SOLVE SYSTEMS OF EQUATIONS - Bloomfield College



SOLVE SYSTEMS OF EQUATIONS

A system of equations refers to "n" unknowns for "n" equations.

Example:        3x - y = 8

                       x + y = 8

Solutions to a system of equations is an ordered pair (x, y) in which when you substitute the values for x and y into both equations it yields a true statement for both equations.

Example:   Is (1, 2) a solution of the system,  y - x =1  and  2x + y = 4?

substitute (1, 2) into y - x = 1:      2 - 1 = 1

                                                     1 = 1          true statement

substitute (1, 2) into 2x + y = 4:    2(1) + 2 = 4

                                                         4 = 4      true statement

Therefore, (1,2) is a solution to the system.

There are many different methods to solve systems of equations.  Three methods will be demonstrated.

SOLVE SYSTEMS OF EQUATIONS BY GRAPHING METHOD.

Solve   x + y = 8

           x - y = 2

Both equations are linear.  Therefore, both their graphs will be lines.  That is why a graph can be used to find the solution.

STEPS:

1)    Graph both linear equations on the same coordinate plane.

2)    The point, (x, y), at which the two lines intersect is the solution to the system of equation.

NOTE:   There are three possible types of solutions to a system.

1)  There is only one point of intersection.  This occurs when the two lines intersect at one point.

2)  There is no solution.  This occurs when the two lines are parallel and don't intersect at any point.

3)  There is an infinite number of solutions.  This occurs when the two lines are graphed on top of each other and so they intersect at every point on the line.

SOLVE SYSTEMS OF EQUATIONS BY SUBSTITUTION METHOD

STEPS:

1)   Choose one equation to solve for one variable. 

2)   Substitute the chosen equation into the unused equation and solve.

3)   Substitute the value found in step 2 into any one of the original equations and solve for the remaining variable.

4)   The solution can be written as an ordered pair, (x, y).

NOTE:   There are three possible types of solutions to a system.

1)  There is a unique solution.  This occurs when a value for both variables are found.

2)  There is no solution.  This occurs when both variables cancel out and a false statement remains.

3)  There is an infinite number of solutions.  This occurs when both variables cancel out and a true statement remains.

EXAMPLES:

1)   Solve    x = 2y

                  3x - y = -5

step 1:    x = 2y is the chosen equation and is already solved for "x"

step 2:    substitute into unused:     3(2y) - y = -5

              solve:             6y - y = -5

                                      5y = -5

                                        y = -1

step 3:    substitute y = -1 into x = 2y   then  x = 2(-1)

                                                                     x = -2

The solution is (-2, -1); a unique solution.

2)   Solve   x - y = -4

                 7x + 5y = -28

step 1:     choose x - y = -4 and solve for x;   x = y - 4.

step 2:     substitute into unused:     7(y - 4) + 5y = -28

               solve:                              7y - 28 + 5y = -28

                                                         12y - 28 = -28

                                                             12y = 0

                                                               y = 0

step 3:      substitute y = 0 into  x = y - 4  then   x = 0 - 4

                                                                           x = -4

The solution is (-4, 0); a unique solution.

3) Solve    [pic]

             2x +  y = 9       

The top equation will be multiplied by the LCD of 12 to yield  4x - 3y = 3.

step 1:    2x + y = 9  is the chosen equation and solve for y, then y = 9 - 2x.

step 2:    substitute into unused:       4x - 3(9 - 2x) = 3

              solve:                                4x - 27 + 6x = 3

                                                           10x - 27 = 3

                                                              10x = 30

                                                                 x = 3

step 3:    substitute x = 3 into y = 9 - 2x  then  y = 9 - 2(3)

                                                                      y =  9 - 6

                                                                      y = 3

The solution is (3, 3); a unique solution.

4)   Solve      3x - 15y = 15

                         x = 5y + 5

step 1:    x = 5y + 5 is the chosen equation and already solved for x.

step 2:    substitute into unused:    3(5y + 5) - 15y = 15

              solve:                             15y + 15 -15y = 15

                                                            15 = 15

The variable canceled out and 15 = 15 is a true statement.  Therefore, the solution is infinite number of solutions.

SOLVE SYSTEMS OF EQUATIONS BY ELIMINATION METHOD

STEPS:

1)   Choose one variable to eliminate when like terms are collected.  

2)   Multiply one or both equations by values so that the coefficients of the chosen variable are additive inverses.  

3)   Collect like terms and solve.

4)   Substitute the value found in step 4 into any one of the original equations and solve for the other variable.

5)   The solution can be written as an ordered pair, (x, y).

NOTE:   There are three possible types of solutions to a system.

1)  There is a unique solution.  This occurs when a value for both variables are found.

2)  There is no solution.  This occurs when both variables cancel out and a false statement remains.

3)  There is an infinite number of solutions.  This occurs when both variables cancel out and a true statement remains.

EXAMPLES:

1)   Solve      3x - y = 4

                    2x + y = 1

step 1:    The variable y has been chosen to eliminate.

step 2:    There is no need to multiply since -y and +y are additive inverses already.

step 3:     collect like terms:    5x = 5

               solve:                       x = 1

step 4:     substitute x =1 into  2x + y = 1,  then   2(1) + y = 1

                                                                            2 + y = 1

                                                                                y = -1

The solution is  (1, -1); a unique solution.

2)     Solve    6x - 10y = 5

                     2x - 3y = 1

step 1:     The variable x is chosen to eliminate.

step 2:     Multiply the bottom equation by -3 so the additive inverse is 6x and -6x.

                        6x - 10y = 5                6x - 10y = 5

                      -3(2x - 3y = 1)             -6x + 9y = -3

step 3:     collect like terms:         -y = 2

                solve:                            y = -2

step 4:     substitute y = -2 into  2x - 3y = 1,  then   2x - 3(-2) = 1

                                                                             2x + 6 = 1

                                                                                2x = 7

                                                                                 x = 7/2

The solution is (7/2, -2); a unique solution.

 3)    Solve       5x - 4y = 1

                        2x + 3y = 5

step 1:    The variable y is chosen to eliminate.

step 2:    Multiply the top equation by 3 and the bottom equation by 4 so the additive inverses are -12y and 12y.

                 3(5x - 4y = 1)          15x - 12y = 3

                  4(2x + 3y = 5)         8x + 12y = 20

step 3:     collect like terms:           23x = 23

               solve:                               x = 1

step 4:     substitute x =1 into 2x + 3y = 5,  then  2(1) + 3y = 5

                                                                            2 + 3y = 5

                                                                                3y = 3

                                                                                  y = 1

The solution is (1, 1); a unique solution.

4)   Solve         2x - 8y = 9

                         x  - 4y = -6

step 1:    The variable x is chosen to eliminate.

step 2:    Multiply the bottom equation by -2 so the additive inverses are 2x and -2x.

                        2x - 8y = 9                        2x - 8y = 9

                       -2(x  - 4y = -6)                -2x  + 8y = 12

step 3:     collect like terms:                         0 = 21

Both variables cancelled out and 0 = 21 is a false statement.  Therefore, the solution is no solution.

Systems of equations prove helpful when solving word problems.

WORD PROBLEM EXAMPLES:

1)  Kerry asked a bank teller to cash a $390 check using $20 bills and $50 bills.  If the teller gave her a total of 15 bills, how many of each type of bill did she receive?

Let    x = amount of $20 bills

         y = amount of $50 bills

Set up a system of two equations.

       total number of bills is  15            then       x + y = 15

       total amount of money is $390     then    20x + 50y = 390

solve using elimination method:     

  -20(x + y) = 15           -20x - 20y = -300

  20x + 50y = 390          20x + 50y =  390        

                                               30y = 90

                                                  y = 3

substitute y = 3 into x + y = 15,  then   x + 3 = 15, then  x = 12                                 

The teller gave Kerry 12-$20 bills and 3-$50 bills.

2)   A financial planner wants to invest $8000, some in stocks earning 15% annually and the rest in bonds earning 6% annually.  How much should be invested at each rate to get a return of $930 annually from the two investments?

Let   x  =   amount invested at 15%

        y  =   amount invested at  6%

Set up a system of equation.

 total amount to be invested is $8000        x + y = 8000

 total amount of interest is $930              0.15x + 0.06y = 930

solve using substitution method:

x = 8000 - y   then    0.15(8000-y) + 0.06y = 930

                                   1200 - 0.15y + 0.06y = 930

                                            -0.09y = -270

                                                 y = 3000

substitute y = 3000 into x = 8000 - y,  then x = 5000.

$5000 should be invested at the 15% rate and $3000 should be invested at the 6% rate.

3)   Matthew flew his ultra light plane to a nearby town against a head wind of 

15 km/h in 2 hours 20 minutes.  The return trip with a tail wind of 15 km/h took 

1 hour 24 minutes.  Find the plane's air speed and the distance to the nearby town.

Definitions:

The air speed refers to the speed of an aircraft in still air  (the speedometer reading).

The wind speed refers to the speed of the wind.

The head wind refers to the wind blowing against the aircraft.

The tail wind refers to the wind blowing with the aircraft.

Recall the formula:    Distance = Rate x Time

NOTE:   The rate of a plane: the air speed plus tail wind

                                             the air speed minus the head wind

Let   d =   the distance to the town and return trip.

        a =   the air speed

set up the equation:

to the town:            d  =   2 1/3 (a - 15)

return trip:             d  =   1 2/5 (a +  15)

solve by substitution:

         7/3(a - 15) = 7/5(a + 15)

           7/3a - 35 = 7/5a + 21

                  14/15a = 56

                       a = 60

The air speed is 60 km/h.

To find the distance:     d = 7/3(60-15)

                                     d = 105

The distance is 105 km.

 

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