If the equation #k(10x-5)=2(3+x)-7# has infinitely many solutions and #k# is a constant, what is the value of #k#?

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Answer 1 · 2018-05-13T17:43:59

#k=1/5#

Given: #k(10x-5)=2(3+x)-7#

For an infinite count of solutions then their graphs are parallel and coincidental over the whole domain/range.

Divide both sides by #(10x-5)#

#k=(2x-1)/(10x-5)#

#k=(2x-1)/(5(2x-1))#

#k= 1/5#

Answer 2 · 2018-05-13T17:46:46

#k=1/5#

#k(10x-5)=2(3+x)-7#
#k(10x-5)=6+2x-7#
#k(10x-5)=2x-1#
#k=(2x-1)/(10x-5)#
#k=(1*cancel(2x-1))/(5*cancel(2x-1))#
#k=1/5#
\0/ here's our answer!