Precalculus, Chapter 7, 7.3, Section 7.3, Problem 31

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) .