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Find the solution to this system using the substitution method. 2x -5y = -1 and x -3y = -3
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2x-5y = -1 2x = -1 + 5y x = \(\large{\frac{-1+5y}{2}}\)
put this value in the second eqn : \[\large{x-3y=-3}\] \[\large{\frac{-1+5y}{2}-3y=-3}\]
form the second equation we have x=3y -3 putting it in the first we have 2(3y-3)-5y=-1 or 6y -6-5y =-1 or y=6-1 or y=5 hence x=3(5) -3=12 x=12 and y=5
\[\large{\frac{-1+5y-6y}{2}=-3}\] \[\large{-1-y=-6}\] \[\large{-1+6=y}\] \[\large{5=y}\]
its too sticky and icky...mmmmmm
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similarly find x x- 3(5) = -3 x-15=-3 x = -3+15
got it @rannsan ?
I think so
Thank you!!
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