[SOLVED] Solving sim equations.

Printable View

Show 40 post(s) from this thread on one page

February 10th 2009, 06:14 AM
ADY

[SOLVED] Solving sim equations.

I need some real help with this please guys, ok so here's the example given to me from tutor:

Example 2x + y = 1, x + 3y = -7

Method 1 - add/subtract

2x + y = 1
x - 3y = 11

You can add or subtract. Adding will enable you to focus on x, subtracting will focus on y.

Perhaps the easiest is to add. BUT first of all you must scale the first expression by 3 to enable the y’s to match, giving the zero count for y, you require.

So 6x + 3y = 3 (Note each part has been x 3)
x - 3y = 11

Adding 7x = 14 (Note the zero count for y - this is what you x = 2 want to happen)

Substituting back into one of the previous expressions -

2(2) + y = 1, 4 + y = 1, y = -3

ANSWER x = 2, y = -3

My first problem is, from the example equations, how can it be re-arranged and make 11?!

I'm confused with this topic, thanks

ADY
February 10th 2009, 07:00 AM
princess_21

i think you got the right answer but what do you mean by this?

Quote:

My first problem is, from the example equations, how can it be re-arranged and make 11?!
February 10th 2009, 07:09 AM
ADY

Its not my answer, its an example.

and i need to understand the methodology of it as i dont get it at all, with reference to the first line, where's " 11" come from?
February 10th 2009, 07:12 AM
princess_21

2x + y = 1
x - 3y = 11

i call it elimination method

you can do that like this way. subtract the first from the second and get x.

$x-3y=11$

$\frac{-2x-y=-1}{-x-4y=10}$

$-x=10+4y$

$x=-10-4y$

now you have x
then use substitution method: you can use either 1st or second equation

$x-3y=11$

$(-10-4y)-3y=11$

$-7y=11+10$

$-7y=21$

$y=-3$

substitute y to get x

$x-3y=11$

$x-3(-3)=11$

$x+9=11$

$x=11-9$

$x=2$

now you have the value of both variables then substitute to check:

$x=2$

$y=-3$

B. $x-3y=11$

$2-3(-3)=11$

$2+9=11$

$11=11$

A. $2x +y = 1$

$2(2)+(-3)=1$

$4-3=1$

$1=1$

Note:11 is just the answer on the equation you are solving but there is no value of x and y so you need to find their values.
February 10th 2009, 07:52 AM
ADY

ok but the example is

Example

2x + y = 1
x + 3y = -7

So where did you get 11 from?
February 10th 2009, 09:26 AM
ADY

ok, i think im there.

so how would you solve

$y=22x + 100, y=25x + 10$
February 10th 2009, 09:33 AM
masters

Quote:

Originally Posted by ADY

I need some real help with this please guys, ok so here's the example given to me from tutor:

Example 2x + y = 1, x + 3y = -7

Method 1 - add/subtract

2x + y = 1
x - 3y = 11 <====Don't know where this came from!

My first problem is, from the example equations, how can it be re-arranged and make 11?!

I'm confused with this topic, thanks

ADY

Quote:

Originally Posted by ADY

ok but the example is

Example

2x + y = 1
x + 3y = -7

So where did you get 11 from?

Hi ADY,

Your second equation x + 3y = -7 cannot be derived from x - 3y = 11. You've either miscopied the example or there's a misprint somewhere.
February 10th 2009, 09:36 AM
masters

Quote:

Originally Posted by ADY

ok, i think im there.

so how would you solve

$y=22x + 100, y=25x + 10$

Hi ADY,

Since y = y , you can equate the right side of the equations as well.

$22x+100=25x+10$

Solve for x. Then substitute the value for x back into either original equation to find y.
February 10th 2009, 09:43 AM
ADY

Quote:

Originally Posted by masters

Hi ADY,

Your second equation x + 3y = -7 cannot be derived from x - 3y = 11. You've either miscopied the example or there's a misprint somewhere.

Thanks the original equation was:

2x + y = 1
x + 3y + -7

Can that work?
February 10th 2009, 09:44 AM
ADY

Quote:

Originally Posted by masters

Hi ADY,

Since y = y , you can equate the right side of the equations as well.

$22x+100=25x+10$

Solve for x. Then substitute the value for x back into either original equation to find y.

Can i have further help with this please, i dont know where im even starting
February 10th 2009, 10:37 AM
masters

Quote:

Originally Posted by masters

Hi ADY,

Since y = y , you can equate the right side of the equations as well.

$22x+100=25x+10$

Solve for x. Then substitute the value for x back into either original equation to find y.

Ok ADY,

$22x+100=25x+10$

First subtract 22x from both sides:

$100=3x+10$

Next, subtract 10 from both sides:

$90=3x$

Finally, divide by 3:

$30=x$

Now, substitute 30 back into y = 22x + 100 to get y = 22(30) + 100, and y = 760
February 10th 2009, 05:58 PM
princess_21

Quote:

Originally Posted by ADY

ok, i think im there.

so how would you solve

$y=22x + 100, y=25x + 10$

so that's what you are asking I didn't realized that the two equations are different one is 7 and one is 11. you can't changed the answer. i'm sorry i didn't look at the first equation. i only saw the one with 11 as the answer.
February 10th 2009, 11:23 PM
princess_21

Quote:

Originally Posted by ADY

Can i have further help with this please, i dont know where im even starting

just like what masters said equate the solution

$y=22x+100$

$y=25x+10$

since y=y it's reflective propery

$22x+100=25x+10$

$100-10=25x-22x$

$90=3x$

$30=x$

to find y, substitute x.

$y=22(30)+100$

$y=660+100$

$y=760$

$25(30)+10$

$y=750+10$

$y=760$

is that what you are asking?
February 11th 2009, 03:00 AM
ADY

Princess - thank you!
February 11th 2009, 04:36 AM
masters

Quote:

Originally Posted by ADY

Princess - thank you!

I'd bet that if my name were "Precious", you'd thank me too!(Giggle)

Show 40 post(s) from this thread on one page

All times are GMT -8. The time now is 08:52 AM.

Copyright © 2005-2013 Math Help Forum. All rights reserved.

Copyright © 2005-2013 Math Help Forum. All rights reserved.