# Thread: system of linear equations

1. ## 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

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

3. 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
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

4. 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

5. 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??
$W_f + W_r =17700$
$(0.847)W_r - W_f =0$

6. Want to learn about them?
(Post #7 and #16)

7. 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

8. ## 0.847?

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??

9. ## ?

Originally Posted by ThePerfectHacker
Want to learn about them?
(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.

10. 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??
$W_f + W_r =17700$
$(0.847)W_r - W_f =0$
You just have to see how can you reduce two equations having two variables to a single equation having one variable.

So, add both the equations.
$W_f$ 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 $W_f$

11. 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: $1.847W_r=1770$

12. ## oh i got it!

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: $1.847W_r=1770$
thanks alot...!!

13. Originally Posted by danielwu
how and why 0.874 is change to 1.874??
you have the system:
$(1)W_r + (1)W_f =17700$
$(.874) W_r + (-1)W_f =0$
add the coefficients to get
$(1+.874)W_r + (1+-1)W_f = 0+17700$
or $(1.874)W_r=17700$
let me know if that made sence...

dan

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

dan
oh ya it made sense.
this is so clear. and its better
now i got the whole idea of it!!

15. 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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