# Math Help - system of linear equations

1. ## 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

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

2. 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

3. 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".

4. 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})
$

5. 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)?

6. 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

7. 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

8. 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?

9. 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

10. 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

11. 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

12. Originally Posted by danielwu
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

13. 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
$

14. 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.

15. 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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