SOLUTION: Solve the systems of equations by using the substitution method. (If the system is dependent, enter a general solution in terms of c. {18x− 12y=24 −6x+ 4y= 12

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Question 1130782: Solve the systems of equations by using the substitution method. (If the system is dependent, enter a general solution in terms of c.
{18x− 12y=24
−6x+ 4y= 12
Found 4 solutions by josgarithmetic, ikleyn, MathTherapy, Alan3354:
Answer by josgarithmetic(39616)   (Show Source): You can put this solution on YOUR website!
Same as the simplified system

A possible substitution:

.
.


What do you think?
Answer by ikleyn(52776)   (Show Source): You can put this solution on YOUR website!
.
18x - 12y = 24     (1)
-6x +  4y = 12     (2)


To reduce coefficients (and make your solution easier), divide eq(1) by 6 (both sides); divide eq(2) by -2 (both sides). 
You will get


3x - 2y =   4      (3)
3x - 2y = -12      (4)


From this point, you can just see that the system is inconsistent:  left sides of equations (3) and (4) are identical,
while their right sides are different.  So, the system has no solution/solutions.


But I will follow instructions and will apply the substitution method.
Let's see what will happen.


From equation (3), I express 3x = 4 + 2y  and substitute it into equation (4). I get


(4 + 2y) - 2y = -12

2y - 2y = -12 - 4

0 = -16.                   (*)


Thus we got self-contradictory equation (*), which means that the original system has no solution/solutions.

It confirms our conclusion, which was made earlier.

Solved.


Answer by MathTherapy(10551)   (Show Source): You can put this solution on YOUR website!
Solve the systems of equations by using the substitution method. (If the system is dependent, enter a general solution in terms of c.
{18x− 12y=24
−6x+ 4y= 12
18x - 12y = 24 ------ eq (i)

- 6x + 4y = 12 ------ eq (ii)

- 18x + 12y = 36 ------- Multiplying eq (ii) by 3 ------ eq (iii)

0 = 60 ------- Adding eqs (iii) & (i)

From above, , so what do you think? 

At least 40 of these have been done for you in the past few days.

You still can't do these on your own?

I really hope that no-one will do any more for you! It's getting awfully boring and annoying doing so many of these for you! 

Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
 If you don't know how to do these by now, find another area of interest.

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

Don't confuse tutoring with giving you the answers.

 

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