How do you solve #4x-12y=5# and #-x+3y=-1# using substitution?

1 Answer
Shwetank Mauria
Apr 27, 2016

No soluion.

Explanation:

In substitution method we get the value of one variable in term of other from one equation and then put it in another equation.

Here in second equation, transposing #x# term to RHS and constant term to LHS gives us #3y+1=x# or #x=3y+1#. Now putting this in first equation we get

#4(3y+1)-12y=5# or

#12y+4-12y=5# or #4=5#, which is not true and hence we have no solution.

Additional fact - This means that if we draw two lines on a Cartesian plane, they lead too two parallel lines and hence no solution. Had it been #0=0#, it would have been two coincident lines and hence infinite solutions.

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