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Find k so that the three equations 3x-y=2, 2x+8=3y, and y=kx have a common solution

1 Answer

3 votes
2x+8=3kx (3k-2)x+8=0
3x-kx=2. (3-k)x-2=0
3k-2)x+10=3-k)x
4k-5=10/x
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3x-y-2=2x+8-3y
X+2y-10=0
X+2kx-10=0
K=(10-x)/2x
User Sbml
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