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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