What is the solution to the system? Line 1 x y –1 9 3 –3 Line 2 x y –1 –1 2 5 ( , )
@hartnn
@zepdrix
@bibby
@mathmath333
These are lines? Just straight lines? So we can use our handy approaches to come up with equations for each line.
yes
Given the data for the first line, \(\Large\rm (-1,9)\) and \(\Large\rm (3,-3)\), Our slope formula gives us:\[\Large\rm m=\frac{y_2-y_1}{x_2-x_1}=\frac{-3-9}{3--1}=\frac{-12}{4}=-3\]So the slope of that first line is -3.\[\Large\rm y=mx+b\]\[\Large\rm y=-3x+b\]To find the y-intercept, let's plug one of the points in and solve for b.\[\Large\rm 9=-3(-1)+b\]Solving for b gives us b=5. So the equation of our first line is, \(\Large\rm y=-3x+5\).
Do you understand the process? :o Think you can apply that to the second set of data?
Woops :) Not b=5, I shoulda said b=6. hehe was moving too fast.
i know how to fid the slope but i dont understand how to find the y intercept
find*
What's the slope of the second line? Take a minute to calculate it :3 imma go grab a drink real quick
the slope is 6/3 for line 2
Mmmm ok very good.\[\Large\rm y=2x+b\]
So we have a line,\[\Large\rm \color{orangered}{y}=2\color{royalblue}{x}+b\]And to find the y-intercept, we need to plug in a point that the line passes through. Let's use this point which was given to us: \(\Large\rm (\color{royalblue}{x},\color{orangered}{y})=(\color{royalblue}{2},\color{orangered}{5})\)
Do you see WHERE we'll plug the values in? I tried to color code them.
5=2(2)
5x=2(2)
We're not plugging anything in for the b, so don't let that part just disappear.\[\Large\rm \color{orangered}{y}=2\color{royalblue}{x}+b\]\[\Large\rm \color{orangered}{5}=2\color{royalblue}{(2)}+b\]
ok
5=4+b
mhm, solve for b.
but how thats what im confused about?
|dw:1420624244569:dw|
To isolate the b, we're just applying a simple algebra step. The 4 is being ADDED to the b, We want to apply the inverse operation to get the b alone. So we'll SUBTRACT 4 from each side, yah?
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