Could someone please explain to me in great detail how to solve the following using cramers rule?

1. 5x + 7y = -3

2x + 3y = -1

2. 5x - 2y = 4

3x + y = 6

3. 2x -y + 3z = 9

x + 2z = 3

3x + 2y + z = 10

4. 3x - 2y = 1

4x + 3y =2

5. 2x + 3y = 7

8x + 12y = 2

Thank You!!

Update:

Listen I got this massive packet of work to do over my christmas break. Its about 16 pages. Neither of my parents took algebra in High School. I am trying to get some help I dont want people to answer them for me just an explantion on how I would arrive at the answer. IF YOU ARE NOT GOING TO BE HELPFUL THEN DO NOT POST. GROW UP!

Relevance

Cramer's Rule relies on finding determinants. So you'll need to find the determinants and work from there.

First, go over what determinants are. Either in your book, or in online tutorials such as this one here:

http://www.mathnotes.com/intermediate/Mchapter04/a...

(note: you can find more tutorials by putting tutorial cramer's rule in your search window, if you don't like this one)

Let's do the first one...

First you'll need the main determinant, formed of the matrix of coeffieicnets. That is,

det | 5..7|

| 2..3| which is (5x3) - (7x2) = 1

This will be the denominator for x and y

Now you need det(x). This is formed by replacing the "x" column with the "answers" column. In other words,

det [-3 .. 7|

x |-1 .. 3| which is (-3x3) - (7x(-1)) = -9 - (-7) = -2

Now put det (x)/det = -2/1 = -2. So x = -2

Now find y. Replace the "y" column with the "asnwers" column

det |5 .. -3|

y | 2...-1| which is (5x(-1)) - ((-3)x(2) = -5 - (-6) =

Now put det(y)/det = 1/1 = 1 So y = 1

Check:

5(-2) + 7(1) = -3

2(-2) + 3(1) = -1

The others would work pretty much the same way. It's very hard typing all tihs in, this online text editor is very limited. Besides, you're the one who needs practice with this, not me. So try the others the same way, and give yourself a treat for learning it.

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• First the implication above is that the two formulas above use an x and a y that have the same values. There are actually multiple ways to solve this problem. I have listed two below. But for Cramer's rule please look at the site listed here for a great explanation: http://library.thinkquest.org/10030/10cramer.htm.

If you look at the formulas 5x + 7y = -3 and 2x + 3y = -1. Cramer's Rule states that for two linear equations a1x + b1y = c1 and a2x + b2y = c2. D = |a1 b1, a2 b2| and this does not equal zero. Remember in matrix math you multiply cross wise. So to solve this equation D = a1*b2 - a2*b1. Dx = |c1 b1, c2 b2| and Dy = |a1 c1, a2 c2|. Finally x = Dx/D and y = Dy/D. ***Note*** On the above web site it shows the correct matrix, but then gives the wrong formula later for Dx. It does matter in which order you write the matrix.***

For these formulas a1 = 5, b1 = 7, c1 = -3, a2 = 2, b2 = 3 and c2 = -1. Solving for D=5*3 - 2*7=1, Dx = -3*3 - -1*7 = -2 and Dy = 5*-1 - 2*-3 = 1. x = -2/1 = -2 and y = 1/1 = 1

Please note that you get the same answer as below and that when you plug the answers for x and y back in the formulas you get the correct answer in both cases. Good luck and study hard.

Let's use formulas 1 for this example. Method 1) Using the first formula 5x + 7y = -3 isolate one of the variables. In this instance the formula would become x = (-3 - 7y)/5. Now plug this back in the second formula for x so that the formula now becomes 2((-3 - 7y)/5) + 3y = -1. Working through this becomes -6/5 -14y/5 + 3y = -1. Multyiplying by five to reduce the fractions to integers it becomes -6 -14y + 15y = -5. Adding like variables this becomes -6 + y = -5. Moving the interger to the right side gives y = 1. Now go back and plug 1 in for y and determine the x. 5x + 7*1 = -3. 5x = -10. x = -10/5 or x = -2. Now go back and prove it in botch equations. 5(-2) + 7(1) = -3. -10 + 7 = -3. And in the second equation 2(-2) + 3(1) = -1. -4 + 3 = -1.

Method 2) Use a common multiplier in both equations so that one variable is the same in both equations. 2x and 5x can become 10 or 3y and 7y can become 21. Multiply the first equation by 2 and the second by 5. This gives:

10x + 14y = -6

10x + 15y = -5

Use subtraction rules on the two formulas. Subtract the first formula from the second. 10x - 10x = 0; 14y - 15y = -1y; -6 -(-5) = -1. The new formula is: -y = -1 or y = 1. Now plug y = 1 back in the formula as was done above and determine x.

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• cramer;s rule is my fav and its easy anyways some find it difficult i will explain u hope u understand it... it;s according to my class method in easier way.. i ll try my level best to make u understand!!! so lets start

1. at first take only the x and y terms i.e

let det = | 5 7 |

| 2 3 |

now calculate the value of det by determinant method i.e

5*3 - 7*2= 15-14=1... therfore det = 1

now u have to calculate the value of det (x) that can be done by replacing the value of x by constant terms which is on the right side and keeping y terms in thier original place i.e

let det(x)= | -3 7 |

| -1 3| now do as before i.e (-3)*3 - 7*(-1)= -9+7= -2

the value of det(x) is -2 similarly find the value of det(y) that can be done by replacing the values of y by constant terms and keeping the value of x in thier original place i.e

let det(y)= |5 -3 |

|2 -1 | now again same method i.e 5*(-1) -(-3)*2= -5+6= 1..the value of det(y) is 1..

now u should find the value of x and y that can be done by this simple formulae i.e

x=det/det(x) ; y = det/det(y)

x=1/-2 ; y = 1/1

therefore x= -2 and y=1

u can also check the answer by replacing x and y terms which u got as answer in the two equations if u get the right side no then its right... i.e 5(-2)+7(1)= -10+7= -3

2(-2)+ 3(1)= -4+3 = -1

similarly u can solve other... good luck

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• Anonymous
4 years ago

you employ Cramer's rule to remedy simultaneous equations, utilising determinants. pcs use Cramer's rule to attend to somewhat complicated issues. concern; 3x + 2y = sixteen 4x + 5y = 33 x = | sixteen 2 | | 33 5 | ___________ | 3 2 | | 4 5 | The denominator is the determinant of the the coefficients of the variables. The numerator is a similar difficulty, yet with the constants changing the coefficients of the variable you're fixing for. in case you have achieved those in the previous, this could make experience; in any different case no risk!

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• 5x + 7y = -3

2x + 3y = - 1

D = 5 7 = 15-14 = 1

2 3

Dx = - 3 7 = -9+7= -2

- 1 3

Dy = 5 - 3 = -5+6 = 1

2 -1

x = Dx / D = - 2/1 = - 2

y = Dy / D = 1/1 = 1

Similarly, you can solve remaining problems

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• May I point out that Cramer's rule is in practice only used to prove that some system has solutions.

There are much much better methods for solving systems than Cramer's rule.

As an exercise for "How to solve lineair equatins" this method is not very usefull. I wonder if you know what exactly you are solving

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• Here is an excelent description of Cramer's rule and it's use to solve systems of linear equations.

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