Which Expression is Equivalent to 5x+10y-15

Which Expression is Equivalent to 5x+10y-15

Systems of Linear Equations


Often it is necessary to look at several functions of the same independent variable.  Consider the prior example where x, the number of items produced and sold, was the independent variable in three functions, the cost function, the revenue function, and the profit function.

            In general there may be:

                        n equations

                        v variables

Solving systems of equations

            There are four methods for solving systems of linear equations:

                        a.  graphical solution

                        b.  algebraic solution

                        c.  elimination method

                        d.  substitution method

Graphical solution

Example 1

            given are the two following linear equations:

                        f(x)  =  y  = 1 +  .5x

                        f(x)  =  y  = 11 –  2x

Graph the
first
equation by finding two data points.  By setting first x and then y equal to zero it is possible to find the y intercept on the vertical axis and the x intercept on the horizontal axis.

            If x = 0, then  f(0)  =  1  + .5(0)  =  1

            If y = 0, then  f(x)  =  0  = 1  +  .5x

                                                -.5x  =  1

                                                     x  =  -2

            The resulting data points are  (0,1)  and  (-2,0)

Graph the
second
 equation by finding two data points.  By setting first x and then y equal to zero it is possible to find the y intercept on the vertical axis and the x intercept on the horizontal axis.

            If x = 0, then  f(0)  =  11  – 2(0)  =  11

            If y = 0, then  f(x)  =  0  = 11  –  2x

                                                2x  =  11

                                                     x  =  5.5

            The resulting data points are  (0,11)  and  (5.5,0)

At the point of intersection of the two equations x and y have the same values.  From the graph these values can be read as x = 4 and y = 3.

Example 2

            given are the two following linear equations:

                        f(x)  =  y  = 15 –  5x

                        f(x)  =  y  = 25 –  5x

Graph the
first
equation by finding two data points.  By setting first x and then y equal to zero it is possible to find the y intercept on the vertical axis and the x intercept on the horizontal axis.

            If x = 0, then  f(0)  =  15  – 5(0)  =  15

            If y = 0, then  f(x)  =  0  = 15  –  5x

                                                5x  =  15

                                                     x  =  3

            The resulting data points are  (0,15)  and  (3,0)

Graph the
second equation by finding two data points.  By setting first x and then y equal to zero it is possible to find the y intercept on the vertical axis and the x intercept on the horizontal axis.

            If x = 0, then  f(0)  =  25  – 5(0)  =  25

            If y = 0, then  f(x)  =  0  = 25  –  5x

                                                5x  =  25

                                                     x  =  5

            The resulting data points are  (0,25)  and  (5,0)

From the graph it can be seen that these lines do not intersect.  They are parallel.  They have the same slope.  There is no unique solution.

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Example 3

            given are the two following linear equations:

                        21x – 7y  =  14

                        -15x  +  5y  =  -10

            Rewrite the equations by putting them into slope intercept form.

            The first equation becomes

                        7y  =  -14  +  21x

                          y  =  -2  +  3x

            The second equation becomes

                        5y  =  -10  +  15x

                          y  =  -2  +  3x

Graph either  equation by finding two data points.  By setting first x and then y equal to zero it is possible to find the y intercept on the vertical axis and the x intercept on the horizontal axis.

            If x = 0, then  f(0)  =  -2  +3(0)  =  -2

            If y = 0, then  f(x)  =  0  = -2  + 3x

                                                3x  =  2

                                                     x  =  2/3

            The resulting data points are  (0,-2)  and  (2/3,0)

From the graph it can be seen that these equations are equivalent.  There are an infinite number of solutions.

Algebraic solution

This method will be illustrated using supply and demand analysis.  This type of analysis is derived from the work of the great English economist Alfred Marshall.

Q  = quantity     and  P  =  price

P (s)= the supply function    and P (d)  = the demand function

When graphing price is placed on the vertical axis.  Thus price is the dependent variable.  It might be more logical to think of quantity as the dependent variable and this was the approach used by the great French economist, Leon Walras.  However by convention economists continue to graph using Marshall’s analysis which is referred to as the Marshallian cross.

The objective is to find an equilibrium price and quantity, i.e. a solution where price and quantity will have the same values in both the supply function and the price function.

            QE  = the equilibrium quantity           PE  = the equilibrium price

            For equilibrium
supply  =  demand
or P (s) = P (d)

Given the following functions

            P (s) =  3Q + 10  and            P (d) =  -1/2Q + 80

Set the equations equal to each other and solve for Q.

            P (s)  =  3Q + 10  =   -1/2Q + 80  = P (d)

                        3.5Q  =  70

                            Q  =  20                  The equilibrium quantity is 20.

Substitute this value for Q in either equation and solve for P.

            P (s)  =  3(20) + 10

            P (s) =  70

            P (d)  =  -1/2(20) + 80

            P (d)  =  70                             The equilibrium price is 70.


Elimination method

This method involves removing variables from the equations.  Variables are removed successively until only a single last  variable is left, i.e. until there is one equation with one unknown.  This equation is then solved for the one unknown.  The solution is then used in finding the second to last variable. The procedure is repeated by adding back variables as their solutions are found.

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Example 1

                        2x  +  3y  =  5

                        -5x   – 2y  = 4

Procedure: eliminate y.  The coefficients of y are not the same in the two equations but if they were  it would possible to add the  two equations and the y terms would cancel out.  However it is possible through multiplication of each equation  to force  the y terms to have the same coefficients in each equation.

            Step 1: Multiply the first equation by 2 and multiply the second equation by 3.  This gives

                        4x  +  6y  =  10

                                    -15x   – 6y  = 12

            Step 2: Add the two equations.  This gives

                                    -11x     =     22

                                    x          =      -2

            Step 3: Solve for y in either of the original equations

2(-2)  +  3y  =  5

3y  =  9

y  =  3              or

-5(-2)  – 2y  = 4

10  – 2y  =     4

2y  =     6

y  =  3

Alternate Procedure: eliminate x.  The coefficients of x are not the same in the two equations but if they were  it would possible to add the  two equations and the y terms would cancel out.  However it is possible through multiplication of each equation  to force  the x terms to have the same coefficients in each equation.

            Step 1: Multiply the first equation by 5 and multiply the second equation by 2.  This gives

                        10x  +  15y  =  25

                        -10x   – 4y     = 8

            Step 2: Add the two equations.  This gives

                                    11y      =     33

                                    y          =      3

            Step 3: Solve for x in either of the original equations

2x  +  3(3)  =  5

2x  =  -4

x  =  -2             or

-5x     – 2(3)  = 4

           – 5x =   10

x  =     -2

Example 2

                        2x1  + 5x2  +  7x3  =  2

                        4x1  – 4x2  –  3x3  =  7

                        3x1  – 3x2  –  2x3  =  5

In this example there are three variables: x1, x2, and x3.  One possible procedure is to eliminate first x1,
to eliminate next x2, and then to solve for x3.  The value obtained for x3
is used to solve for x2
and finally the values obtained for x3
and x2
are used to solve for x1.

Procedure     Part A  First eliminate x1.

            Step 1 Multiply the first equation by 2 and subtract  the second equation from the first equation.  This gives

                        4x1  + 10x2  +  14x3  =  4                    first equation

                        4x1  – 4x2  –  3x3        =  7                   second equation

                               14x2   + 17x3     = -3                    second equation subtracted from  the first

            Step 2 Multiply the first equation by 3, multiply the third equation by 2, and subtract  the third  equation from the first equation.  This gives

                        6x1  + 15x2  +  21x3  =  6                    first equation

                        6x1  – 6x2  –  4x3       =  10                   third equation

                               21x2   + 25x3     = -4                    third equation subtracted from  the first

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Procedure     Part B  Second eliminate x2.   From Part A there are two equations left.  From these two equations eliminate x2.

                               14x2   + 17x3     = -3                    first equation

                               21x2   + 25x3     = -4                   second equation

            Step 1 Multiply the first equation by 21, multiply the second equation by 14. and subtract  the second equation from the first equation.  This gives

                           294x2   + 357x3     = -63                  first equation

                           294x2   + 350x3     = -56                  second equation

                                                  7x3  =   -7                 second equation subtracted from first

                                                    x3  =   -1

Part C             Solve for x2
by inserting the value obtained for x3
in either equation from Part B.

                             14x2   + 17(-1)     = -3

                                              1 4x2   =     14

                                                   x2   =     1                or

                             21x2   + 25(-1)    = -4

                                              21x2    =   21

                                                   x2   =     1

Part D             Solve for x1
by inserting the values obtained  x2
andx3
in any of the three original equations.

                            2x1  + 5x2  +  7x3  =  2                    first original equation

                         2x1  + 5(1) +  7(-1)  =  2

                                                2x1    =  4

                                                  x1    =  2                    or

                             4x1  – 4x2  –  3x3  =   7                    second original equation

                         4x1  – 4(1)  –  3(-1)  =  7

                                                4x1    =  8

                                                  x1    =  2             or

                             3x1  – 3x2  –  2x3  =  5                     third original equation

                           3x1  – 3(1)  -2(-1)  =  5

                                                3x1    =  6

                                                  x1    =  2

Substitution method

This involves expressing one variable in terms of another until there is a single equation in one unknown.  This equation is then solved for that one unknown.  The result is then used to solve for the variable which was expressed in terms of the variable whose solution has just been found.

Example

            12x  – 7y  =  106                                 first equation

            8x    +  y    =  82                                 second equation

            Solve the second equation for y and then substitute the value obtained for y in the first equation.

            y  =   82   –   8x                                   second equation solved for y

            12x  – 7(82 – 8x) =  106                     first equation rewritten in terms of x

            12x  – 574   +56x =  106

            68x   =   680

            x    =   10

Substitute the value obtained for x in either of the original equatioins.

            12x  – 7y  =  106                                 first equation

            12(10)  – 7y  =  106

            7y  =  14

            y  =  2

            8(10)    +  y    =  82                            second equation

            y    =  2

[Index]


Which Expression is Equivalent to 5x+10y-15

Sumber: http://www.columbia.edu/itc/sipa/math/systems_linear.html

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