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

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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)`



`+`   `-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`



                        `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.


`+`     `30x + 10z=0`



Then, isolate the x.



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

EQ5: `3x+z=0`


And, solve for z.




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

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








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`





`1=1`     `:. True`


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





`0=0`      `:. True`


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





`0=0`     `:. True`


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

Approved by eNotes Editorial Team
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