Find conditions on a amd b such that the system of linear equations has (a) no solutions, b) unique solution c) infinite many solutions
x+2y= 3 and ax + by = -9
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I still didn't understand the question dear
do you want to know what will a = ? and b=2? to satisfy the given conditions?
b = ?*
lool, it's like for example, for a), if the two linear equations are parallel to each other, than there's no solutions at all, and the questions is to find the a and b, which value should we take in order to have *no solutions* lol = )
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alright then first write the equation in slope intercept form :
y = -x/2 + 3
in this case m = -1/2 so (a) should be -1/2 , parallel = means same slope.
Perpendicular = means the reciprocal of the slope.
hmmm, not sure about (b) though >_< lol
all right lol thanks = )
np :) sorry for the mess >_<
it's all good ^_^
yes the slope is -1/2 and to have no solution they need to have same slope
however, a does not represent the slope
put 2nd equation in slope-intercept form
y = (-a/b)x -9/b
-a/b is slope
-a/b = -1/2
a/b = 1/2
a=1 and b=2
a=2, b=4 ...
infinite solutions means they are the same line
so same slope of -1/2 again
but now the y_intercepts must be the same
y = -x/2 +3/2 , y = (-a/b)x -9/b
-9/b = 3/2
3b = -18
b = -6
a/b = 1/2
a/-6 = 1/2
a = -3
a and b can be any other numbers except those used in part a and c
if 2 lines are distinct and not parallel then they must intersect at some point
how abt: x+2y= 3----> y=(3-x)/2
\[x \neq3 cause 3-3/2=0\]
let two equations be a1x+b1y=c1 & a2x+b2y=c2.
they have uniq solutions if \[a1/a2\neq b1/b2\]has no solutions if \[a1/a2=b1/b2\neq c1/c2\]has infinite solutions if \[a1/a2=b1/b2=c1/c2\]
so from your given equations for infinite solutions 1/a=2/b=3/-9
for no solutions \[1/a =2/b \neq3/-9\] \[b/a=2 \] \[a \neq -3 , b\neq -6\]
for unique solutions \[a/b \neq 1/2\]