Question 128354
I'll do the first two to get you started





Start with the given system of equations:


{{{system(10m-9n=15,5m-4n=10)}}}




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





So let's isolate n in the first equation


{{{10m-9n=15}}} Start with the first equation



{{{-9n=15-10m}}}  Subtract {{{10m}}} from both sides



{{{-9n=-10m+15}}} Rearrange the equation



{{{n=(-10m+15)/(-9)}}} Divide both sides by {{{-9}}}



{{{n=((-10)/(-9))m+(15)/(-9)}}} Break up the fraction



{{{n=(10/9)m-5/3}}} Reduce




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


Since {{{n=(10/9)m-5/3}}}, we can now replace each {{{n}}} in the second equation with {{{(10/9)m-5/3}}} to solve for {{{m}}}




{{{5m-4highlight(((10/9)m-5/3))=10}}} Plug in {{{n=(10/9)m-5/3}}} into the first equation. In other words, replace each {{{n}}} with {{{(10/9)m-5/3}}}. Notice we've eliminated the {{{n}}} variables. So we now have a simple equation with one unknown.




{{{5m+(-4)(10/9)m+(-4)(-5/3)=10}}} Distribute {{{-4}}} to {{{(10/9)m-5/3}}}



{{{5m-(40/9)m+20/3=10}}} Multiply



{{{(9)(5m-(40/9)m+20/3)=(9)(10)}}} Multiply both sides by the LCM of 9. 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>)




{{{45m-40m+60=90}}} Distribute and multiply the LCM to each side




{{{5m+60=90}}} Combine like terms on the left side



{{{5m=90-60}}}Subtract 60 from both sides



{{{5m=30}}} Combine like terms on the right side



{{{m=(30)/(5)}}} Divide both sides by 5 to isolate m




{{{m=6}}} Divide






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



So the first part of our answer is: {{{m=6}}}










Since we know that {{{m=6}}} we can plug it into the equation {{{n=(10/9)m-5/3}}} (remember we previously solved for {{{n}}} in the first equation).




{{{n=(10/9)m-5/3}}} Start with the equation where {{{n}}} was previously isolated.



{{{n=(10/9)(6)-5/3}}} Plug in {{{m=6}}}



{{{n=60/9-5/3}}} Multiply



{{{n=5}}} 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: {{{n=5}}}










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


So our answers are:


{{{m=6}}} and {{{n=5}}}







<hr>



# 27







Start with the given system of equations:


{{{system(3c-7d=-3,2c+6d=-34)}}}




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





So let's isolate d in the first equation


{{{3c-7d=-3}}} Start with the first equation



{{{-7d=-3-3c}}}  Subtract {{{3c}}} from both sides



{{{-7d=-3c-3}}} Rearrange the equation



{{{d=(-3c-3)/(-7)}}} Divide both sides by {{{-7}}}



{{{d=((-3)/(-7))c+(-3)/(-7)}}} Break up the fraction



{{{d=(3/7)c+3/7}}} Reduce




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


Since {{{d=(3/7)c+3/7}}}, we can now replace each {{{d}}} in the second equation with {{{(3/7)c+3/7}}} to solve for {{{c}}}




{{{2c+6highlight(((3/7)c+3/7))=-34}}} Plug in {{{d=(3/7)c+3/7}}} into the first equation. In other words, replace each {{{d}}} with {{{(3/7)c+3/7}}}. Notice we've eliminated the {{{d}}} variables. So we now have a simple equation with one unknown.




{{{2c+(6)(3/7)c+(6)(3/7)=-34}}} Distribute {{{6}}} to {{{(3/7)c+3/7}}}



{{{2c+(18/7)c+18/7=-34}}} Multiply



{{{(7)(2c+(18/7)c+18/7)=(7)(-34)}}} Multiply both sides by the LCM of 7. 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>)




{{{14c+18c+18=-238}}} Distribute and multiply the LCM to each side




{{{32c+18=-238}}} Combine like terms on the left side



{{{32c=-238-18}}}Subtract 18 from both sides



{{{32c=-256}}} Combine like terms on the right side



{{{c=(-256)/(32)}}} Divide both sides by 32 to isolate c




{{{c=-8}}} Divide






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



So the first part of our answer is: {{{c=-8}}}










Since we know that {{{c=-8}}} we can plug it into the equation {{{d=(3/7)c+3/7}}} (remember we previously solved for {{{d}}} in the first equation).




{{{d=(3/7)c+3/7}}} Start with the equation where {{{d}}} was previously isolated.



{{{d=(3/7)(-8)+3/7}}} Plug in {{{c=-8}}}



{{{d=-24/7+3/7}}} Multiply



{{{d=-3}}} 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: {{{d=-3}}}










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


So our answers are:


{{{c=-8}}} and {{{d=-3}}}