Solving Systems of Equations



Solving Systems of Equations

In order to solve for two variables, you need to have two equations.  If you only have one equation there are an infinite amount of ordered pairs (x,y) that will work.  For example:

4x – 2y = 16 you can have x = 4 and y = 0 (4,0) and (2, -2) and  (0, -4) and an infinite amount of others.  To be able to solve for a single ordered pair, you need a second equation. 

 When we introduce the second equation, we will be able to solve for a single ordered pair that will work in both equations.  There are two ways to solve a system of equations (algebraically and graphically).  We will focus on solving algebraically.  There are two methods of solving algebraically (substitution and elimination).  The key to both of them is changing one (or both) equations so there is only one variable to solve for.   Then you follow all the rules of solving for the one variable.  Then plug the value back into one of the original equations to find the value of the second variable.  Always state your answer as an ordered pair.

SUBSTITUTION       

       Example:  x = 3y + 8

                      5x + 2y = 6            

                   5(3y + 8) + 2y = 6 Substitute 3y +8 for the x in the 2nd equation                             

                   15y + 40 + 2y = 6  Distribute and solve.

                           17y + 40 = 6

                                   17y = -34

                                        y = -2

                                     x = 3(-2) + 8           substitute the value for y back in to find x.

                                        x= -6 + 8

                                         x = 2

                                        (2, -2) Check in BOTH ORIGINAL EQUATIONS!

                                                 

           

Solve each system and check (in both equations):                     

            a) x = 2y + 1                            b) y = 3x + 4                            c) 5x – y = 7

               5x – 6y = 13                             9x + 2y = -37                           4x + 2y = 28

 

            d) x + 3y = 11                          e) 7x + 9y = -74                           f) 10x – y = 1

                6x – 5y = 20                            4x + y = -5                                  8x + 3y = 35

 

 

Solving Systems with Linear Combinations (“Elimination”):

 

Sometimes solving a system of equations using substitution can be very difficult. For these problems we solve using Linear Combinations (or Elimination). With elimination you solve by eliminating one of the variables. This is accomplished by adding the 2 equations together. Before you can add the equations together, you need one of the two variables to have two things:

 

1) Same Coefficient

2) Different Signs (one positive and one negative)

 

When you add terms with the same coefficient and different signs, the term drops out. You then solve for the variable that is left. After you have solved for one variable, you plug the value into one of the original equations and solve for the 2nd variable (just like Substitution). Then, you check the solution in both original equations. The only difference between Substitution and Elimination is how you solve for the 1st variable. After that they are the same.

Examples:

A) Sometimes it works out that the 2 equations already have a variable with the same coefficient and different signs. You can then just add the equations:

 

3x + 4y = 10 (The +4y and -4y cancel out Plug x = -6 in:

5x – 4y = -58 leaving you with just 8x.) 3(-6) + 4y = 10

8x = -48 -18 + 4y = 10

8 8 +18 +18

4y = 28

x = -6 4 4

y = 7

 

Final Solution: (-6, 7) CHECK IN BOTH!!!!

B) Sometimes (usually) the equations do not have same coefficient and different signs, so we have a little bit of manipulating to do.

3x + 8y = 25 With this system, nothing will drop out if we just add the

5x + 4y = 23 equations. So we will multiply the bottom one by (-2).

-2(5x + 4y = 23) Now the y’s have the same coefficient with different signs.

- 10x -8y = -46

3x + 8y = 25 Now plug x = 3 in:

- 10x -8y = -46 3(3) + 8y = 25

- 7x = -21 9 + 8y = 25

-7 - 7 -9 -9

8y = 16

x = 3 8 8

y = 2

Final Solution: (3,2) CHECK IN BOTH!!!!

 

C. Sometimes we need to manipulate both equations. We can do this by “criss crossing the coefficients.”

6x + 7y = 11 This is different than Example B, because no coeffcient

5x – 6y = -50 goes into another evenly.

-5(6x + 7y = 11) You need the negative sign to change the 6x to negative

6(5x – 6y = -50) so the signs will be different.

                                                 You can also use 5 and -6. You can also “criss cross”

the y coefficients.

 

-30x – 35y = -55

30x – 36y = -300 Plug in y = 5

- 71y = -355 5x – 6(5) = -50

-71 -71 5x – 30 = -50

+30 +30

5x = -20

y = 5 5 5

x = -4

Final Solution: (-4, 5) CHECK IN BOTH!!!!

 

Practice:

1) 7x + 3y = 10 2) 11x + 5y = 27 3) 9x + 7y = 126

5x – 6y = 56 4x + 6y = 60 7x – 9y = -32 

 

4) 12x – 5y = 63 5) 5x + 9y = 14 6) 10x – 9y = 36

8x + 3y = 23 6x + 11y = 18 4x + 3y = -12 

 

7) 5x + 6y = 42 8) 7x – 5y = -42 9) 4x – 3y = 19

3x + 14y = 20 8x + 3y = -48 8x + 5y = 159

Solve each system algebraically:

1) 5x -  2y  = -9                        2)  -4x + 2y = -16

    7x + 2y  = -27                            5x – 3y  = 19

3)  x = 2y -6                             4) 5x – 6y = -74                                  

5y –3x = 11                                7x + 5y = 17   

5) 4x – 5 = y                          6) 7x + 4y = -11                                          

    7x + 5y = 83                             5x + 2y = - 13

7) 5x – 6y = -17                       8) x = 6 + 2y               

    3x + 8y = -16                         6x – 5y = 15    

9) 6x + 5y = 23 10) y = 3x + 4

11x + 4y = 6 8x – 9y = 59

11) 12x – 7y = 46 12) 9x – 4y = -88

4x + 3y = -6 2x + 5y = 4

13) 24x – 6y = -66 14) 5x – 6y = 42

12x – 3y = -33 15x – 18y = 54

15) 7x + 6y = -12 16) 13x – 3y = 78

5x + 2y = -20 4x + 6y = -66

17) 2y – 5 = x 18) 3x – 7y = -10

4x – 11y = -38 5x + 12y = -64

19) 6x – 17y = -104 20) 9x – 5y = -43

4x – 7y = -39 3x + 11y = 87

21) 9x = 11y + 25 22) 6y = 5x - 38

5x – 12y = 8 7x + 9y = 1

23) 6x + 5y = 33 24) y = 3x + 5

5x + 37 = 3y 12x – 7y = 1

Answer Key to Algebraic Systems (page 4):

1) (-3,-3) 7) (-4, -.5) 13) many sol. 19) (2.5, 7)

2) (5,2) 8) (0,-3) 14) no sol. 20) (-1/3, 8)

3) (8,7) 9) (-2,7) 15) (-6,5) 21) (4,1)

4) (-4,9) 10) (-5,-11) 16) (3,-13) 22) (4,-3)

5) (4,11) 11) (1.5, -4) 17) (7,6) 23) (-2,9)

6) (-5,6) 12)(-8, 4) 18) (-8, -2) 24) (-4,-7)

1) 6x – 5y = -7 2) 5x + 4y = -69

11x + 5y = 58 5x -7y = 52

3) 6x + 7y = -28 4) 11x – 4y = 53

5x – 14y = -182 7x – 8y = 1

5) 3x – 7y = 42 6) 9x – 4y = 177

2x + 5y = 57 6x – 5y = 111

7) 8x – 11y = 77 8) 13x – 2y = 72

6x + 4y = -28 9x + 5y = -14

9) 12x = 20- 8y 10) 5y = 8x + 97

5x – 6y = -57 10x + 7y = 51

Answer Key for this sheet:

1) (3, 5) 2) (-5, -11)

3) (-14, 8) 4) (7,6)

5) (21,3) 6) (21, 3)

7) (0, -7) 8) (4, -10)

9) (-3,7) 10) (-4, 13)

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