Thread: system

I go to the first class of lyceum and I have to solve the systems in algebra:

x+y=-1
2x+2y=-2

and another one:

x+2y=13
3x+y=4



Please help me with the mathematics.

Thanks a lot.




Pigaino stin proti lukeiou kai prepei na luso ta sustimata stin algebra:

x+y=-1
2x+2y=-2

kai akomi ena:

x+2y=13
3x+y=4

Parakalo boithiste me sta mathimatika.

Euxaristo polu.

Hello,

x+y=-1
2x+2y=-2

There is an infinity of solutions for this one as one is the double of the other...

x+2y=13
3x+y=4

from the first one, we can say that x=13-2y

thus 3(13-2y)+y=4
39-6y+y=4
-5y=-35
y=7

x=13-2*7=-1

(-1,7) is solution

Thanks.

And what about the other system?


x+y=-1
2x+2y=-2

Originally Posted by gdespina

Thanks.

And what about the other system?


x+y=-1
2x+2y=-2

Originally Posted by Moo

Hello,

x+y=-1
2x+2y=-2

There is an infinity of solutions for this one as one is the double of the other...

That was your answer.

-Dan

There is an infinity of solutions for this one as one is the double of the other...

x=-1-y

And it works for any couple that verify it.

For example :

(1,-2)
(2,-3)

etc...

Yes..

x+y=-1
2x+2y=-2

the first -> x=-1-y

and the second:

-> 2(-1-y)+2y=-2
-2-2y+2y=-2
-2y+2y=-2+2
0=0

is this correct?


how could this system solve?

If you divide your second equation by 2, it turns out to be the same as the first equation....thus, the same line. If two lines coincide with each other (one on top of the other), then there are an infinite number of solutions.

Remember to check the slopes of the lines in a system. If the slopes are equal, but the y-intercepts are different, then there are no solutions. The lines are parallel.

Dale