15x + 22y = 5
40x + 55y = 13
Solve for X and Y
( Elimination method )
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15x+22y=5 (1) x5
40x+55y =13 (2)x2
subtract in eq(1)&eq(2)
:) 75x+110y =25
80x+110y =26
- - -
----------------------------------------
-5x = -1
x = -1/-5
x = 1/5
now put the value of X in eq(1)
15x+22y=5
15×1/5+22y =5
1/3+22y = 5
now, 22y = 5 - 1/3
22y = (15-1)/3
y = 14/ 3 x 22
y = 7 /33.
so x = 1/5&y = 7/33
40x+55y =13 (2)x2
subtract in eq(1)&eq(2)
:) 75x+110y =25
80x+110y =26
- - -
----------------------------------------
-5x = -1
x = -1/-5
x = 1/5
now put the value of X in eq(1)
15x+22y=5
15×1/5+22y =5
1/3+22y = 5
now, 22y = 5 - 1/3
22y = (15-1)/3
y = 14/ 3 x 22
y = 7 /33.
so x = 1/5&y = 7/33
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