SOLUTION: 26. 10m-9n=15 27. 3c-7d= -3 28. 6g-8h=50 30. 3x= -31+2y 5m-4n=10 2c+6d= -34 4g+6h=22 5x+6y=23

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: 26. 10m-9n=15 27. 3c-7d= -3 28. 6g-8h=50 30. 3x= -31+2y 5m-4n=10 2c+6d= -34 4g+6h=22 5x+6y=23      Log On


   

Question 128354This question is from textbook
: 26. 10m-9n=15 27. 3c-7d= -3 28. 6g-8h=50 30. 3x= -31+2y
5m-4n=10 2c+6d= -34 4g+6h=22 5x+6y=23 This question is from textbook

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
I'll do the first two to get you started




Start with the given system of equations:

system%2810m-9n=15%2C5m-4n=10%29



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%2B15 Rearrange the equation


n=%28-10m%2B15%29%2F%28-9%29 Divide both sides by -9


n=%28%28-10%29%2F%28-9%29%29m%2B%2815%29%2F%28-9%29 Break up the fraction


n=%2810%2F9%29m-5%2F3 Reduce



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

Since n=%2810%2F9%29m-5%2F3, we can now replace each n in the second equation with %2810%2F9%29m-5%2F3 to solve for m



5m-4highlight%28%28%2810%2F9%29m-5%2F3%29%29=10 Plug in n=%2810%2F9%29m-5%2F3 into the first equation. In other words, replace each n with %2810%2F9%29m-5%2F3. Notice we've eliminated the n variables. So we now have a simple equation with one unknown.



5m%2B%28-4%29%2810%2F9%29m%2B%28-4%29%28-5%2F3%29=10 Distribute -4 to %2810%2F9%29m-5%2F3


5m-%2840%2F9%29m%2B20%2F3=10 Multiply


%289%29%285m-%2840%2F9%29m%2B20%2F3%29=%289%29%2810%29 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 solver)



45m-40m%2B60=90 Distribute and multiply the LCM to each side



5m%2B60=90 Combine like terms on the left side


5m=90-60Subtract 60 from both sides


5m=30 Combine like terms on the right side


m=%2830%29%2F%285%29 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=%2810%2F9%29m-5%2F3 (remember we previously solved for n in the first equation).



n=%2810%2F9%29m-5%2F3 Start with the equation where n was previously isolated.


n=%2810%2F9%29%286%29-5%2F3 Plug in m=6


n=60%2F9-5%2F3 Multiply


n=5 Combine like terms and reduce. (note: if you need help with fractions, check out this solver)



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


So the second part of our answer is: n=5









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

So our answers are:

m=6 and n=5









# 27






Start with the given system of equations:

system%283c-7d=-3%2C2c%2B6d=-34%29



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=%28-3c-3%29%2F%28-7%29 Divide both sides by -7


d=%28%28-3%29%2F%28-7%29%29c%2B%28-3%29%2F%28-7%29 Break up the fraction


d=%283%2F7%29c%2B3%2F7 Reduce



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

Since d=%283%2F7%29c%2B3%2F7, we can now replace each d in the second equation with %283%2F7%29c%2B3%2F7 to solve for c



2c%2B6highlight%28%28%283%2F7%29c%2B3%2F7%29%29=-34 Plug in d=%283%2F7%29c%2B3%2F7 into the first equation. In other words, replace each d with %283%2F7%29c%2B3%2F7. Notice we've eliminated the d variables. So we now have a simple equation with one unknown.



2c%2B%286%29%283%2F7%29c%2B%286%29%283%2F7%29=-34 Distribute 6 to %283%2F7%29c%2B3%2F7


2c%2B%2818%2F7%29c%2B18%2F7=-34 Multiply


%287%29%282c%2B%2818%2F7%29c%2B18%2F7%29=%287%29%28-34%29 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 solver)



14c%2B18c%2B18=-238 Distribute and multiply the LCM to each side



32c%2B18=-238 Combine like terms on the left side


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


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


c=%28-256%29%2F%2832%29 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=%283%2F7%29c%2B3%2F7 (remember we previously solved for d in the first equation).



d=%283%2F7%29c%2B3%2F7 Start with the equation where d was previously isolated.


d=%283%2F7%29%28-8%29%2B3%2F7 Plug in c=-8


d=-24%2F7%2B3%2F7 Multiply


d=-3 Combine like terms and reduce. (note: if you need help with fractions, check out this solver)



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


So the second part of our answer is: d=-3









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

So our answers are:

c=-8 and d=-3