anonymous
  • anonymous
3x+y=1/3 2x-3y=8/3 solve for x & y
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Do you know how to use the substitution method?
anonymous
  • anonymous
no
anonymous
  • anonymous
OK use the first equation and get y by itself... This will be the definition for y that we will substitute into the second equation. I'll show you.

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anonymous
  • anonymous
ok
anonymous
  • anonymous
\[y=-3x +\frac{1}{3}\]
anonymous
  • anonymous
Substituting... \[2x + (-3)\big(-3x+\frac{1}{3}\big)=\frac{8}{3}\] Do you see where I substituted the right half of equation #1 for the "y" term in equation 2????
anonymous
  • anonymous
yes got it so far
anonymous
  • anonymous
OK can you take it from here? Solve Eqn #2 for x. Then take that number and put in for the "x" in equation #1 to get the number value for "y". Post your answer and I'll check.
anonymous
  • anonymous
y=-2/3, x=1/3 but not sure if there was an easier way to get 1/3
anonymous
  • anonymous
let me show you.
anonymous
  • anonymous
OK (that's the answer I got).
anonymous
  • anonymous
2x+9x+(-1)=8/3 11x+(-1)=8/3 11x=1 8/3 11x=3 2/3 which then is 3 2/3 divided by 11.
anonymous
  • anonymous
11/3 divided by 11
anonymous
  • anonymous
=1/3
anonymous
  • anonymous
Looking....
anonymous
  • anonymous
Right... There is no reason to use a mixed fraction on the third line. It looks like you went back to an improper fraction when you cleared the 11 from the left anyway. I always recommended that my students left fractions as improper fractions - much easier to divide by multiplying the inverse of that fraction on both sides.
anonymous
  • anonymous
so should I have made -1 to be -3/3 ?
anonymous
  • anonymous
Right - I try to keep things as fractions. You'll get pretty good at it and won't need to write the intermediate step. Eg \[11x + (-1)=8/3 \Rightarrow11x=8/3+1\Rightarrow11x=\frac{(8+3)}{3}\] etc.... You'll get good enough to skip steps mentally (careful!)
anonymous
  • anonymous
\[\frac{1}{11}*\frac{11x}{1}=\frac{11}{3}*\frac{1}{11}\]
anonymous
  • anonymous
thank you very much! :)
anonymous
  • anonymous
No problem. Good luck!

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