SOLUTION: How do u solve systems of equations like this 3x+2y+5z=33 4y-3z=-11 z=5
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Question 33094
:
How do u solve systems of equations like this 3x+2y+5z=33
4y-3z=-11
z=5
Answer by
longjonsilver(2297)
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you know z=5
so, from 4y-3z=-11 we get
4y-3(5)=-11
4y-15=-11
4y=4
y = 1
And finally from 3x+2y+5z=33 we get
3x+2(1)+5(5)=33
3x+2+25=33
3x+27=33
3x = 6
x = 2
So, solution is:
x=2
y=1
z=5
jon