You are not logged in.
Pages: 1
hey can any body help my switch these equation to an slope intercept form
x + 2y = -4
x + y = 3
x + 3y = -6
Slope intercept form is the form y=mx+b.
This means that in each equation, we must solve for y in terms of x and a constant, b.
To solve for y, we need to have it alone on one side. In order to do that, we must move x to the other side. In all three equations, x is being added to y on the left side, so in order to move x over, we must subtract x from the left side. But whatever we do to the left side we must do to the right side. So after one step, you should have
x + 2y - x= -4 - x
x + y - x= 3 -x
x + 3y - x= -6 - x
which is the same as
2y= -4 - x
y= 3 -x
3y= -6 - x
Now, the second equation is in slope-intercept form, but the first and third are not. We must apply the same algebraic strategy that we did in the first step to get y alone. For example, in the first equation, y is being multiplied by 2. So in order to get rid of the two, we must divide the y by 2. But we must also divide the WHOLE right side by two, not just the -4. Once you do that for the first and third equation, you should end up with
y= -2 - x/2
y= 3 - x
y= -2 -x/3
Does that help?
Offline
Pages: 1