Simultaneous equations unique, infinite solutions

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January 28th 2009, 01:47 AM
sinners

Simultaneous equations unique, infinite solutions

hello all, frantic rush to get all the h/w done before school starts again lol.

here it is i have absolutely no idea what i am suppose to do here, the book provides little to no help at all.

This is the example they give which i need help figuring out as well as the actual problem.

Consider the simultaneous linear equations
(m-2)x + y = 2 and mx+2y = k
Find the values of m and k such that the system of equations has

a) a unique solution
b) no solution
c) infinitely many solutions

Now for the question......

Find the value of m for which the simultaneous equations

(m+3)x +my = 12
(m-1)x + (m-3)y = 7

have no solution.

i no that when they want the equations to have no solutions it means the lines are parallel and when the infinite solutions they are the same line however i have no idea what to do about it.

Thanks

Josh
January 28th 2009, 05:59 AM
red_dog

1) From the first equation $y=2-(m-2)x$

Replacing y in the second equation we have $x(4-m)=k-4$
This equation has:

a) a unique solution if $4-m\neq 0\Leftrightarrow m\neq 4$
b) infinetely many solutions if $4-m=0, \ k-4=0\Leftrightarrow m=k=4$
c) no solution if $4-m=0, \ k-4\neq 0\Leftrightarrow m=4, \ k\neq 4$
January 28th 2009, 12:17 PM
sinners

hello, thanks for the reply. i dont understand how you interpreted a) b) c) from the two equations?

thanks

Josh
January 28th 2009, 01:33 PM
mr fantastic

Quote:

Originally Posted by red_dog

1) From the first equation $y=2-(m-2)x$

Replacing y in the second equation we have $x(4-m)=k-4$
This equation has:

a) a unique solution if $4-m\neq 0\Leftrightarrow m\neq 4$
b) infinetely many solutions if $4-m=0, \ k-4=0\Leftrightarrow m=k=4$
c) no solution if $4-m=0, \ k-4\neq 0\Leftrightarrow m=4, \ k\neq 4$

Quote:

Originally Posted by sinners

hello, thanks for the reply. i dont understand how you interpreted a) b) c) from the two equations?

thanks

Josh

It should be clear where $x(4-m)=k-4$ has come from.

a) It should be clear that from the solution red_dog gives here that you can solve for a unique value of x (and hence y) eg. m = 2 and k = anything (3, say).

b) What happens when you substitute the solution red_dog gives here ....? Can you get a unique value of x or can x be any value ....?

c) What happens if m = 4 and k = 2, say ....?
January 28th 2009, 01:57 PM
sinners

ohh i see thanks guys i understand now