Ask your own question, for FREE!
Algebra 19 Online
OpenStudy (anonymous):

how many solutions does the following system have ? 2x+3y=1 -3x-2y=-1 A.infinitely many solutions b.two solutions c.one solution d.no solutions

OpenStudy (freckles):

Try to set the system up for elimination. lets look just the x's we have 2x in the first and -3x in the second the lcm of 2 and 3 is 6 so if we get one of the coeifficients to be 6 and the other to be -6 then all is left is adding the equations to eliminate the x part and solve for the y part then go back and solve for the x when we have the y anyways multiply the first equation by 3 and multiply the second equation by 2

OpenStudy (freckles):

then add the equations after doing that

OpenStudy (anonymous):

@candy13106 you're wrong

OpenStudy (candy13106):

how

OpenStudy (anonymous):

only one solution

OpenStudy (anonymous):

because it would be a solution , it would be one

OpenStudy (candy13106):

but how when i graphed it , it wouldn't intersect

OpenStudy (anonymous):

check the determinant. it is non singular

OpenStudy (candy13106):

and when line dont intersect they r no solution

OpenStudy (anonymous):

but i just took the test and it said the correct answer was one solution

OpenStudy (candy13106):

i sure believe u look at your score and look at mine

OpenStudy (anonymous):

OpenStudy (freckles):

@bluerobinbird Have you tried doing what I suggested?

OpenStudy (anonymous):

check my attachmnet

OpenStudy (anonymous):

*suppose

OpenStudy (candy13106):

ok im just not going to make a big deal out of it

OpenStudy (anonymous):

i just did @princer ...um and yes i did *@freckles &it doesn't matter i only got one wrong anyway so @candy13106

OpenStudy (candy13106):

ok so good bye and have a nice day people

OpenStudy (anonymous):

Check this one, it has a general rule.

OpenStudy (anonymous):

ok ,so my question didn't have no solution im kind of confused

OpenStudy (anonymous):

no it has only one solution, here ad-bc is non zero and the rhs of both are equations are non zero.

OpenStudy (anonymous):

ok this time you wont have any confusion, check this out

OpenStudy (anonymous):

OpenStudy (anonymous):

kind of what i was trying to say , my computer said the correct one was 1 solution i clicked no solution , but what you just said "no it has only one solution, here ad-bc is non zero and the rhs of both are equations are non zero." just confused ,me because you're saying that it has 1 solution but also no solution (basically )

OpenStudy (anonymous):

Put x=1/5 and y=1/5 on the lhs of both the equations. see you will get their corresponding rhs, hence (1/5,1/5) is a solution. and this solution is unique, i.e. only one solution

OpenStudy (anonymous):

i never said no solution.

OpenStudy (anonymous):

oh it make sense now so what do you mean by non zero then ? what's a non zero ? never heard of that

OpenStudy (anonymous):

non zero means a number other than 0

OpenStudy (anonymous):

oh duh ... lol

OpenStudy (anonymous):

common sense right .. but thanks for your help

OpenStudy (anonymous):

you know how to do MATRIX ?

OpenStudy (anonymous):

Yeah.

OpenStudy (anonymous):

can you help me with one ...

OpenStudy (anonymous):

ok.. sure

OpenStudy (anonymous):

ok hold on

OpenStudy (anonymous):

8 7 + 4 -3 -6 5 + 1 13 @princer_jones

OpenStudy (anonymous):

What is the question ? to find [8 7]+[4 -3]?

OpenStudy (anonymous):

And [-6 5]+[1 13]?

OpenStudy (anonymous):

OpenStudy (anonymous):

@bluerobinbird

OpenStudy (kiakat):

Just to clarify this, to check how many solutions an equation has, you look at the highest exponent. So for this one it's only one solution because the x is only raised to the 1st power. if x or y happened to be x^3 or y^3 there would be three solutions.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!