Log in

View Full Version : 9th grade algebra: elimination


vannaboo
Feb 7, 2008, 05:51 PM
Im in 9th algebra && I need help with elimination..

Does anyone know how to do this?

Thanks

:)

galactus
Feb 7, 2008, 05:54 PM
No, no one knows how. Can you please explain?

Wondergirl
Feb 7, 2008, 05:58 PM
Yes. It's lots of fun--one of the funnest things in algebra. Elimination by addition and subtraction or by multiplication?

Wondergirl
Feb 7, 2008, 05:59 PM
See this --

https://www.askmehelpdesk.com/math-sciences/algebra-ii-elimination-substitution-20603.html

vannaboo
Feb 7, 2008, 06:01 PM
its elimination by addition && subtraction.

the problem I need help with is:

3x+4=19
3x+6y=33

Wondergirl
Feb 7, 2008, 06:03 PM
its elimination by addition && subtraction.

the problem i need help with is:

3x+4=19
3x+6y=33

Are you sure you copied that correctly?

vannaboo
Feb 7, 2008, 06:18 PM
Hahaha

Sorry.

I left out the y by the 4

:)

mama na bojana
Feb 9, 2008, 10:24 AM
Just imagine that any equation is like a balance
x- box with some not known own weight
y- basket with some not known weight

So you are sure that 3 boxes and 4 baskets are exactly 19 kg weight
And you know that 3 boxes and 6 baskets are giving 33 kg

Imagine now
You can take out from the balance same weight from the left and from the right side, the balance will still stay in equilibrium
And do it

Take from the left side of the second balance 3 boxes and 4 baskets, and take from the right 19 kg

What is remaining?
2 baskets on left equal to 14 kg
It is logical any basket is exactly 7 kg…….
And ooooops do we have a problem?
4 baskets are 28 kg?
What about the boxes?
They are obviously foul with Helium, like balloons, they are not pushing the left side of the equations down, they are lifting, just like any quantity with negative sign.

Any basket is lifting with force equal to?

3*x+4*7=19
3*x=19-28
3*x=-9
X=-3

Each box is lifting with force equal by intensity to 3kg, but because this is lifting force the sign is –

Think
You can find similar illustrations for any system.
You can work with prices instead of weights….

All the best
And more work

simoneaugie
Feb 9, 2008, 09:37 PM
3x + 4 = 19

Subtract 4 from both sides The 3x would stay
The 4 disapears
19 - 4 = 15

Now you have 3x = 15 All you have to do now is to divide both sides by 3

Answer: x = 5 That one is easy

If x is the same in both problems,

3x + 6y = 33 becomes 15 + 6y = 33


To solve that one, subtract 15 from both sides 15 - 15 =0

33 - 15 = 18

The new equation is 6y = 18 What is y? It is 3.

saoco
Feb 12, 2008, 06:32 PM
For the following set of numbers, find the average. The median, and the mode. 22,16,23,17,25,25?

morgaine300
Feb 12, 2008, 08:24 PM
for the following set of numbers, find the average. the median, and the mode. 22,16,23,17,25,25?

It would be good to start your own thread with this, especially since no one expects to find this under a thread entitled "elimination."

Do you remember learning how to do an average clear back in elementary school? Same thing. Only usually we call it a "mean" in statistics.

To get the medium, you will need to "order" the numbers, that is, put them from smallest to largest. Then find the middle. If you have an odd number of elements, it's literally the middle. (i.e. if you have 7 items then it's the 4th one over, the middle one.) If it's an even number, like you have, it's the average of the two center ones. (i.e. the average of the 3rd and 4th one.) That'll be easier to see once you have them ordered.

A mode is which thing appears most often in the set.

dh664
Apr 10, 2010, 09:05 PM
I assume you mean 3x+4y=19 so
3x+4y=19
3x+6y=33
subtract the equations
0x-2y=-14
divide -14 by 2
y=-7
put -7 in for y in one of the equations
3x+4(-7)=19
3x-28=19
add 28 to both sides
3x=47
take 47/3
x=15.6666
(15and2/3,-7)

dh664
Apr 10, 2010, 09:08 PM
sorry I meant y=7 not -7