Question 176933


Start with the given system of equations:

{{{system(3w-2z=42,2w-z=26)}}}



{{{-2(2w-z)=-2(26)}}} Multiply the both sides of the second equation by -2.



{{{-4w+2z=-52}}} Distribute and multiply.



So we have the new system of equations:

{{{system(3w-2z=42,-4w+2z=-52)}}}



Now add the equations together. You can do this by simply adding the two left sides and the two right sides separately like this:



{{{(3w-2z)+(-4w+2z)=(42)+(-52)}}}



{{{(3w+-4w)+(-2z+2z)=42+-52}}} Group like terms.



{{{-w+0z=-10}}} Combine like terms.



{{{-w=-10}}} Simplify.



{{{w=(-10)/(-1)}}} Divide both sides by {{{-1}}} to isolate {{{w}}}.



{{{w=10}}} Reduce.



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



{{{3w-2z=42}}} Now go back to the first equation.



{{{3(10)-2z=42}}} Plug in {{{w=10}}}.



{{{30-2z=42}}} Multiply.



{{{-2z=42-30}}} Subtract {{{30}}} from both sides.



{{{-2z=12}}} Combine like terms on the right side.



{{{z=(12)/(-2)}}} Divide both sides by {{{-2}}} to isolate {{{z}}}.



{{{z=-6}}} Reduce.



So our answer is {{{w=10}}} and {{{z=-6}}}.



So the point of intersection is *[Tex \LARGE \left(10,-6\right)].



This means that the system is consistent and independent.