# system of linear equations

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• September 4th 2006, 09:01 AM
danielwu
solving systems of two linear equations in two unknows algebraically
Set up appropriate systems of two linear equations and slove the systems algebraically. all data are accurate to at least two significant digits.

the weight Wf supported by the front wheels of a certain car and the weight Wr supported by the rear wheels together equal the weight of the car, 17700N. also , the ratio of Wr to Wf is 0.847. what are the weights supported by each set of wheels??

can any kind hearted help me out with this word problems :confused:
• September 4th 2006, 09:02 AM
dan
do you know anything about matrices??? the cramer's rule is pretty simple if you know matrix algebra...
dan
• September 4th 2006, 09:08 AM
dan
Quote:

Originally Posted by danielwu
Set up appropriate systems of two linear equations and slove the systems algebraically. all data are accurate to at least two significant digits.

the weight Wf supported by the front wheels of a certain car and the weight Wr supported by the rear wheels together equal the weight of the car, 17700N. also , the ratio of Wr to Wf is 0.847. what are the weights supported by each set of wheels??

can any kind hearted help me out with this word problems :confused:

here i think you have...someone correct me if i'm wrong...
$W_f + W_r =17700$
$(0.847)W_r - W_f =0$
~dan
• September 4th 2006, 09:15 AM
danielwu
Quote:

Originally Posted by dan
do you know anything about matrices??? the cramer's rule is pretty simple if you know matrix algebra...
dan

nope i am sorry i dont know about matrix algebra :confused:
• September 4th 2006, 09:20 AM
danielwu
Quote:

Originally Posted by dan
here i think you have...someone correct me if i'm wrong...
$W_f + W_r =17700$
$(0.847)W_r - W_f =0$
~dan

$
I have the answer as W_r = 8120N, W_f = 9580N
$

As for this. how should i go about to solve?? :confused:
$W_f + W_r =17700$
$(0.847)W_r - W_f =0$
• September 4th 2006, 09:22 AM
ThePerfectHacker
(Post #7 and #16)
• September 4th 2006, 09:25 AM
dan
Quote:

Originally Posted by dan
here i think you have...someone correct me if i'm wrong...
$W_f + W_r =17700$
$(0.847)W_r - W_f =0$
~dan

then by adding you wold have $(1.847)W_r =17700$
$W_r = \frac{17700}{1.847} =9583.12$
and $W_r(0.874)=W_f$

~dan
• September 4th 2006, 09:39 AM
danielwu
0.847?
Quote:

Originally Posted by dan
then by adding you wold have $(1.847)W_r =17700$
$W_r = \frac{17700}{1.847} =9583.12$
and $W_r(0.874)=W_f$

~dan

how and why 0.874 is change to 1.874??
• September 4th 2006, 09:46 AM
danielwu
?
Quote:

Originally Posted by ThePerfectHacker
(Post #7 and #16)

those are determinants method

i mean the question i post was solving by algebraically.
which than i am confused in doing so!
i can use cramer's rule and i am quite okay with it, just that algebraically i am real bad at it. ;)
• September 4th 2006, 10:22 AM
Quote:

Originally Posted by danielwu
$
I have the answer as W_r = 8120N, W_f = 9580N
$

As for this. how should i go about to solve?? :confused:
$W_f + W_r =17700$
$(0.847)W_r - W_f =0$

You just have to see how can you reduce two equations having two variables to a single equation having one variable.

$W_f$ goes out and the resultant equation contains only one variable, i.e., $W_r$. The resulting equations is
$W_r+0.847W_r=17700+0$
$1.847W_r=17700$
$W_r=\frac{17700}{1.847}\approx9583.11$
Then substitute the value of $W_r$ in one of the previous equations to find $W_f$
• September 4th 2006, 10:22 AM
Quick
Quote:

Originally Posted by danielwu
how and why 0.874 is change to 1.874??

You know that $W_f=1770-W_r$

you also know that $0.847W_r=W_f$

subsitute: $0.847W_r=1770-W_r$

add: $0.847W_r+1W_r=1770$

add: $1.847W_r=1770$
• September 4th 2006, 10:35 AM
danielwu
oh i got it!
Quote:

Originally Posted by Quick
You know that $W_f=1770-W_r$

you also know that $0.847W_r=W_f$

subsitute: $0.847W_r=1770-W_r$

add: $0.847W_r+1W_r=1770$

add: $1.847W_r=1770$

:D thanks alot...!!
• September 4th 2006, 10:39 AM
dan
Quote:

Originally Posted by danielwu
how and why 0.874 is change to 1.874??

you have the system:
$(1)W_r + (1)W_f =17700$
$(.874) W_r + (-1)W_f =0$
$(1+.874)W_r + (1+-1)W_f = 0+17700$
or $(1.874)W_r=17700$
let me know if that made sence...

dan
• September 4th 2006, 11:25 AM
danielwu
Quote:

Originally Posted by dan
you have the system:
$(1)W_r + (1)W_f =17700$
$(.874) W_r + (-1)W_f =0$
$(1+.874)W_r + (1+-1)W_f = 0+17700$
or $(1.874)W_r=17700$
let me know if that made sence...

dan

this is so clear. and its better
now i got the whole idea of it!! ;)
• September 4th 2006, 11:42 AM
danielwu
hi there i have problem with this word problems

Set up appropriate systems of two linear equations in two unknowns and then solve the systems by determinants. all numbers are accurate to at least two significant digits.

(A) A shipment of 320 cell phones and radar detectors was destroyed due to a truck accident. on the insurance claim, the shipper stated that each phone was worth $110, each detector was worth$160, and their total value was $40700. how many of each were in the shipment? (1)$110 + $160 =$40700<--------is the first equation right??
(2) ??????????????????????

was wondering is there a way to master these word problem?
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