SOLUTION: hi, i tried solving this system and it just confuses me so much.
1/a + 1/b - 2/c = 9
3/a - 2/b + 1/c + -1
2/a - 1/b + 3/c = 7
i tried modifying it to
a + b - 2c = 9
Algebra.Com
Question 241679: hi, i tried solving this system and it just confuses me so much.
1/a + 1/b - 2/c = 9
3/a - 2/b + 1/c + -1
2/a - 1/b + 3/c = 7
i tried modifying it to
a + b - 2c = 9
3a - 2b + c = -1
2a - b + 3c + 7
i tried moving the variables into the numerator and switching the signs after i did that but it doesnt seem to work because i eliminated and substituted and the triplet i got was (65, 119, 29)
so please help me
Found 2 solutions by scott8148, checkley77:
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
multiply 1st eqn by -3 and add to 2nd eqn ___ -5/b + 7/c = -28 (I)
multiply 1st eqn by -2 and add to 3rd eqn ___ -3/b + 7/c = -11 (II)
subtract eqn (I) from (II) ___ 2/b = 17 ___ 2/17 = b
substituting ___ -3/(2/17) + 7/c = -11
-51/2 + 7/c = -11 ___ 7/c = 29/2 ___ c = 14/29
substituting ___ 1/a + 1/(2/17) - 2/(14/29) = 9 ___ 1/a + 17/2 - 58/14 = 9
1/a + 119/14 - 58/14 = 126/14 ___ 1/a = 65/14
a = 14/65
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
The first thing you did wrong is eliminate the numerators.
a + b - 2c = 9
3a - 2b + c = -1
2a - b + 3c + 7
---------------------------
You need to eliminate one of the terms.
1/a+1/b-2/c=9 add this equation to the third equation
3/a-2/b+1/c[=]-1
2/a-1/b+3/c=7
1/a+1/b-2/c=9
-------------------
3/a+1/c=16 # now subtract equation 2
-3a+2/b-1/c=+1
------------------
2/b=17
17b=2
b=2/17 ans.
1/a+1/(2/17)-2/c=9
1/a+1*17/2-2/c=9
1/a-2/c=9+17/2
1/a-2/c=18/2-17/2
1/a-2/c=1/2 multiply the equation marked with a # by 2 & add.
6/a+2/c=32
------------
7/a=31.5
32.5a=7
a=7/32.5
a=14/65 ans.
2/(14/65)-1/(2/17)+3/c=7
2*65/14-1*17/2+3/c=7
130/14-17/2+3/c=7
3/c=7-130/14+17/2
3/c=(98-130+119)/14
3/c=87/14
87c=3*14
87c=42
c=42/87
c=14/29 ans.
I'll leave the proof up to you.
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