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anonymous

  • 5 years ago

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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  1. anonymous
    • 5 years ago
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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?

  2. anonymous
    • 5 years ago
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    b = ?*

  3. anonymous
    • 5 years ago
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    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 = )

  4. anonymous
    • 5 years ago
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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.

  5. anonymous
    • 5 years ago
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    hmmm, not sure about (b) though >_< lol

  6. anonymous
    • 5 years ago
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    all right lol thanks = )

  7. anonymous
    • 5 years ago
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    np :) sorry for the mess >_<

  8. anonymous
    • 5 years ago
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    it's all good ^_^

  9. dumbcow
    • 5 years ago
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    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 or a=2, b=4 ... part c) 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 and a/b = 1/2 a/-6 = 1/2 a = -3 part b) 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 so a=3 b=5

  10. angela210793
    • 5 years ago
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    how abt: x+2y= 3----> y=(3-x)/2 \[x \neq3 cause 3-3/2=0\]

  11. anonymous
    • 5 years ago
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    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 \[a1/a2=b1/b2=c1/c2\]

  12. anonymous
    • 5 years ago
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    so from your given equations for infinite solutions 1/a=2/b=3/-9 a=-3, b=-6. for no solutions \[1/a =2/b \neq3/-9\] \[b/a=2 \] \[a \neq -3 , b\neq -6\] for unique solutions \[a/b \neq 1/2\]

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