Answer:
[tex]x=4\\y=3\\z=1[/tex]
Step-by-step explanation:
[tex](a) x+y+z=8\\(b)x+3y+3z=16\\(c)2x+y-z=10[/tex]
Let's add equation a and c to make a new equation in which z is not a variable.
[tex]x+y+z=8\\2x+y-z=10[/tex]
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[tex](d)3x+2y=18[/tex]
Now let's add equation b and c, but first, multiply equation c by 3, so that the z's will be eliminated.
[tex](3)(2x+y-z=30)\\(e)6x+3y-3z=30[/tex]
Add b and e.
[tex]x+3y+3z=16\\6x+3y-3z=30[/tex]
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[tex](f)7x+6y=46[/tex]
Now you have a two variable system of equations (d and f)
[tex]3x+2y=18\\7x+6y=46[/tex]
You can solve this by multiplying by equation d by 6 and equation f by -2
[tex](6)(3x+2y=18)\\(-2)(7x+6y=46)[/tex]
Leaving our equations like;
[tex]18x+12y=108\\-14x-12y=-92[/tex]
Add them to eliminate y
[tex]18x+12y=108\\-14x-12y=-92[/tex]
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[tex]4x=16[/tex]
solve for x;
[tex]x=\frac{16}{4} \\x=4[/tex]
Replace x in either d or f, to find y
[tex]3x+2y=18[/tex]
[tex]3(4)+2y=18\\12+2y=18\\2y=18-12\\2y=6\\y=\frac{6}{2}\\ y=3[/tex]
Now that you have found x and y, replace them in either a, b, or c, to find z
[tex]x+y+z=8\\4+3+z=8\\7+z=8\\z=8-7\\z=1[/tex]
To make sure that you have found the right values, replace all three variables in any of the equations and it should be equal.
[tex]x+3y+3z=16\\4+3(3)+3(1)=16\\4+9+3=16\\16=16[/tex]
