Thursday, June 15, 2017

`3x - 5y + 5z = 1, 5x - 2y + 3z = 0, 7x - y + 3z = 0` Solve the system of linear equations and check any solutions algebraically.

EQ1:  `3x-5y+5z=1`


EQ2:  `5x-2y+3z=0`


EQ3:  `7x-y+3z=0`


To solve this system of equations, let's use elimination method. In elimination method,  a variable or variables should be eliminated to get the value of the other variable.


Let's eliminate y by multiply EQ3 by -5. Then add it with EQ1.


EQ1: `3x-5y+5z=1`


EQ3: `(7x-y+3z=0)*(-5)`



             `3x-5y+5z=1`


`+`   `-35x+5y-15z=0`


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


               `-32x - 10z=1`    ...

EQ1:  `3x-5y+5z=1`


EQ2:  `5x-2y+3z=0`


EQ3:  `7x-y+3z=0`


To solve this system of equations, let's use elimination method. In elimination method,  a variable or variables should be eliminated to get the value of the other variable.


Let's eliminate y by multiply EQ3 by -5. Then add it with EQ1.


EQ1: `3x-5y+5z=1`


EQ3: `(7x-y+3z=0)*(-5)`



             `3x-5y+5z=1`


`+`   `-35x+5y-15z=0`


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


               `-32x - 10z=1`       Let this be EQ4.         


Eliminate y again by multiplying EQ3 by -2. And add it with EQ2.


EQ2: `5x-2y+3z=0`


EQ3: `(7x-y+3z=0)*(-2)`



              `5x - 2y+3z=0`


`+`      `-14x+2y-6z=0`


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


                   `-9x-3z=0`


                        `3x+z=0`       Let this be EQ5.


Then, consider two new equations.


EQ4:  `-32x-10z=1`


EQ5: `3x + z=0`


Eliminate the z in these two equations by multiplying EQ5 with 10. And,  add them.


     `-32x-10z=1`


`+`     `30x + 10z=0`


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


                      `-2x=1`


Then, isolate the x.


`(-2x)/(-2)=1/(-2)`


`x=-1/2`


Plug-in this value of x to either EQ4 or EQ5.


EQ5: `3x+z=0`


`3(-1/2)+z=0`


And, solve for z.


`-3/2+z=0`


`-3/2+3/2+z=0+3/2`


`z=3/2`


Then, plug-in the values of x and z to either of the original equations.


EQ3: `7x-y+3z=0`


`7(-1/2)-y+3(3/2)=0`


`-7/2-y+9/2=0`


`1-y=0`


`1-1-y=0-1`


`-y=-1`


`(-y)/(-1)=(-1)/(-1)`


`y=1`


To check, plug-in the values of x, y and z to the three original equations. If the resulting conditions are all true, then, it verifies it is the solution of the given system of equations.


EQ1: `3x-5y+5z=1`


`3(-1/2)-5(1)+5(3/2)=1`


`-3/2-5+15/2=1`


`-3/2-10/2+15/2=1`


`2/2=1`


`1=1`     `:. True`



EQ2: `5x-2y+3z=0`


`5(-1/2)-2(1)+3(3/2)=0`


`-5/2-2+9/2=0`


`-5/2-4/2+9/2=0`


`0/2=0`


`0=0`      `:. True`



EQ3: `7x-y+3z=0`


`7(-1/2)-1+3(3/2)=0`


`-7/2-1+9/2=0`


`-7/2-2/2+9/2=0`


`0/2=0`


`0=0`     `:. True`



Therefore, the solution is   `(-1/2,1,3/2)` .

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