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OCW Scholar - Single Variable Calculus
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For some reason I simply cannot find the solution to Problem Set 1 Question 1A-5 part a). Can someone explain the solution for me in steps. Part b) is no problem but I just don't see how y=(x-1)/(2x+3)gets to x=(3y+1)/(1-2y). Probably an easy answer but I AM STUCK !:'( Thnx !!
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Substitute xes for ys in the original equation. So you'll have x=(y-1)/(2y+3). Multiply both sides by 2x+3. 2yx+3x=y-1. Rearrange so all y terms are on one side. y-2yx=3x+1. Factor out a y on the left. (y)(1-2x)=3x+1. Divide and you get y=3x+1/(1-2x) .
\[y=(x-1)/(2x+3)\]\[y(2x+3)=(x-1)\]\[2yx+3y=(x-1)\]\[x-2yx=3y-1\]\[3y+1=x-2yx\]\[3y+1=x(1-2y)\]\[x=(3y+1)/(1-2y)\]
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