Question 132472


Start with the given system of equations:


{{{system(3b+2a=2,-2b+a=8)}}}




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





So let's isolate a in the first equation


{{{3b+2a=2}}} Start with the first equation



{{{2a=2-3b}}}  Subtract {{{3b}}} from both sides



{{{2a=-3b+2}}} Rearrange the equation



{{{a=(-3b+2)/(2)}}} Divide both sides by {{{2}}}



{{{a=((-3)/(2))b+(2)/(2)}}} Break up the fraction



{{{a=(-3/2)b+1}}} Reduce




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Since {{{a=(-3/2)b+1}}}, we can now replace each {{{a}}} in the second equation with {{{(-3/2)b+1}}} to solve for {{{b}}}




{{{-2b+highlight(((-3/2)b+1))=8}}} Plug in {{{a=(-3/2)b+1}}} into the first equation. In other words, replace each {{{a}}} with {{{(-3/2)b+1}}}. Notice we've eliminated the {{{a}}} variables. So we now have a simple equation with one unknown.




{{{(2)(-2b-(3/2)b+1)=(2)(8)}}} Multiply both sides by the LCM of 2. 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>)




{{{-4b-3b+2=16}}} Distribute and multiply the LCM to each side




{{{-7b+2=16}}} Combine like terms on the left side



{{{-7b=16-2}}}Subtract 2 from both sides



{{{-7b=14}}} Combine like terms on the right side



{{{b=(14)/(-7)}}} Divide both sides by -7 to isolate b




{{{b=-2}}} Divide






-----------------First Answer------------------------------



So the first part of our answer is: {{{b=-2}}}










Since we know that {{{b=-2}}} we can plug it into the equation {{{a=(-3/2)b+1}}} (remember we previously solved for {{{a}}} in the first equation).




{{{a=(-3/2)b+1}}} Start with the equation where {{{a}}} was previously isolated.



{{{a=(-3/2)(-2)+1}}} Plug in {{{b=-2}}}



{{{a=6/2+1}}} Multiply



{{{a=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>)




-----------------Second Answer------------------------------



So the second part of our answer is: {{{a=4}}}










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


So our answers are:


{{{b=-2}}} and {{{a=4}}}