5)

And solve from there

6x - 8 = 10x - 25

Subtract 6x from both sides and add 25 to both sides to get

17 = 4x

And finally,

dfrac{17}{4} = x

Assignment [handouts attached]

  • Equations worksheet
  • Mathcounts worksheet


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