Question 174321
First, let's solve the second equation {{{2y+3x=4}}} for y



{{{2y+3x=4}}} Start with the second equation.



{{{2y=4-3x}}} Subtract {{{3x}}} from both sides.



{{{2y=-3x+4}}} Rearrange the terms.



{{{y=(-3x+4)/(2)}}} Divide both sides by {{{2}}} to isolate y.



{{{y=((-3)/(2))x+(4)/(2)}}} Break up the fraction.



{{{y=-(3/2)x+2}}} Reduce.



Notice how this equation is NOT equivalent to any other equation (so far at least).


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Now, let's solve the third equation {{{(y-5)=(3/2)(x-2)}}} for y



{{{y-5=(3/2)(x-2)}}} Start with the third equation



{{{y-5=(3/2)x+(3/2)(-2)}}} Distribute {{{3/2}}}



{{{y-5=(3/2)x-3}}} Multiply {{{3/2}}} and {{{-2}}} to get {{{-3}}}



{{{y=(3/2)x-3+5}}} Add 5 to  both sides to isolate y



{{{y=(3/2)x+2}}} Combine like terms {{{-3}}} and {{{5}}} to get {{{2}}} 




So this equation is identical to the first equation in part a)




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Now, let's solve the fourth equation {{{2y-3x=4}}} for y 



{{{2y-3x=4}}} Start with the fourth equation.



{{{2y=4+3x}}} Add {{{3x}}} to both sides.



{{{2y=3x+4}}} Rearrange the terms.



{{{y=(3x+4)/(2)}}} Divide both sides by {{{2}}} to isolate y.



{{{y=((3)/(2))x+(4)/(2)}}} Break up the fraction.



{{{y=(3/2)x+2}}} Reduce.




So this equation is identical to the first equation in part a) (and in part c)


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So the only equation that is NOT identical to the equation in part a) is the equation in part b)


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Answer:


The only different equation is {{{2y+3x=4}}}