SOLUTION: Solve the following system algebraically: {{{ 3x+5y-z=2 }}} {{{ -x-8y+3z=-6 }}} {{{ 3z=2x }}} What I have tried: {{{ 3z = 2x }}} -> {{{ 3z/2=x }}}; {{{ 3x + 5y – z = 2 }}} -> {

Algebra ->  Expressions-with-variables -> SOLUTION: Solve the following system algebraically: {{{ 3x+5y-z=2 }}} {{{ -x-8y+3z=-6 }}} {{{ 3z=2x }}} What I have tried: {{{ 3z = 2x }}} -> {{{ 3z/2=x }}}; {{{ 3x + 5y – z = 2 }}} -> {      Log On


   

Question 632866: Solve the following system algebraically: +3x%2B5y-z=2+ +-x-8y%2B3z=-6+ +3z=2x+
What I have tried: +3z+=+2x+ -> +3z%2F2=x+; +3x+%2B+5y+%96+z+=+2+ -> +3%283z%2F2%29%2B5y-z=2+ -> +y=%28-9z%2F2%2Bz%2B2%29%2F5%29+; +-x-8y%2B3z=-6+ -> +-%283z%2F2%29+-8%28%28-9z%2F2%2Bz%2B2%29%2F5%29%2B3z=-6+ -> i multiplied every thing by 2 and got -> +-3z%2B%2872-16z-32%2F5%29%2B6z=-12+ -> i multiplied every thing by 5 and got -> +-15z%2B72-16z-32%2B30z=-60+ -> +-z%2B40=-60+ -> +z=100+ -> i plugged this number into the third top equation and got x=150, and plugged z and x into the second equation and got y=18. But when i plug these numbers into the first equation, the answer is wrong. I cannot figure out what I am doing wrong?
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

Note:  general Procedure is to first the eliminate one of the variables using a pair of EQs

+3x%2B5y-z=2+  

+-x-8y%2B3z=-6+     |Multiplying 2nd Eq by 3 and adding to the first

     y+=+%28-8%2F7%29x

+3z=2x+  z+=+%282%2F3%29x

+3x%2B5y-z=2+      | Substituting for y and z 

    x =  -.5915

    y = %28-8%2F7%29%28-.5915%29+=+.6761

    z =  %282%2F3%29%28-.5915%29+=+-.3944