Question 151298
{{{system(3x+3y+z=5,x-3y+2z=-13,8x-2y+3z=-8)}}}



Add equations 1 and 2. Let's call the result equation 4:


{{{3x+3y+z=5}}}
+{{{x-3y+2z=-13}}}
-----------------
{{{4x+0y+3z=-8}}} 



Equation 4: {{{4x+3z=-8}}}



{{{2(3x+3y+z)=2(5)}}} Multiply both sides of equation 1 by 2



{{{6x+6y+2z=10}}} Multiply. Let's call this equation 5


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{{{3(8x-2y+3z)=3(-8)}}} Multiply both sides of equation 3 by 3



{{{24x-6y+9z=-24}}} Multiply. Let's call this equation 6




Now add equations 5 and 6. Let's call the result equation 7:

{{{6x+6y+2z=10}}}
+{{{24x-6y+9z=-24}}}
-------------------
{{{30x+0y+11z=-14}}}


Equation 7: {{{30x+11z=-14}}}



===================================


So we now have the two equations equation 4 and equation 7:


equation 4: {{{4x+3z=-8}}} 
equation 7: {{{30x+11z=-14}}}





Start with the given system of equations:


{{{system(4x+3z=-8,30x+11z=-14)}}}




Now in order to solve this system by using substitution, we need to solve (or isolate) one variable. I'm going to solve for z.





So let's isolate z in the first equation


{{{4x+3z=-8}}} Start with the first equation



{{{3z=-8-4x}}}  Subtract {{{4x}}} from both sides



{{{3z=-4x-8}}} Rearrange the equation



{{{z=(-4x-8)/(3)}}} Divide both sides by {{{3}}}



{{{z=((-4)/(3))x+(-8)/(3)}}} Break up the fraction



{{{z=(-4/3)x-8/3}}} Reduce




---------------------


Since {{{z=(-4/3)x-8/3}}}, we can now replace each {{{z}}} in the second equation with {{{(-4/3)x-8/3}}} to solve for {{{x}}}




{{{30x+11highlight(((-4/3)x-8/3))=-14}}} Plug in {{{z=(-4/3)x-8/3}}} into the first equation. In other words, replace each {{{z}}} with {{{(-4/3)x-8/3}}}. Notice we've eliminated the {{{z}}} variables. So we now have a simple equation with one unknown.




{{{30x+(11)(-4/3)x+(11)(-8/3)=-14}}} Distribute {{{11}}} to {{{(-4/3)x-8/3}}}



{{{30x-(44/3)x-88/3=-14}}} Multiply



{{{(3)(30x-(44/3)x-88/3)=(3)(-14)}}} Multiply both sides by the LCM of 3. This will eliminate the fractions  (note: if you need help with finding the LCM, check out this <a href=http://www.algebra.com/algebra/homework/divisibility/least-common-multiple.solver>solver</a>)




{{{90x-44x-88=-42}}} Distribute and multiply the LCM to each side




{{{46x-88=-42}}} Combine like terms on the left side



{{{46x=-42+88}}}Add 88 to both sides



{{{46x=46}}} Combine like terms on the right side



{{{x=(46)/(46)}}} Divide both sides by 46 to isolate x




{{{x=1}}} Divide




So the first part of our answer is: {{{x=1}}}







Since we know that {{{x=1}}} we can plug it into the equation {{{z=(-4/3)x-8/3}}} (remember we previously solved for {{{z}}} in the first equation).




{{{z=(-4/3)x-8/3}}} Start with the equation where {{{z}}} was previously isolated.



{{{z=(-4/3)(1)-8/3}}} Plug in {{{x=1}}}



{{{z=-4/3-8/3}}} Multiply



{{{z=-4}}} Combine like terms and reduce.  (note: if you need help with fractions, check out this <a href="http://www.algebra.com/algebra/homework/NumericFractions/fractions-solver.solver">solver</a>)




So the third part of our answer is: {{{z=-4}}}





{{{3x+3y+z=5}}} Go back to the first equation 



{{{3(1)+3y+(-4)=5}}} Plug in  {{{x=1}}} and  {{{z=-4}}}



{{{3+3y-4=5}}} Multiply



{{{3y-1=5}}} Combine like terms.


{{{3y=5+1}}} Add {{{1}}} to both sides.



{{{3y=6}}} Combine like terms on the right side.



{{{y=(6)/(3)}}} Divide both sides by {{{3}}} to isolate {{{y}}}.



{{{y=2}}} Reduce.




So the second part of the answer is {{{y=2}}} 



-----------------Summary------------------------------


So our answers are:


{{{x=1}}}, {{{y=2}}}, and {{{z=-4}}}



They form the point (1,2,-4)