Question

College Algebra: I'm doing 2X2 Linear Systems and I'm not so well with fractions. Can someone solve this problem step by step for me? I'm reviewing for my chapter test coming up next week. I need to solve the system and determine whether it's independent, dependent, or inconsistent.
1/2x-1/3y=12
1/4x-1/2y=1
I'll rewrite the problem again below.

Answer 1

\[\frac{ 1 }{ 2 }x-\frac{ 1 }{ 3 }y=12\]
\[\frac{ 1 }{ 4 }x-\frac{ 1 }{ 2 }y=1\]

Answer 2

to get rid of the fraction multiply each term by the LCM (lowest common multiple) of the denominator

Answer 3

for equation 1 1/2 and 1/3 the lcm is 2*3 = 6
so multiply by 6 to get
3 x - 2 y = 72
now do equation 2 before trying to solve simultaneously

Answer 4

what do you have for equation 2?

Answer 5

I'm sorry, but I'm entirely lost. That doesn't seem like anything you're suppose to do. I need to try to combine the equations and solve.

Answer 6

Then determine if it's independent, dependent, or inconsistent

Answer 7

I think getting rid of the fractions like @triciaal as suggested would help you a ton especially since you said you don't like dealing with them.
The cool thing about fractions in equations is you manipulate the equation so you see no fractions.

Answer 8

Just whatever you do to one side an equation you can do the other.

Answer 9

yes but it's easier to work with whole numbers so the first step is to get rid of the fraction

Answer 10

to get rid of the fraction multiply each term in the equation by the lcm

Answer 11

Okay, how did you get 3x-2y=72?

Answer 12

I'm getting something else.

Answer 13

I did equation 1 for you and now you need to do the same for equation 2

Answer 14

I can'y do equation 2 if I don't know how you got equation 1.

Answer 15

*can't

Answer 16

repost
for equation 1 1/2 and 1/3 the lcm is 2*3 = 6 so multiply by 6 to get 3 x - 2 y = 72 now do equation 2 before trying to solve simultaneously

Answer 17

\[\frac{ 1 }{ 2 }x-\frac{ 1 }{ 3 }y=12 \text{ so I see 2 and 3 on bottom } \\ \text{ so I will multiply both sides by 2*3 } \\ 2 \cdot 3 \frac{1}{2}x-2 \cdot 3 \frac{1}{3}y=2 \cdot 3 (12) \\\cancel{2} \cdot 3 \frac{1}{\cancel{2}}x-2 \cdot \cancel{3} \frac{1}{\cancel{3}}y=2 \cdot 3 (12) \]
since 2/2=1 and 3/3=1
so you have for the first equation can be written as
3x-2y=72

Answer 18

Okay, now I see how you did that. Thank you. Now I can do equation 2.

Answer 19

yep you try

Answer 20

2x-4y=8

Answer 21

Alight, so now I got the equations 3x-2y=72 and 2x-4y=8.
What do I do next?

Answer 22

for equation 2 do you see that you can divide by 2 it will be the same as
x-2y = 4

Answer 23

Okay, I see.

Answer 24

there is nothing wrong with what you have you just didn't use the lowest common denominator for 2 and 4 which would be 4

Answer 25

Oh, so you move the -2y on the other side. So it would be x=2y+4 right

Answer 26

to solve by substitution yes

Answer 27

That's what I'm doing at the moment. :) Awesome!

Answer 28

there are 4 ways to solve simultaneous equations. substitution is just one of them

Answer 29

3(2y+4)-2y=72 is what I got. Then I got 6y+12-2y=72. Then combine like-terms and it will be 4y+12=72. Subtract on both sides and it's 4y=60. Divide on both sides and it's y=15

Answer 30

now you substitute for x in equation 1 and solve for y
use this value in the expression for x to get the value for x