Find out the value of x and y #xy=6# #2x-y=3# ???

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Answer 1 · 2017-10-12T12:58:33

If would use substitution. If we rearrange the second equation for #y#, we get;

#2x - 3 = y#

Accordingly:

#x(2x -3) = 6#

#2x^2 - 3x - 6 =0#

#x = (-(-3) +- sqrt(3^2 - 4 * 2 * -6))/(2 * 2)#

#x= (3 +-sqrt(57))/4#

We can now solve for #y#.

#y = 2((3 + sqrt(57))/4) - 3 #

#y = (3 + sqrt(57) - 6)/2#

#y = (sqrt(57) - 3)/2#

Now for the other solution, we get:

#y = (-sqrt(57) - 3)/2#

So our solutions are #((3 - sqrt(57))/4, (-sqrt(57) - 3)/2)# and #((3 + sqrt(57))/4, (sqrt(57) - 3)/2)#

Hopefully this helps!