Find the variable: Find 'a' so ax+y=-5, -1/3x-2y=-1 has no sol'n.

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Sk.yyy

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ax+y=-5
-1/3x-2y=-1


If the system of equation above has no solution, and 'a' is a constant, what is the value of 'a'?
 
Sk.yyy said:
ax+y=-5
-1/3x-2y=-1


If the system of equation above has no solution, and 'a' is a constant, what is the value of 'a'?
Click to expand...

What methods have you been taught to solve linear simultaneous equations? Have you been taught "matrix method"?

What are your thoughts?

Please share your work with us ...even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/announcement.php?f=33
 
Sk.yyy said:
ax+y=-5
-1/3x-2y=-1
If the system of equation above has no solution, and 'a' is a constant, what is the value of 'a'?
Click to expand...
Can you solve this determinate?
\(\displaystyle \left| {\begin{array}{*{20}{c}}a&1\\{\tfrac{{ - 1}}{3}}&{ - 2}\end{array}} \right| = 0\)
 
Sk.yyy said:
ax+y=-5
-1/3x-2y=-1


If the system of equation above has no solution, and 'a' is a constant, what is the value of 'a'?
Click to expand...

Those are equations of two straight lines. Their intersection point will be the solution of the system.

Under what condition, two straight lines will not intersect each other?
 
Sk.yyy said:
ax+y=-5
-1/3x-2y=-1

If the system of equation above has no solution, and 'a' is a constant, what is the value of 'a'?
Click to expand...
What does it mean, graphically, for a system of two straight lines to "have a solution"? What then does it mean, graphically, for a system of two straight lines not to have a solution? What must be true of the two lines?

If you solve each of these equations for "y=", what values (in "y = mx + b") must match for the lines to have no solution? Setting these parts of the "y=" equations equal, what must be the value of "a"?

If you get stuck, please reply with a clear listing of your answers to my questions above. Thank you! ;)
 
As some of colleagues have mentioned, you can solve this task by looking at those equations like equations of lines, which they are actually. You would write them as: y=−ax−5\displaystyle y=-ax-5 and y=−16x+12\displaystyle y=-\dfrac{1}{6}x+\dfrac{1}{2} . This system won't have any solution when those two lines are parallel( they do not intersect). Two lines that are parallel have something equal.... you can use that fact. You just have to set the equation from what I've told you and you will be able to get a value of 'a'.
 
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