what are the three methods used to solve a system of linear equations?

I can think of at least four ways:
1) elimination (add or subtract the equations to eliminate a variable)
2)  substitution (rearrange one equation to get one equation by itself, then substitute into the other equation
3) Kramer's Rule (matrix operations)
4) Graphically (look for the intersection)

Yes, a system can have more than one solution-- if the lines are all the same line, there are an infinite number of solutions (every point on the line is a solution).  Algebraically, you end up with something that looks like 0 = 0 or 3 = 3.

A system may also have no solutions (if the lines are parallel, or if more than two lines intersect do not intersect in one point).  Algebraically, you end up with something that looks like 0 = 3, or 9 = 12.

The way the equations are presented often determine which is easier.  ex:
y = 3x - 5
2x + 3y = 18 ==> since the first equation is already solved for y, this would be easier to use substitution
(the second equation would become : 2x + 3(3x - 5) = 18

4x - 5y = -8
2x + 5y = 26 ==> would be easier to solve by elimination, since adding the equtions would yield 6x = 18

note: sometimes, if the first solution you calculate is a "messy fraction", it would then be easier to go back and eliminate the other variable rather than substitute the fraction back in to solve for the other

However, done properly, every method will yield the same solution.  Graphically (by hand) is the trickiest to arrive at the proper answer-- especially if it is not an integer, or the coefficients are not integers.  A calculator takes care of this.

The three methods are:
1. elimination
2. substitution
3. Cramer's Rule & matrices

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Examples:
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ELIMINATION EXAMPLE
~~~~~~~~~~~~~~~~~~~

You want to multiply the equations if needed so that when you add them together, one of the variables will cancel.  Continue solving it and plugging the value of the solved variable into one of the equations to find the value of the missing variable.

3x - 2y = 22
2x + 3y = 19

Multiply the first equation by 3.
3(3x - 2y) = 3(22)
3(3x) + 3(-2y) = 66
9x - 6y = 66

Multiply the second equation by 2.
2(2x + 3y) = 2(19)
2(2x) + 2(3y) = 38
4x + 6y = 38

Your two equations are now:
9x - 6y = 66
4x + 6y = 38

Add them together.
9x - 6y = 66
4x + 6y = 38
------------------------
13x = 104

Divide both sides by 13.
13x / 13 = 104 / 13
x = 8

Plug this back into one of the equations and solve for y.
3x - 2y = 22
3(8) - 2y = 22
24 - 2y = 22
24 - 2y - 24 = 22 - 24
-2y = -2
-2y / -2 = -2 / -2
y = 1

ANSWER: (8, 1)

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SUBSTITUTION EXAMPLE
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Substitution involves solving for one of the terms and plugging into the other other equation, solving it, and plugging the value of the solved variable into one of the equations to find the value of the missing variable.

3x - 2y = 22
2x + 3y = 19

Solve the first one for x.
3x - 2y = 22
3x - 2y + 2y = 22 + 2y
3x = 22 + 2y
3x / 3 = (22 + 2y) / 3
x = (22 + 2y) / 3

Substitute x with (22 + 2y) / 3 in the second equation.
2x + 3y = 19
2[(22 + 2y) / 3] + 3y = 19
[2(22 + 2y) / 3] + 3y = 19

Multiply both sides by the LCD 3.
3{[2(22 + 2y) / 3] + 3y} = 3(19)
3[2(22 + 2y) / 3] + 3(3y) = 57
2(22 + 2y) + 9y = 57
2(22) + 2(2y) + 9y = 57
44 + 4y + 9y = 57
44 + 13y = 57
44 + 13y - 44 = 57 - 44
13y = 13
13y / 13 = 13 / 13
y = 1

Plug this into one of the equations and solve for x.
3x - 2y = 22
3x - 2(1) = 22
3x - 2 = 22
3x - 2 + 2 = 22 + 2
3x = 24
3x / 3 = 24 / 3
x = 8

ANSWER: (8, 1)

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CRAMER'S RULE & MATRICES EXAMPLE
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Cramer's Rule involves setting up matrices for determinants and the other variables and evaluating them.

The matrix for D is:
......[3   -2]
D = [2   3]

Evaluate for D.
......[3   -2]
D = [2   3]

D = (3)(3) - (2)(-2) = 9 - -4 = 13

The matrix for Dx is:
........[22   -2]
Dx = [19   3]

Evaluate for Dx.
........[22   -2]
Dx = [19   3]

Dx = (22)(3) - (19)(-2) = 66 - -38 = 104

The matrix for Dy is:
........[3   22]
Dy = [2   19]

Evaluate for Dy.
........[3   22]
Dy = [2   19]

Dy = (3)(19) - (2)(22) = 57 - 44 = 13

Evalute for x and y.
x = Dx / D = 104 / 13 = 8
y = Dy / D = 13 / 13 = 1

ANSWER: (8, 1)

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Is it possible for a system of linear equations to have more than one answer? How and Why?

Yes, it is possible.  If one of the variables in one of the equations has a power of a multiple of 2, the equation can have more than one answer because the answer can be positive or negative.  The square root of a term with an even exponent yields two results.

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Are there times when one method of solving is better than other methods?

I prefer using Cramer's Rule and matrices if there are 3 or more variables to avoid having to keep track of all the of modified equations while using one of the other methods.

The three methods are the elimination method, the substitution method, and matrices.  I know absolutely nothing about matrices.  The substitution method is when you manipulate one equation so that a variable has a value, then you use that value and substitute it for the variable in the other equation.  For example:

2x+y=16
x-3y= -20

I might take the first equation and get the y by itself:
2x+y=16 subtract 2x from each side:
y=(16-2x) then use that as the value for y in the second equation:
x-3(16-2x)= -20 and distribute the -3 to the terms in the parentheses:
x-48+6x=-20 combine like terms:
7x-48= -20 add 48 to each side
7x=28 divide by 7
x=4

Then use that value for x in one of my original equations and solve for y:
2x+y=16 substitute
2(4)+y=16 multiply
8+y=16 subtract 8 from each side
y=8

The elimination method involves manipulating one or both equations to get the coefficients of one of the variables to be the same but opposite in value, so that when you add the equations vertically, the variables cancel out:

2x+3y=32
5x-2y= 4

I made this example a little bit complicated. None of the coefficients of x or y are common, so we're going to have to multiply both equations by something.  If I want to eliminate the y, I have to multiply the first equation by 2 and the second one by 3 so that I have coefficients in front of the y's of 6 and -6.  Or, I can multiply the first equation by 5 and the second one by -2 so that the x's have coefficients of 10 and -10.  Let's eliminate the y's:

2x+3y=40 Multiply this equation by 2:
2(2x+3y)=2(32)
4x+6y=64

5x-2y= 4 Multiply this equation by 3:
3(5x-2y)=3(4)
15x-6y=12

Now, write the new equations, one on top of the other:

4x+6y=64
15x-6y=12
----------------- add/subtract vertically
19x=76  divide both sides by 19
x=4

And again, take that value back to one of the original equations and solve for y.

When you have a system of equations, you have to check your solutions in all of the equations.  Sometimes, they don't solve all of the equations and when that happens, your solutions are not the solution to the system, and the solution to the system is the empty set.

For the third method, matrices, here's a link that I hope will explain how to do it.  I never learned it, but one of these days I plan to:

http://www.purplemath.com/modules/matrices.htm

wpf.

Hi,

You can solve a system of equations by:

1) Graphing

Solve both linear equations for y to get slope-intercept form, y = mx + b. (If there is no y, solve for x to get a vertical line.) Graph the 2 lines. Their point of intersection is the solution to the system.

y = -x + 6
y = x + 2

\| . . . . /
 |\ . . ./
 | .\ ./
 | . X <=(2,4) is the graphical solution
 | / . .\
/| . . . .\

2) Substitution

Solve for one variable in either equation. If possible, solve for a variable with a coefficient of 1 or -1. Substitute the expression you got into the other equation in place of the letter you got alone. Solve the equation for that variable. Plug the answer back into either equation and solve for the second variable.

2x + 3y = 12
x + 5y = -1

Solve the second equation for x.
x = -5y - 1

Substitute -5y - 1 in place of x in the first equation.
2(-5y - 1) + 3y = 12
-10y - 2 + 3y = 12
-7y - 2 = 12
-7y = 14
y = -2

If y = -2 and x = -5y - 1, then:
x = -5(-2) - 1
x = 10 - 1
x = 9

(9,-2) is the answer.

3) Elimination or Addition/Subtraction

Choose which letter to eliminate. Multiply either or both equations so that that letter has opposite coefficients, like 12x and -12x. Add the equations together. Solve the resulting equation and plug the answer back into either equation and solve for the second variable.

4x - 7y = 3
3x + 2y = 53

Eliminate y by multiplying the first equation by 2 and the second equation by 7.

2(4x - 7y = 3)
7(3x + 2y = 53)

8x - 14y = 6
21x + 14y = 371
------------------------
29x = 377
x = 13

If x = 13 and 4x - 7y = 3, then:
4(13) - 7y = 3
52 - 7y = 3
-7y = -49
y = 7

(13,7) is the answer

4) Determinants

Write both equations in the form ax + by = c.

Take these equations:

ax + by = c
dx + ey = f

into these determinants to solve:

|c b|
|f e|
------ = x
|a b|
|d e|

|a c|
|d f|
------ = y
|a b|
|d e|

Each determinant's value is found by taking the product of the first diagonal minus the other diagonal, like this:

|a b| = ae - bd
|d e|

Then divide or reduce each fraction.

5x + 8y = 21
2x - 4y = 3

This gives the determinants:

|21 8|
|3 -4|
-------- = x
|5. 8|
|2 -4|

|5 21|
|2 . 3|
-------- = y
|5. 8|
|2 -4|

Solve the determinants.

|21 8|
|3 -4| = 21(-4) - 8(3) . . -84 - 24 . . -108
-------- . ----------------- = ------------ = ------- = 3 = x
|5. 8| = 5(-4) - 8(2) . . . -20 - 16 . . -36
|2 -4|

|5 21|
|2 . 3| . .5(3) - 21(2) . .15 - 42 . . . -27 . . .3
-------- = --------------- = ------------- = ------ = --- = y
|5. 8| =  5(-4) - 8(2) . . -20 - 16 . . -36 . . . 4

(3, 3/4) <==ANSWER


|2 -4|

5) Augmented Matrices

Put 2 equations in a 2 x 3 matrix [A] on a graphing calculator. Using the calculator, find the matrix command rref([A]) and solve to get the reduced row echelon form to get your solutions. This can be done by hand but is longer and more tedious.

3x + 4y = 8
5x + 6y = 6

[3 . 4 . 8]
[5 . 6 . 6] = [A]

rref([A]) =

[1 . 0 .-12]
[0 . 1 . 11]

(-12,11) is the answer.


Graphing can be nice if the answer is integer values and not too large, but if the answer is a decimal or fraction or is a very large number, graphing is not always a good way to solve a system.

Substitution is easy if a variable has a coefficient of 1 or -1, but it is harder if you must divide by a number that creates fractions. This means your equation may have fractions that make it harder to solve.

Addition/subtraction or elimination is frequently one of the easier ways to solve a system, with very few drawbacks.

Determinants are not hard, but care must be taken with setup.

Augmented matrices are easy if you have a graphing calculator  and know how to use it. It is especially nice if you have 3 equations with 3 unknowns.

I hope that helps!! :-)