Solve the following system of equations for a,b,c and d :
2a + 5c = 3
3a + 8c = 43
2b + 5d = 0
3b + 8d = 22
Thank You.
Answers
5d=-2b. (from iii)
d=-2b/5
putting this in iv) we get
3b+8×-2b/5=22
3b-16b/5=22
-b/5=22
b=-110
putting this value in iv) we get
-330-22=-8d
-352=8d
d=44
to get a and b multiply first by 3 and second by 2
6a+15c=9
6a+16c=86
-c=-77
c=77
putting this in first we get
2a=3-385
2a=-382
a=-191
2a + 5c = 3. ----------(I)
3a + 8c = 43. --------(ii)
I)×8 => 16a + 40c = 24
ii) × 5 => (-) 15a + 40c = 215
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a. = -191
put the value of a in (I)
2a + 5c = 3.
=> 2×-191 +5c = 3
=> -382 + 5c = 3
=> 5c = 3+382
=> c = 385/5
= 77
and ,. 2b + 5d = 0 ---------(iii)
3b + 8d = 22 -------(iv)
(iii) ×8 =>. 16b + 40d = 0
(iv) × 5 => (-) 15b + 40d = 22
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b = -22
put the value of b in (iii)
2b + 5d = 0
=> 2 × -22 + 5d = 0
=> -44 +5d =0
=> 5d = 44
=> d = 44/5