Question #e44f3

1 Answer
Jul 18, 2017

Make y the subject and then substitute.

Explanation:

  1. Make y the subject.
    #7y=-6x-6#
    #y=(-6/7)x-(6/7)#

  2. Using #y=mx+c# we can see that the coefficient of #x# must remain the same for the gradient to be the same and therefore the lines to be parallel. We only need to calculate what the integer value will be now. To do this we substitute and we will call the integer value #c#.
    #y=(-6/7)x+c#
    #4=(-6/7)3+c#
    #4=(-18/7)+c#
    #4+(18/7)=c#
    #c=(28/7)+(18/7)#
    #c=46/7#

  3. We then put the value for #c# back in the equation to get our answer which would be:
    #y=(-6/7)x+46/7#

Hope this has helped.