SOLUTION: solve for y in 6x-5y=-5. determine if the line is parallel to y=6/5x+5/9

Question 115771: solve for y in 6x-5y=-5. determine if the line is parallel to y=6/5x+5/9
Found 2 solutions by edjones, bucky:
Answer by edjones(8007) (Show Source): You can put this solution on YOUR website!
y=mx+b m=slope
y=6x/5+5/9 m=6/5
If the next equation has the same slope they are parallel.
6x-5y=-5
6x-6x-5y=-6x-5
-5y=-6x-5
-5y/-5=-6x/-5 -5/-5
y=6x/5+1 m=6/5
They are parallel.
.
Ed

Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
One step at a time. First, solve the equation for y:
.
Begin by getting rid of the term 6x on the left side so that you just have the term containing
the y alone on the left side. Do this by subtracting 6x from both sides to get:
.

.
You are trying to solve for +y so at this point you may want to change the sign of -5y to +5y.
You can do that by multiplying both sides of the equation (all terms) by -1 to change the
equation to:
.

.
Finally, solve for y by dividing both sides of this equation by 5 ... the multiplier of
y to get:
.

.
Notice that the equation we now have is:
.

.
and this is in the slope-intercept form:
.

.
in which m, the multiplier of x, is the slope of the graph and b is the value on the y-axis
where the graph crosses the y-axis. By comparing your equation with the slope intercept form
you can see that the graph of your equation has a slope of and it crosses the
y-axis at the value of +1 on the y-axis.
.
Now look at the other equation you were given ... namely:
.

.
Comparing this equation to the slope intercept form you will see that it also has a slope
of but its graph crosses the y-axis at .
.
Now recognize that two graphs having the same slope but different crossing points on the
y-axis are parallel lines that are always separated in vertical distance by an amount equal
to the difference on the y-axis equal to the crossing points. The graph of the two equations
shows this. The "red" graph is the graph of the equation and the green
graph is the graph of the equation
.

.
Hope this helps you to understand the problem and shows you that lines given by the two equations
are actually parallel.
.

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