anonymous
  • anonymous
Can someone help with these 2 problems please!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
UnkleRhaukus
  • UnkleRhaukus
first one \[\frac{a-1}{7}+\frac{a-2}{4}=\frac{6}{7}\]
UnkleRhaukus
  • UnkleRhaukus
get rid of the denominators ie multiply by each term by 28 \[4\times({a-1})+7\times({a-2})=4\times{6}\]

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UnkleRhaukus
  • UnkleRhaukus
expand brackets\[4a-4+7a-14=24\]and simplify\[3a-18=24\]add eighteen to both sides\[3a=6\]divide by three\[a=2\]
UnkleRhaukus
  • UnkleRhaukus
i made a mistake
UnkleRhaukus
  • UnkleRhaukus
\[4a+7a=11a≠3a\]
anonymous
  • anonymous
I think it works out to 11a=42, so a=42/11
UnkleRhaukus
  • UnkleRhaukus
i actually made two mistakes AnimalAin is correct \[24+18=42≠6\]\[\cdots \]a\[11a-18=24\]\[11a=42\]\[a=42/11\]
UnkleRhaukus
  • UnkleRhaukus
second one (ill try to be more careful) \[\frac{3r-3}{r^2+6r-7}+\frac{3}{r+7}=\frac{3}{r-1}\] first step is to factorize the denominator of the first term
UnkleRhaukus
  • UnkleRhaukus
\[{r^2+6r-7}=(r+...)(r-...)\]
UnkleRhaukus
  • UnkleRhaukus
so we need two numbers that have a sum of 6 and a product of -7
anonymous
  • anonymous
So it would be 7 and -1
UnkleRhaukus
  • UnkleRhaukus
correct
UnkleRhaukus
  • UnkleRhaukus
\[\frac{3r-3}{(r+7)(r-1)}+\frac{3}{r+7}=\frac{3}{r-1}\] now we have to get rig of the denominators to do this multiply all terms by \[(r+7)(r-1)\]
anonymous
  • anonymous
Would the bottom ones cancel out ?

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