Solve the system to find how many athletes finished in each place.

Lawrence High prevailed in Saturdays track meet with the help of 20 individual-event placers earning a combined 68 points. A first place finish earns 5 points, a second place earns 3 points, and third place earns 1 point. Lawrence had a strong second-place showing, with as many second-place finshes as first and third place finishers combined.

Solve the system to find how many athletes finished in each place.

Lawrence High prevailed in Saturdays track meet with the help of 20 individual-event placers earning a combined 68 points. A first place finish earns 5 points, a second place earns 3 points, and third place earns 1 point. Lawrence had a strong second-place showing, with as many second-place finshes as first and third place finishers combined.


Step One - Translate word problem into formula(e):

First one is :

Let:
a=First place
b=Second
c=Third

Then:
68=5a+3b+c

Your turn for the other two, and I'll follow up with you then. I'll need to know whether we'll be using Cramer's rule or substituting and simplifying.

Fianchetto

Solve the system to find how many athletes finished in each place.

Lawrence High prevailed in Saturdays track meet with the help of 20 individual-event placers earning a combined 68 points. A first place finish earns 5 points, a second place earns 3 points, and third place earns 1 point. Lawrence had a strong second-place showing, with as many second-place finshes as first and third place finishers combined.




Solve the system to find out how many athletes finished in each place.

Lawrence H.S. prevailed in Saturday's track meet with the help of 20 individual-event placers earning a combined 68 points. A first place finish earns 5 points, a second place earns 3 points and a third place earns 1 point. Lawrence had a strong second-place showing, with as many second-place finishes as first and third place finishers combined.

First write your equations:

Let:
a = First place finishes
b = Second
c = Third
Then:

A. 68 = 5a + 3b + c
B. 20 = a + b + c
C. b = a + c
-----------------------------------------------------------

Substitute and Simplify

Since "C." is true, let?s plug it in for b in "A.":

68 = 5a + 3 (a + c) + c
68 = 5a + 3a + 3c + c
68 = 8a + 4c
17 = 2a + c

Now let's do the same for b in "B."

20 = a + (a + c) + c
20 = 2a + 2c
10 = a + c

Looking back at "C.":

b=10

From earlier,
Since:

10 = a + c
c = 10 - a

And:

17 = 2a + c

Substituting again, we get:

17 = 2a + c
17 = 2a + (10 - a)
17 = a + 10
a = 7

Last substitution yields:

b = a + c
10 = 7 + c
c = 3

So:

a = 7
b = 10
c = 3
-----------------------------------------------------------

Cramer's Rule


First write your equations:

Let:
a = First place finishes
b = Second
c = Third
Then:

A. 68 = 5a + 3b + c
B. 20 = a + b + c
C. b = a + c
-----------------------------------------------------------






| 5 3 1 |
D = | 1 1 1 | = 5 | 1 1 | - 3 | 1 1 | + 1 | 1 1 |
| 1 -1 1 | |-1 1 | | 1 1 | | 1 -1 |




= 5(-1) - 3(1) + 1(-1)
= -5 - 3 - 1
= -9


Next, replace each column with the "solutions" of your equations, and , one at a time, as above, lather, rinse, repeat. You will get Dx, Dy, and Dz.


x = Dx/D; y = Dy/D; and z = Dz/D



(for a better picture/description, see www.purplemath.com - type cramer's in the site search engine)

Solve the system to find how many athletes finished in each place.

Lawrence High prevailed in Saturdays track meet with the help of 20 individual-event placers earning a combined 68 points. A first place finish earns 5 points, a second place earns 3 points, and third place earns 1 point. Lawrence had a strong second-place showing, with as many second-place finshes as first and third place finishers combined.



Solve the system to find out how many athletes finished in each place.

Lawrence H.S. prevailed in Saturday's track meet with the help of 20 individual-event placers earning a combined 68 points. A first place finish earns 5 points, a second place earns 3 points and a third place earns 1 point. Lawrence had a strong second-place showing, with as many second-place finishes as first and third place finishers combined.

First write your equations:

Let:
a = First place finishes
b = Second
c = Third
Then:

A. 68 = 5a + 3b + c
B. 20 = a + b + c
C. b = a + c
-----------------------------------------------------------

Substitute and Simplify

Since "C." is true, let?s plug it in for b in "A.":

68 = 5a + 3 (a + c) + c
68 = 5a + 3a + 3c + c
68 = 8a + 4c
17 = 2a + c

Now let's do the same for b in "B."

20 = a + (a + c) + c
20 = 2a + 2c
10 = a + c

Looking back at "C.":

b=10

From earlier,
Since:

10 = a + c
c = 10 - a

And:

17 = 2a + c

Substituting again, we get:

17 = 2a + c
17 = 2a + (10 - a)
17 = a + 10
a = 7

Last substitution yields:

b = a + c
10 = 7 + c
c = 3

So:

a = 7
b = 10
c = 3
-----------------------------------------------------------

Cramer's Rule


First write your equations:

Let:
a = First place finishes
b = Second
c = Third
Then:

A. 68 = 5a + 3b + c
B. 20 = a + b + c
C. b = a + c OR 0 = a ? B + c
-----------------------------------------------------------

5 3 1
1 1 1
1 ?1 1




D = \left| \begin{matrix} 5 & 3 & 1 \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{matrix} \right|



D = \left| \begin{matrix} 1 & 1 \\ -1 & 1 \end{matrix} \right| - \left| \begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix} \right| + \left| \begin{matrix} 1 & 1 \\ 1 & -1 \end{matrix} \right|



Next, replace each column with the "solutions" of your equations:,



D_s = \left| \begin{matrix} 68 \\ 20 \\ 0 \end{matrix} \right|



and, one at a time, as above, lather, rinse, repeat. You will get Dx, Dy, and Dz.



D_x = \left| \begin{matrix} 68 & 3 & 1 \\ 20 & 1 & 1 \\ 0 & -1 & 1 \end{matrix} \right|




D_x = \left| \begin{matrix} 1 & 1 \\ -1 & 1 \end{matrix} \right| - \left| \begin{matrix} 3 & 1 \\ -1 & 1 \end{matrix} \right| + \left| \begin{matrix} 3 & 1 \\ 1 & 1 \end{matrix} \right|



x = Dx/D; y = Dy/D; and z = Dz/D



(for a better picture/description, see www.purplemath.com - type cramer's in the site search engine)
__________________
Fianchetto

"Fianchetto" is a chess term describing the position of a bishop in front of a castled king (one file from the edge of the board) with clear attack diagonally across the center of the board. Literally, fianchetto means (Italian) "by the flank", or "by the side", (where a sword or dagger is carried).

Solve the system to find how many athletes finished in each place.

Lawrence High prevailed in Saturdays track meet with the help of 20 individual-event placers earning a combined 68 points. A first place finish earns 5 points, a second place earns 3 points, and third place earns 1 point. Lawrence had a strong second-place showing, with as many second-place finshes as first and third place finishers combined.


More cramer's rule: attached