# system of linear equations

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• September 2nd 2006, 12:13 PM
danielwu
system of linear equations
Hi guys after the lessons i have problem understanding and solving system of linear equations. I need help badly to understand how it actually work outs???

pls help me :confused:

solve the given systems of equations by the method of elimination by substitution

33x + 2y = 34
40y = 9x + 11

by the method of elimination by addition or subtraction

2x - 3y - 4 = 0
3x + 2 = 2y
• September 2nd 2006, 12:47 PM
dan
Quote:

Originally Posted by danielwu

solve the given systems of equations by the method of elimination by substitution

33x + 2y = 34
40y = 9x + 11

ok sove for x in terms of y in #1:
$x=\frac{34-2y}{33}$
then substitute that value in for x in #2
$40y=9(\frac{34-2y}{33})+11$
does that help???
dan
• September 2nd 2006, 04:20 PM
ThePerfectHacker
Quote:

Originally Posted by danielwu

by the method of elimination by addition or subtraction

2x - 3y - 4 = 0
3x + 2 = 2y

You have,
$2x-3y=4$ (1)
$3x-2y=-2$ (2)
Multiply (1) by 2 and (2) by 3 to get,
$4x-6y=8$ (3)
$9x-6y=-6$ (4)
From (4) subtract (3),
$5x=-14$
Thus,
$x=-2.8$
From here you can solve for "y".
• September 3rd 2006, 10:48 AM
danielwu
Quote:

Originally Posted by dan
ok sove for x in terms of y in #1:
$x=\frac{34-2y}{33}$
then substitute that value in for x in #2
$40y=9(\frac{34-2y}{33})+11$
does that help???
dan

Hi i do know how to do the steps but i don't know how to slove it!

how to slove this?-----> $
9(\frac{34-2y}{33})
$
:)
• September 3rd 2006, 10:58 AM
danielwu
Quote:

Originally Posted by ThePerfectHacker
You have,
$2x-3y=4$ (1)
$3x-2y=-2$ (2)
Multiply (1) by 2 and (2) by 3 to get,
$4x-6y=8$ (3)
$9x-6y=-6$ (4)
From (4) subtract (3),
$5x=-14$
Thus,
$x=-2.8$
From here you can solve for "y".

why do i have to multiply by 2 in (1)?
&
why do i have to multiply by 3 in (2)?
• September 3rd 2006, 11:55 AM
CaptainBlack
Quote:

Originally Posted by danielwu
why do i have to multiply by 2 in (1)?
&
why do i have to multiply by 3 in (2)?

Because you are going to subtract the two resulting equations and eliminate
the term in y. So you need to make the coefficient of y in each equation the
same.

In the first equation the coefficient of y is -3, and in the second is -2.
So multiplying these equations by 2 and 3 makes the coefficient of y -6
in both equations, so when the equations are subtracted the y terms is
removed.

RonL
• September 3rd 2006, 12:49 PM
Quick
Quote:

Originally Posted by danielwu
Hi i do know how to do the steps but i don't know how to slove it!

how to slove this?-----> $
9(\frac{34-2y}{33})
$
:)

divide by nine and then multiply bt 33 :)
• September 3rd 2006, 08:45 PM
danielwu
Quote:

Originally Posted by Quick
divide by nine and then multiply bt 33 :)

my ghosh i still don't get it :)

what divide by nine? then multiply by 33?
• September 3rd 2006, 10:00 PM
danielwu
hi i am still quite confuse.
it says solve the given systems of equations by either method of this section?

30p = 55 - Q
19p + 14Q + 3 = 0

so in becomes
30p + Q = 55
19p + 14Q = -3
and from equation (1) do i multiply by 14? to make the coefficient of Q in (1) the same as equation 2?

420p + 14Q = 770
• September 3rd 2006, 11:07 PM
CaptainBlack
Quote:

Originally Posted by danielwu
hi i am still quite confuse.
it says solve the given systems of equations by either method of this section?

30p = 55 - Q
19p + 14Q + 3 = 0

so in becomes
30p + Q = 55
19p + 14Q = -3
and from equation (1) do i multiply by 14? to make the coefficient of Q in (1) the same as equation 2?

420p + 14Q = 770

That is correct, now subtract the second equation fron this to get:

420p - 19p + 14Q - 14Q = 770 + 3,

simplifying:

401p = 773,

p = 773/401.

Now going back to your first equation:

30p + Q = 55,

so:

Q=55-30p=55-30.773/401=-1135/401

RonL
• September 3rd 2006, 11:31 PM
danielwu
Quote:

Originally Posted by CaptainBlack
That is correct, now subtract the second equation fron this to get:

420p - 19p + 14Q - 14Q = 770 + 3,

simplifying:

401p = 773,

p = 773/401.

Now going back to your first equation:

30p + Q = 55,

so:

Q=55-30p=55-30.773/401=-1135/401

RonL

Hi can i know how you get
=-1135/401
• September 4th 2006, 12:43 AM
CaptainBlack
Quote:

Originally Posted by danielwu
Quote:

Originally Posted by CaptainBlack
...

Q=55-30p=55-30.773/401=-1135/401

RonL

Hi can i know how you get
=-1135/401

We have found that $p=773/401$.

$Q=55-30p$ is just a rearrangement of one of the equations that you were
given.

So:

$Q=55-30\times \frac{773}{401}$.

Now $55=\frac{55 \times 401}{401}=\frac{22055}{401}$, so:

$Q=55-30\times \frac{773}{401}=\frac{22055}{401}-\frac{23190}{401}=-1135/401$.

RonL
• September 4th 2006, 07:34 AM
danielwu
Hi was hoping someone could help me with this :)

show that the given systems of equations have either an unlimited number of solutions or no solution. if there is an unlimited number of solutions, find one of them.

$
x - 2y - 3z = 2
$

$
x - 4y -13z = 14
$

$
-3x + 5y +4z = 0
$
• September 4th 2006, 08:00 AM
ThePerfectHacker
Quote:

Originally Posted by danielwu
Hi was hoping someone could help me with this :)

show that the given systems of equations have either an unlimited number of solutions or no solution. if there is an unlimited number of solutions, find one of them.

$
x - 2y - 3z = 2
$

$
x - 4y -13z = 14
$

$
-3x + 5y +4z = 0
$

You look at its determinant which is zero.
Then you use Cramer's rule and see if each $D_x,D_y,D_z$ is also zero or not.
Since each is zero there system has infinitely many solutions.
• September 4th 2006, 08:46 AM
danielwu
Quote:

Originally Posted by ThePerfectHacker
You look at its determinant which is zero.
Then you use Cramer's rule and see if each $D_x,D_y,D_z$ is also zero or not.
Since each is zero there system has infinitely many solutions.

i have understand the cramer's rule.
but this equation i have to solve it algebraically.
thus i am really worst at algebra.
:(
can help me futher?
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