Math, asked by Anonymous, 8 months ago

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

Answered by Rajshuklakld
3

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

Answered by rumig0720
3

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

――――――――――――––―――

b = -22

put the value of b in (iii)

2b + 5d = 0

=> 2 × -22 + 5d = 0

=> -44 +5d =0

=> 5d = 44

=> d = 44/5

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