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 A.0 B.exactly 1 C.exactly 2 D.infinitely many The coordinates below represent two linear equations. How many solutions does this system of equations have? Line 1 x y –4 3 –2 4 Line 2 x y 0 8 2 6 A.0 B.exactly 1 C.exactly 2 D.infinitely many i rlly dont get this so could someone help? will fan and medal
Find the equations of both lines, then compare their slopes and y-intercepts. I). Infinitely many solutions if both lines have the same slope and the same intercept(s). II). No solution if they have the same slope but different intercepts. III). One solution if they have different slopes.
@SolomonZelman can you help?
can you tell me, what is the slope of the first line? { knowing that it goes through points (-2,7) and (0,6) }
for the first one i got y=1/2x+5 but im not sure if its right
this is the equation of the first line, if so, then you got only the slope correctly.
the y-intercept is (0,6) so your first line should say not +5, but ...?
-5?
y=mx+b, where m is the slope (which you correctly found to be 1/2) and b is the y-intercept, which in this case is 6, since when x is 0, the y is 6. [this is given to us, by a point (0,6) ]
so the equation for the first line, should say?
y=1/2x+6?
yes.
now, the second line is going through points (3,0) and (0,-3)
can you find the slope?
y=x+-3?
+- ?
the y-intercept is when x=0... we have a point (0,-3) {this is when x=0} so the equation of your line is: y=x..... (plus or minus 3? )
plus 3?
-3.
because if y-intercept is -3, then filling in the y=mx+b we get y=x (because we know that the slope is 1, as you have already said) and then, y=x-3 (because y-intercept, the b, is -3)
okay, so your 2 equations are: y=1/2 x+6 y= x - 3
can you solve for x?
or for y
I would: ~ multiply the 1st equation times -1. ~ add the (new version of the) first equation and the second equation. ~ add 9 to both sides ~ multiply the 1st equation times -1. ~ multiply both sides times 2. this would give you the x.
( first equation: y=1/2 x+6 second equation: y= x - 3 )
so in the first equation i multiply -1 by 1/2 or 6?
you multiply each thing in the first equation times -1 (for the first step)
~~~~~~~~~~~~~~ y times -1, 1/2 x times -1, +6 times -1. ~~~~~~~~~~~~~~
im confused with solving for x or y
\(\large\color{black}{ b<c }\) multiply this: y=1/2 x+6 ,times -1. you get: -y=-1/2x-6 so now your system is: -y= -1/2x-6 y = x - 3
b<c shuldn't be there, ignore that
so for the second equation it would be -y=-x+3?
you are only multiplying the first equation times -1. the second equation, leave it the way it is.
oh ok
-y = -1/2 x - 6 (<< is the result of multiplying times -1) y = x - 3 (<< we didn't change it) add the equations, -y = -1/2 x - 6 y = x - 3 ---------------
y=-1/3x-9? idk im not good with adding equations like these
when you add them, the Ys cancel
y+(-y)=? -1/2x + x =? -6 + (-3) = ?
y+(-y)=0? -1/2x+x= -x/2? and -6+(-3)=-9
yes.
besides that -1/2 x + x= 1/2x (not -1/2x)
so the addition gives us: 0=1/2 x - 9.
i have to go cause i have a basketball game. I'll be back tomorrow. thanks for the help though
okay, sure. yw
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