|
Question 36286: solve
4a-5b-c=-26
5a+b+2c=-5
2a+4b+c=6
show work
Answer by vidhyak(98)   (Show Source):
You can put this solution on YOUR website! 4a - 5b - c = -26 ------- equation 1
5a + b + 2c = - 5 ------- equation 2
2a + 4b + c = 6 ------- equation 3
Add eq1 and eq3 to eliminate variable c
4a - 5b - c = -26
+(2a + 4b + c = 6)
4a - 5b - c = -26
2a + 4b + c = 6
We get
6a - b = -20 --------equation 4
Lets compare eq2 and eq3 to eliminate variable c
Multiply eq3 by 2 and subtract it from eq2
5a + b + 2c = - 5
-2(2a + 4b + c = 6)
5a + b + 2c = - 5
-4a - 8b - 2c = - 12
Solving we get
a - 7b = -17 --------equation 5
Having eliminated variable c we have got 2 equations 4 and 5
Solve 4 and 5 to get the values of a and b
6a - b = -20 --------equation 4
a - 7b = -17 --------equation 5
Multiply eq5 by 6 and subtract eq4 and eq5
6a - b = -20
-6( a - 7b = -17 )
6a - b = -20
-6a + 42b = +102
-----------------------
41b = 82
b = 82/41 = 2 --- solution 1
Substitute solution 1 in either equation 4 or equation 5 to get the value of "a"
I have considered equation 4
6a - b = -20
6a - 2 = -20
6a = -20 +2
6a = -18
a = -3 --- solution 1
Substitute soln 1 and soln 2 in either eq 1 or eq 2 or eq 3 to get the value of "c"
I have considered equation 1
a = -3 , b = 2
4a - 5b - c = -26
4(-3) - 5(2) - c = -26
-12 - 10 - c = -26
-22 - c = -26
-c = -26 + 22
-c = -4
c = 4
So the answers are
a = -3 , b = 2, c = 4
substitute in the other 2 equations to verify
|
|
|
| |
