The coordinates below represent two linear equations. How many solutions does this system of equations have?
Line 1 x y -2 7 0 6
Line 2 x y 3 0 0 -3
i think there's 2 solution
two lines will never have 2 solutions. They can be coincident (lay on top of each other), and that would mean \(\large\color{slate}{ \infty}\) of solutions. They can be parallel, and have no solutions because they won't intersect. They can have different slopes, and intersect at 1 point.
lets find the slope of the first line, can you do that?
umm they tell me how to do it like this-3-(-6) -------- -3-(-4)
like: \(\large\color{black}{ \frac{\LARGE y_1-y_2 }{\LARGE x_1-x_2 } }\)
in your case: your points are not \(\large\color{black}{ (x_1,y_1) }\) and \(\large\color{black}{ (x_2,y_2) }\) but, \(\large\color{black}{ (-2,7) }\) and \(\large\color{black}{ (0,6) }\)
So your slope would be equivalent to?
ya
can you calculate the slope? I will plug in the point for you: \(\large\color{black}{ (7-6)/(-2-0) }\) and you give me the answer (for the slope of the first line)
ya and i got -1/2
yes, correct.
and -3/-3
3/3
And you are given the y-intercept to be 0,6. So the first line would be; \(\large\color{black}{ y=\frac{\Large 1}{\Large 2}x+6}\)
The slope of the second line is ? (you wrote it correctly, but didn't simplify it completely)
oh ya ur right it is infinity
3/3 is equal to what?
a/a=1 4/4=1 (-7)/(-7)=1 and any number divided by itself is 1. so 3/3 = ?
I didn't say there are infinity of soltions here, btw, I said that it is a possible case when dealing with 2 lines
3/3 is equal to
1
sorry took so long
yes, so the slope of your second line is 1.
the y-intercept of your second line is -3, so, you can write the equation for a second line: \(\large\color{black}{ y=x-3}\)
A solution means point(s) where the lines intersect: line 1, \(\large\color{black}{ y=\frac{\Large 1}{\Large 2}x+6}\) line 2, \(\large\color{black}{ y=x-3}\) \(\large\color{black}{ \frac{\Large 1}{\Large 2}x+6=x-3}\)
solve for x, can you do that?
uymm
no.
tip to solve for x. \(\large\color{black}{ \frac{\Large 1}{\Large 2}x+6\color{red}{-\frac{\Large 1}{\Large 2}x+3}=x-3\color{red}{-\frac{\Large 1}{\Large 2}x+3}}\)
simplify on both sides.
oh.
yes, so what do you get on both sides after simplfying?
|dw:1420511332445:dw|
\(\large\color{black}{ \frac{\Large 1}{\Large 2}x+6\color{red}{-\frac{\Large 1}{\Large 2}x+3}=x-3\color{red}{-\frac{\Large 1}{\Large 2}x+3}}\) \(\large\color{black}{ \cancel{\frac{\Large 1}{\Large 2}x~}+6\color{red}{\cancel{-\frac{\Large 1}{\Large 2}x}+3}=x-3\color{red}{-\frac{\Large 1}{\Large 2}x+3}}\) \(\large\color{black}{ 6\color{red}{+3}=x-3\color{red}{-\frac{\Large 1}{\Large 2}x+3}}\)
And from, \(\large\color{black}{ 6\color{red}{+3}=x-3\color{red}{-\frac{\Large 1}{\Large 2}x+3}}\) \(\large\color{black}{ 6\color{red}{+3}=x\cancel{-3}\color{red}{-\frac{\Large 1}{\Large 2}x\cancel{+3}}}\) \(\large\color{black}{ 6\color{red}{+3}=x\color{red}{-\frac{\Large 1}{\Large 2}x}}\)
can you simplify on both sides, now?
9 = x -1/2
you got the 9 correctly
forgot the x
you can still simplify the right side, after you have: \(\large\color{black}{ 9=x\color{red}{-\frac{\Large 1}{\Large 2}x}}\)
-.- aww man.....
tell me, 1- (1/2) is how much?
hmm, les see 1/1?
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