Question

Please help me :). Will fan.

Use Cramer’s rule to solve the system below. Show your work.
2x+4y=6
-3x+y+5

Answer 1

This is really just about plugging in numbers. The following link lists the general formula for the solutions to a linear 2-by-2 system like yours.
https://en.wikipedia.org/wiki/Cramer%27s_rule#Explicit_formulas_for_small_systems
Let me know if you're having trouble with any of those expressions.

Answer 2

So i looked at that and it confused me even more!
@arthur326

Answer 3

Well it gives a general 2 by two system, in the same form as the one you have, and it says what \(x\) and \(y\) have to be, which are your solutions.

Answer 4

Can you point out where you get confused? For example, is it where it mentions determinants, or earlier? Otherwise I don't know what you need help with.

Answer 5

The whole thing. i never been good at this kind of stuff. Do i need like step by step. Or something. because i have no idea how to do this.

Answer 6

Okay, let's walk through this then.

Answer 7

okay sounds good

Answer 8

The page says to consider the system \[a_1x+b_1y=c_1\] \[a_2x +b_2 y = c_2.\] What are all the \(a\)'s, \(b\)'s and \(c\)'s in your case?

Answer 9

2 4 6

Answer 10

3 y 5

Answer 11

Careful; \(b_2\) is not \(y\).

Answer 12

ohok

Answer 13

What number is really in front of \(y\)?

Answer 14

there isnt one ? or is it 3

Answer 15

It's definitely not 3. Think about it: the expression \(7y\) for example, means there are 7 \(y\)'s, and the expression \(2y\) mean there are 2 \(y's\). If we just have \(y\), how many \(y's\) do we have?

Answer 16

1 ? or 5 ?

Answer 17

1 :)

Answer 18

so y turns into 1 ?

Answer 19

\(y\) is the same thing as \(1y\), so \(b_2=1\). Does that make sense?

Answer 20

so what

Answer 21

some

Answer 22

Well we know what all the \(a\)s, \(b\)'s and \(c\)'s are, and if you look at the page, it says what \(x\) and \(y\) are in terms of the \(a\)s, \(b\)'s and \(c\)'s. For example, \[x=\dfrac{c_1b_2-b_1c_2}{a_1b_2-b_1a_2}.\] Now you just plug in the numbers.

Answer 23

So how do i plug the numbers ?

Answer 24

Look at \(c_1b_2\). Since \(c_1 = 6\), and \(b_1 = 1\), the term \(c_1b_2\) is \(6\cdot 1 = 6\).

Answer 25

(Sorry I meant \(b_2 = 1\).)

Answer 26

Here's a helpful page
http://www.1728.org/cramer.htm
and let's work it out.

Answer 27

I never been so confused !

Answer 28

ax + by = e
cx + dy = f

2x+4y=6
-3x+y+5

a=2 b=4 e= 6
c=-3 d=1 f=5


denom = |a b|
|cd| = a*d -c*b

denom =
|2 4|
|-3 1|
= 2*1 --3*4
denom =14

Answer 29

x =|e b|
|f d|
divided by denominator
x =
|6 4|
|5 1|
divided by 14

x = (6*1 - 5*4) / 14
x = (6 -20) / 14
x = -14/ 14
x = -1

Answer 30

-1 is correct (I already solved the equations)

Answer 31

WOW thank you so much, i understand all of this now !

Answer 32

Wait , so i write down all of that ?

Answer 33

I'm still working on "y" but when you use Cramer's rule, you would write all of that down.

Answer 34

ok

Answer 35

and thats all that has to be done ?

Answer 36

Yes, but I'm still working on "y"
THEN you'll be finished
and y does equal 2

Answer 37

okay y = 2

Answer 38

y=
|a e|
|c f|
divided by denom

y=
|2 6|
|-3 5|
divided by denom

y = (2 *5 --3*6) / 14
y = (10 + 18) / 14
y = 28 / 14
y = 2
YAY it's just that simple LOL

Answer 39

My mistake was in the definition of "y" (in my first calculation) which I think I'll delete.

Answer 40

your amazing lol

Answer 41

Thank you
and I think I'll have to change my page for that correction.
http://www.1728.org/cramer.htm
(That page explains it rather well, except for the definition of "y")

Answer 42

thank you !

Answer 43

u r welcome