Chris M. answered • 11/30/17
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There's a couple of ways to solve this problem short of graphing.
Since the first equation already has y isolated we can substitute it into the second equation:
6x=2(3x-5)+10
6x=6x-10+10
6x=6x
x=x
1=1
0=0
Since since this is true for all possible values of x we can say that there are an infinite number of solutions.
Note if the first equation had been y=3x-6, then when we plugged it into the second equation we would have gotten
6x=2(3x-6)+10
6x=6x-12+10
6x=6x-2
0=-2
This isn't true (for any value of x). So if this were the case then there would be no solutions to the system of equations.
The last case would be a single solution. For example let the first equation be y=2x-4. Substituting this back into the second equation we'd have
6x=2(2x-4)+10
6x=4x-8+10
6x=4x+2
2x=2
x=1
Substituting x=1 back into the first equation would yield y=2(1)-4=-2
So the solution would be (1,-2). Remember you can always validate that the solution is correct by plugging it into both equations.
Another method to quickly assess the number of solutions to a system of linear equations is to put both equations into slope-intercept form. In your original case the first equation is already there, so lets manipulate the second equation:
6x=2y+10
6x-10=2y
3x-5=y
But this is the same as the first equation, y=3x-5. Since they are the SAME line they will have infinite solutions.
In the second scenario I setup (y=3x-6) note that with both equations in slope-intercept form they both have the same slope (3) but different y intercepts. So these are two parallel (same slope) lines that never intersect, hence they have no solutions.
In the third case (a single solution) the two lines will have a different slope.
Good Luck.
Cheers
-Chris
