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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