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Question 73126: can i have help solving this system using the elimination method and separately help solving it using Cramer's rule?...
s + p + c = 40
c = 4p
s = p - 2
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! By elimination:
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s + p + c = 40 <---equation 1
c = 4p <---equation 2
s = p - 2 <---equation 3
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Using equation 2, eliminate the c in equation 1 by replacing c with its equal of 4p. When you
do, equation 1 becomes  and the equation set is:
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s + p + 4p = 40 <---equation 1
c = 4p <---equation 2
s = p - 2 <---equation 3
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Using equation 3, eliminate the s in equation 1 by replacing s with its equal of p-2. When you
do, equation 1 now becomes  . By combining the terms containing p
this further reduces to  . Next, add +2 to both sides and you have  .
And as the final step you divide both sides of this by 6 to get that  .
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Now that you know p = 7, you can back solve for c using equation 2. When you do you get
 .
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Furthermore, you can use equation 3 to solve for s by substituting p = 7 into the right
side to find that  .
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In summary we found by elimination that s = 5, p = 7, and c = 28.
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So much for elimination. On to Cramer's rule. The first thing we need to do is get the
equations into a standard alignment ... variables on the left and constants on the right.
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In doing this, equation 1 remains unchanged.
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For equation 2 we had  . Subtracting 4p from both sides results in 
and this can also be written as 
.
For equation 3 we had  . Subtract p from both sides to get  .
Then we can add 0c as a placeholder to get the equation 
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So now the 3 equations are:
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s + p +c = 40 <---equation 1
0s-4p+c = 0 <---equation 2
s - p+0c = -2 <---equation 3
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In solving this make the denominator determinant by making an array of the coefficients of
s, p, and c as follows:
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| +1 +1 +1|
| 0 -4 +1|
| +1 -1 +0|
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Expand by cofactors down the first column to get the denominator for Cramer's rule:
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1*((-4*0)-(+1*-1))- 0*((+1*0)-(+1*-1))+1*((+1*+1)-(+1*-4))
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This reduces to:
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1*(+1) - 0 + 1*(1+4)
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and further reduces to 1 + 5 = 6
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To solve for s, form an array that is the same as the denominator array except that you
replace the column containing the s coefficients (the first column) with the numbers from
the right side of the three equations. When you do this you get:
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| +40 +1 +1|
| 0 -4 +1|
| -2 -1 +0|
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Because there is a zero in the first column, let's again expand by cofactors down the
first column to get:
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+40*((-4*0)-(1*-1)) - 0 + (-2)*((1*1) - (1*-4))
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This simplifies to:
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40*1 - 0 - 2*5 = 40 - 10 = 30
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This determinant divided by the denominator determinant is 30/6 and it tells us that
s = 5. This agrees with what we found s to be by the elimination method.
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On to solving for p. Start with the denominator determinant only this time replace the
column containing the p coefficients (the middle column) with the right side numbers of the
original 3 equations. When you do that your determinant array for the numerator is:
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| +1 40 +1|
| 0 0 +1|
| +1 -2 +0|
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This time there are two zeros in the middle row. That will help save us some work.
So let's expand by cofactors along the middle row to get:
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-0 + 0 -(+1)*((1*-2)-(40*1)) = -1*(-2 - 40) = -1*(-42) = +42
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Divide this by the denominator determinant of 6 to get p = 42/6 = 7. This also agrees with
the answer we got for p when we solved by elimination.
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To solve for c by Cramer's rule, start with the denominator determinant and replace the
coefficients in the c column with the numbers on the right sides of the three equations to get:
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| +1 +1 +40|
| 0 -4 +0|
| +1 -1 -2|
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The middle row again has two zeros so let's expand by cofactors along it to save some work.
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-0 +(-4)*((1*-2)-(40*1)) -0
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This simplifies to:
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-4*(-2 - 40) = -4*(-42) = +168
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Then find c by dividing this result by the denominator determinant value. When you do you
get:
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c = 168/6 = 28. Again, this agrees with what we got by elimination.
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This has been a pretty long run with Cramer's rule and you will probably have trouble
following what I did unless you have studied how this rule works and the rules for working
with cofactors. (It's probably hard to follow even if you have studied Cramer's rule.) So be
patient with yourself and see if you can slog your way through it. At least you know the
answers you should get for this problem and maybe you can use the work to check what you do.
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Remember to watch your signs on the cofactors. They start with a + for the upper left corner
of the array and alternate for each vertical or horizontal change that you make.
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There is another way to work 3 by 3 determinants but it only works for 3 by 3's. Cofactor
expansion works for 3 by 3's, 4 by 4's and so on. For high school math books generally
do not contain problems that require 4 by 4 or larger arrays because they are a lot more work
than the above problem.
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If you have more questions about determinants, you can send them to me at:
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guezwho@msn.com
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I don't normally check that very often, but I will for the next couple of days. Use the word
math in the subject line so that I don't throw it out with all the spam that I normally get
at that address.
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Cheers and so on.
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