SOLUTION: solve by elimination method: 0.3x-0.2y=4 0.4x+0.3y=73/19

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Question 167543: solve by elimination method:
0.3x-0.2y=4
0.4x+0.3y=73/19
Found 2 solutions by aswathytony, sowmya:
Answer by aswathytony(47) (Show Source):
You can put this solution on YOUR website!
0.3x -0.2y =4 ...........(1)
0.4 x + 0.3 y = 73/19 ...........(2)
(1) * 0.3 & (2) * 0.2
0.3 x * 0.3 - 0.2y * 0.3 = 4 *0.3
0.09x -0.06y = 1.2 .............(3)
0.4x * 0.2 + 0.3 y * 0.2 = 73/19 * 0.2
0.08x + 0.06 y = 0.768 ..........(4)
(3) + (4)
0.17x = 1.968
x = 1.968/0.17= 11.58
substituting x = 11.58 in (1)
0. 3 * 11.58 -0.2 y = 4
3.474 -0.2y = 4
- 0.2y = 4 - 3.474
- 0.2 y = 0.526
y = 0.526 / -0.2
= -2.63

Answer by sowmya(32) (Show Source):
You can put this solution on YOUR website!
solve by elimination method:
0.3x-0.2y=4
0.4x+0.3y=73/19
eq 1:
3x-2y = 40
eq 2:
4x-3y = 730/19
from eq 1
x = (40+2y)/3
sub. in eq2,
4(40+2y)/3 -3y =730/19
(160+8y)/3 - 3y = 730/19
160+8y - 9y = 2190/19
y = 160-2190/19
=3040-2190 /19
y =850/19
x = 2460/57