Question

How do I find the solution to this system? Step by step, please :)
{3x + 3y = 10
{-9x + 9y = -30
Please help!

Answer 1

You can use cramers rule, matrix row operations or substitution. You have a preference?

Answer 2

Not entirely. Algebra is my weakest subject, you see. So...no.

Answer 3

Ok, substitution is the easiest conceptually. The idea is you solve for one variable in one equation then plug it into the other. If you divide the top equation by three you get:

\[\frac{ 3x }{ 3 }+\frac{ 3y }{ 3 }=\frac{ 10 }{ 3 }\]

Answer 4

\[x+y=\frac{ 10 }{ 3 }\]

Answer 5

RIghteo boss, I'm with you so far.

Answer 6

solve for x: \[x=\frac{ 10 }{ 3 }-y\]

Answer 7

okee doki

Answer 8

now plug your value for x into the other equation. We can first divide by a -9, to make the numbers a little smaller.

Answer 9

Do I have an x factor?

Answer 10

\[\frac{ -9x }{-9 }+\frac{ 9y }{ -9 }=\frac{ -30 }{ -9 }\]

Answer 11

alright

Answer 12

\[x-y=\frac{ 30 }{ 9 }\]

Answer 13

Ok

Answer 14

now remember \[x=\frac{ 10 }{ 3 }-y\] right

Answer 15

so substitute: \[(\frac{ 10 }{3 }-y)-y=\frac{ 30 }{ 9 }\]

Answer 16

you with me?

Answer 17

yeah, I think so

Answer 18

then combine terms and solve:

\[\frac{ 10 }{3}-2y=\frac{ 30 }{ 9 }\]

Answer 19

Um, but I'm still unsure of what y is. I am really bad at this stuff. (thank you sooo much for your help by the way)

Answer 20

Also, how do I find Y?

Answer 21

Are you sure you copied the problem correctly, with the negatives in the right places?

Answer 22

Yep. Why? Is there something wrong?

Answer 23

Yes, you get zero like this, let me double check everything.

Answer 24

Okee doki.

Answer 25

I double checked, y has to be zero for that system. Which makes x = 3 1/3

Answer 26

oh, okee doki!

Answer 27

@JakeV8 check this for me would ya

Answer 28

Oh, um, hi!

Answer 29

lots to read, but checking it now.

Elimination method might have been easier, but lemme look...

Answer 30

Please help, I'm a little confused

Answer 31

Oh, okee doki!

Answer 32

you get 18y=0?

Answer 33

me? or Jake?

Answer 34

Jake, sorry, I just wanted him to double check, its odd you were given a problem with a zero and an irrational answer.

Answer 35

It's perfectly alright

Answer 36

Equation 1: 3x + 3y = 10
Equation 2: -9x + 9y = -30

Multiply equation 1 by 3 all the way across...

Eq 1: 9x + 9y = 30
Eq 2: -9x + 9y = -30

Then add both equations:
9x - 9x + 9y + 9y = 30 - 30

Simplify:
18y = 0

Solve for y:
y = 0

Substitute y = 0 into either equation to solve for x:
Eq 1: 3x + 3y = 10
3x + 3(0) = 10
3x = 10
x = 10/3

Final check: put both x = 10/3 and y = 0 into the other equation:
Eq 2: -9x + 9y = -30
-9(10/3) + 9(0) = -30
-30 + 0 = -30 <<<--- TRUE, so we got it right :)

Answer 37

Righteo!

Answer 38

Sorry, that's a lot all at once.. I can explain if you have questions about anything.

Do you understand how I started it off? I wanted to multiply Eq 1 by something so that when I added the equations, one variable would be eliminated.

Answer 39

Well done Jake. I think the problem was set up to cancel just like you did.

Answer 40

Thank you both!

Answer 41

yep, me too.

Substitution would work fine too.. I just do whatever seems easiest. I find it harder when they tell you to use a certain method that doesn't seem easiest to me.

Answer 42

@CyberCierra, does this make sense? Algebra doesn't have to be your favorite or best class, but make sure you're learning enough to get through it :)

Answer 43

I understand it now! Certainly! If you wouldn't mind, could I bother you for a bit more help?

Answer 44

Can you post it in a new question? @p505 can continue to help to if desired :)

Answer 45

Righteo boss! Thank you both!