Sunday 21 February 2010

Simultaneous Equations

Simultaneous equations - two or more equations that have one or more points (solutions) in common.


There are three ways to solve simultaneous equations: graphing, substitution and elimination.

Graphing:

To solve for a common solution to two equations using graphing, you graph each line either on paper or on a graphing calculator.

The point at which they intersect is the solution to both equations.


Example: Graph the solution for
y = x + 1 and y = (-1/2)x + 4


The solution is the point at which the lines cross.


Some lines have more than one solution.


y = 2x - 2 and 3y = 6x - 6


Lines with the same slope and same y-intercept will be graphed one on top of the other so that all points coincide. All points on the lines are solutions (
infinite solutions).



Lines with the same slope but different y-intercepts are parallel and have no points in common. In this situation there is no solution.


Substitution:

To use substitution you must have at least one equation in simplest form that is equal to one of the variables
.

Example:

x + 2y = 8
and y = x + 1

Since the second equation is in simplest form (y =), we will use it to substitute.


If y =
x + 1 we can substitute the value for y into the other equation:

x + 2(
x + 1) = 8

Simplify using distribution:


x + 2x + 2 = 8

Combine like terms:


3x + 2 = 8


Simplify:


3x + 2 - 2 = 8 - 2


3x = 6


Divide both sides by 3.


x = 2


Now we need to solve for y too!


Use either of the original equations: x + 2y= 8 and y = x + 1


y = 2 + 1


y = 3


The solutions to this system of equations is
( 2, 3)




Elimination:

In this form we will eliminate one of the variable to solve for one at a time, similar to substitution but using addition and subtraction.


We will start with the same equations we used in the first two forms but written in Standard Form:


y = x + 1 becomes -x + y = 1


y = -1/2x +4 becomes 2y = -x = 8 when we clear the fraction,

then x + 2y = 8 in Standard Form.


Line the two equations up so that like terms are in a column:


-x + y = 1
x + 2y = 8


Combine like terms vertically.


-x and x cancel each other out eliminating x from the equation.


y + 2y = 3y


1 + 8 = 9


3y = 9


y = 3


Substitute the value for y back into either original equations:

-x + y = 1 or x + 2y = 8

x + 2(3) = 8


x + 6 = 8


x + 6 - 6 = 8 - 6


x = 2


The solution for these two equations is
(2, 3).



For more complicated equations we may need to multiply one of the equations to make the coefficient inverse.


2x + 3y = 11

2x + y = 3


We can eliminate the x by multiplying the second equation by -1


2x + 3y = 11

-2x - y = - 3


The x is eliminated leaving:


3y = 11

- y = - 3

Combining those terms gives us


2y = 8


y = 4


Substituting y back in either of the original equations:


2x + 3(4) = 11


2x + 12 = 11

2x + 12 - 12 = 11 - 12


2x = -1


x = - 1/2

Solution (-1/2, 4)



Some equations are even more complicated and you must multiply to get one variable ready for elimination.


2x + 3y = 7

3x - y = 5


Multiply the second equation by 3 to eliminate the y.


2x + 3y = 7

9x -3y = 15


y is eliminated leaving

11x = 22


x = 2


Solve for y.


2(2) + 3y = 7


4 + 3y = 7


4 - 4 + 3y = 7 - 4


3y = 3


y = 1

Solution (2, 1)



For even more complicated equations you may need to multiply both equations.

5x + 2y = 1

2x - 3y = -11


We can multiply the first equation by
2 and the second equation by -5.

2(5x + 2y = 1)

-5(2x - 3y = -11)


10x + 4y = 2

-10x + 15y = 55


The x is eliminated leaving:


4y = 2

15y = 55


19y = 57


divide by 19

y = 3

Substitute the value for y back in either original equation.


5x + 2y = 1


5x + 2(3) = 1


5x + 6 = 1


5x + 6 - 6 = 1 - 6


5x = -5


x = -1

Solution
(-1, 3)

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