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Mathematics 10 Online
OpenStudy (anonymous):

Give one value of t for which the system of equations x +3y=ty 3x+y =tx has infinitely many solutions. Please provide me with a worked example ...kind regards.

OpenStudy (anonymous):

these are the equations of two lines. They will have infinitely many solutions when they are in fact, the same line. Your task is to find the value of t which makes the two equations identical.

OpenStudy (anonymous):

(x-3y)/y=(3x+y)/x x(x-3y)=y(3x+y) x^2-3xy=3xy+y^2 x^2-y^2-6xy .

OpenStudy (turingtest):

this problem can be done pretty fast with cramer's rule if you know it

OpenStudy (turingtest):

@WONDEMU I think you have misunderstood the problme @unseenoceans do you know Cramer's rule?

OpenStudy (turingtest):

problem*

OpenStudy (anonymous):

is that ?

OpenStudy (anonymous):

It can be done even faster by subtraction and comparison ;)

OpenStudy (anonymous):

I'll have to revise cramer's rule. Thanks for pointing me in that direction. I was lost for a bit. What is you method jemirray? What confused me is the terms tx and ty. I'm used to numbers only.

OpenStudy (anonymous):

If you subtract, that gives you \[(3-t)x + y = 0\] \[x + (3-t)y = 0\] and you're just trying to make these two equations look the same...

OpenStudy (anonymous):

your lines are \[y=\frac{1}{t-3}x\]and,\[y=(t-3)x\]For these to be equations of the same line, they must have the same slope and go through an identical point (actually infinitely many identical points). You're only asked to find one value of t which satisfies this, so pick an arbitrary point for these lines to go through, such as (1,1). Now, this implies,\[\frac{1}{t-3}=t-3\]or,\[t^2-6t+8=0\]Solve the quadratic to get a value of t.

OpenStudy (anonymous):

Notice that we could have picked any other point for the lines to go through and it wouldn't change the quadratic we get. So there are only two possible values of t for this system to have infinitely many solutions. t=2 and t=4

OpenStudy (anonymous):

Wow, openstudy, is amazing. Didn't expect a flood of replies. Thanks guys.

OpenStudy (turingtest):

just to finish my way using Cramer's rule (though @Jemurray3 's way is simpler), what you would do is set either \(D_x\) or \(D_y\) equal to zero, then solve for t.

OpenStudy (anonymous):

eseidl . In the 1st step , do you rewrite the system in terms of y? Correct?

OpenStudy (anonymous):

There is probably a more elegant solution to this but it's nice to see different approaches to the same problem :)

OpenStudy (anonymous):

@unseenoceans yep :)

OpenStudy (anonymous):

Okay. I'm going to study these solutions and then make a decision. What am I supposed to do at the end. Choose a best response?

OpenStudy (anonymous):

yeah

OpenStudy (anonymous):

eseidl "Solve the quadratic to get a value of t. " The quadratic has two solutions 2 and 4 . Are both correct?

OpenStudy (anonymous):

Both t=2 and t=4 are correct; both these values of t give infinitely many solutions to the system. If t=2 then both lines are y=-x; Since the lines are identical, there are infinitely many solutions to the system. If t=4, then both lines are given by y=x; again, if two linear equation describe the same line, there are infinitely many solutions for the system.

OpenStudy (anonymous):

Wow. Much appreciated y'all. Special thanks yo eseidl for the best respone.

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