Question 516236: How do I solve 0.3x-0.2y=4 and 0.2x + 0.3y=29/19 using the elimination method? Found 2 solutions by mananth, Maths68:Answer by mananth(16949) (Show Source):
You can put this solution on YOUR website! 0.3x-0.2y=4
multiply by 10
3x-2y=40.............1
and 0.2x + 0.3y=29/19
multiply by 190
38x+57y=290.............2
3 x -2 y = 40 .............1
38 x + 57 y = 290 .............2
Eliminate y
multiply (1)by 57
Multiply (2) by 2
171 x -114 y = 2280
76 x 114 y = 580
Add the two equations
247 x = 2860
/ 247
x = 11.58
plug value of x in (1)
3 x + -2 y = 40
34.74 + -2 y = 40
-2 y = 40-34.74
-2 y = 5.26
y = -2.63
You can put this solution on YOUR website! 0.3x-0.2y=4...............(1)
0.2x + 0.3y=29/19.........(2)
Make above equations more simple
3x/10+2y/10=4
Multiply by 10 both sides of above equation
3x+2y=40.................(3)
2x/10+3y/10=29/19
Multiply by 10 both sides of above equation
2x+3y=290/19.............(4)
Multiply (3) by -2 and (4) by 3 and add the resultant equations
-6x-4y=-80
6x+9y=870/19
------------------
5y=870/19-80
5y=(870-1520)/19
5y=-650/19
5y/5=-650/19/5
y=-130/19
Put the value of y in (3)
3x+2y=40.................(3)
3x+2(-130/19)=40
3x-260/19=40
Multiply by 19 both sides
57x-260=760
57x=760+260
57x=1020
57x/57=1020/57
x=1020/57
x, y =1020/57,-130/19
Check
Put the values of x and y in (4)
2x+3y=290/19.............(4)
2(1020/57)+3(-130/19)=290/19
2040/57-390/19=290/19
(2040-1170)/57=290/19
870/57=290/19
290/19=290/19
