solve for x and y ,3x+5y=15,12x+20y=60 using elimination method
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Given :
3x+5y=15,
12x+20y=60
use elimination method
To Find : solve for x and y
Solution:
3x+5y=15 Eq1
12x+20y=60 Eq2
4 * Eq1 - Eq2
=> 4(3x + 5y) - (12x + 20y) = 4(15) - 60
=> 12x + 20y - (12x + 20y) = 60 - 60
=> 0 = 0
Which is always true independent of x and y
Hence x, y ∈ R
Infinite solution exists.
As 3x+5y=15 and 12x+20y=60 represent same line.
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