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So, I'm trying to solve this: \(\begin{pmatrix} 4\\ 3 \end{pmatrix} (2x-1)=-12\)

Do I multiply both sides by the recriprical of 4/3 or do I distribute first?

 

Thanks!

Guest Oct 13, 2018
 #1
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Multiply both sides by 3/4 to get 2x-1=-9. Add 1 from both sides to get 2x=-8. Dividing both sides by 2 you get x as -4.

Guest Oct 13, 2018
 #2
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Its [4!/3!(4-3)!](2x-1)=-12 <=>

<=> 24/6(2x-1)=-12 <=>

<=> 4(2x-1)=-12 <=>

<=> 8x-4=-12 <=>

<=> 8x=-8 so x=-1

I hope to helped you!

Dimitristhym  Oct 13, 2018
 #3
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Did you mean that as 4/3 or 4 choose 3

Guest Oct 13, 2018
 #4
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The 1st guest who answered was correct, the solution is \(x=-4\) (I checked). Thanks for the quick help!

Guest Oct 13, 2018
 #5
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Oh im sorry i understand 4 choose 3 because you write it like this :( Im sorry !

Dimitristhym  Oct 13, 2018
edited by Dimitristhym  Oct 13, 2018
 #6
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Thats OK, for whatever reason LaTeX does NOT have an option for fractions, so I had to make-do with what looked like a fraction :)

Guest Oct 13, 2018
 #7
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Ok I don't know this!

Dimitristhym  Oct 13, 2018
 #8
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No hard feelings dude smiley

Guest Oct 13, 2018
 #9
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At least you learnt it now, but good explanation still

NerdyKid  Oct 13, 2018

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