Saturday, April 4, 2015

`2x + y + 3z = 1, 2x + 6y + 8z = 3, 6x + 8y + 18z = 5` Solve the system of linear equations and check any solutions algebraically.

EQ1:  `2x+y+3z=1`


EQ2:  `2x+6y+8z=3`


EQ3: `6x+8y+18z=5`


To solve this system of equation, let's apply elimination method. In elimination method, a variable/variables should be removed in order to come up with another equation.


In this system of equation, let's eliminate variable x using EQ1 and EQ2.


`2x + y + 3z=1`


`2x +6y + 8z=3`


To do so, multiply EQ2 by -1.


`-1 * (2x + 6y+8z)=3*(-1)`


`-2x - 6y-8z=-3`


And add this to EQ1.


 ...

EQ1:  `2x+y+3z=1`


EQ2:  `2x+6y+8z=3`


EQ3: `6x+8y+18z=5`


To solve this system of equation, let's apply elimination method. In elimination method, a variable/variables should be removed in order to come up with another equation.


In this system of equation, let's eliminate variable x using EQ1 and EQ2.


`2x + y + 3z=1`


`2x +6y + 8z=3`


To do so, multiply EQ2 by -1.


`-1 * (2x + 6y+8z)=3*(-1)`


`-2x - 6y-8z=-3`


And add this to EQ1.


           `2x + y + 3z = 1`


`+`     `-2x - 6y -8z=-3`


`-----------------`


               `-5y - 5z=-2 `        Let this be EQ4.


Eliminate x variable again to come up with another equation that has y and z only. Use EQ1 and EQ3.


`2x+y+3z=1`


`6x+8y+18z=5`


So multiply EQ1 by -3.


`-3(2x+y+3z)=1*(-3)`


`-6x-3y-9z=-3`


Then, add this to EQ3.


        `-6x - 3y-9z=-3`


`+`       `6x + 8y+18z=5`


`-----------------`


                    `5y + 9z = 2 `          Let this be EQ5.   


So the two new equations are:


EQ4:  `-5y-5z = -2`


EQ5:   `5y+ 9z = 2`


Since there two equations still contain more than two variables, let's eliminate another variable. Let's remove the y. To do so, add these two equations.


        `-5y - 5z =-2`


`+`       `5y +9z=2`


`------------`


                  `4z=0`


Then, solve for z.


`4z=0`


`z=0/4`


`z=0`


Then, plug-in this to either EQ4 or EQ5 to get the value of y.


`5y+9z=2`


`5y+9(0)=2`


`5y=2`


`y=2/5`


Now that the values of y and z are known, solve for x. To do so, plug-in these two values to either EQ1, EQ2 or EQ3.


`2x+y+3z=1`


`2x+2/5+3*0=1`


`2x+2/5=1`


`2x=1-2/5`


`2x=3/5`


`x=(3/5)/2`


`x=3/10`


Therefore, the solution is `(3/10, 2/5,0)` .


-6x-3y-9z=-3

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