Question 922638: solving 3 equations with elimination.
-9a+3b-2c=61
8a+7b+5c=-138
5a-5b+8c=-45
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! solving 3 equations with elimination.
-9a+3b-2c=61
8a+7b+5c=-138
5a-5b+8c=-45
------------------
Pick one of the 3 variables to eliminate first.
I would pick c since Eqn 1 times 4 would give the same coefficients for c as Eqn 3.
-----
-9a+3b-2c=61 *4 --> -36a+12b-8c=244
8a+7b+5c=-138
5a-5b+8c=-45
-------------
-36a+12b-8c=244 Eqn 1 times 4
5a - 5b+ 8c=-45 Eqn 3
---------------------- Add
-31a + 7b = 199 Eqn A
-----
Now it's necessary to eliminate c using another pair of the original 3 eqns
--
-9a+3b-2c=61 Eqn 1
8a+7b+5c=-138 Eqn 2
-------------
Make the coefficients of the c terms match:
-9a+3b-2c=61 *5 --> -45a + 15b - 10c = 305
8a+7b+5c=-138 *2 --> 16a + 14b + 10c = -276
---
-45a + 15b - 10c = 305
16a + 14b + 10c = -276
------------------------------- Add
-29a + 29b = 29 Eqn B
-31a + 7b = 199 Eqn A
-----
Divide eqn B by 29, then multiply it by 7 so the coefficients of b are the same
-7a + 7b = 7 Eqn B /29, times 7
-31a + 7b = 199 Eqn A
------------------------------ Subtract
24a = - 192
a = -8
--------------
Sub for a in -7a + 7b = 7, or b - a = 7
--> b = -1
---------------
Sub for a & b in any eqn to find c
|
|
|
