How do you solve 3r-5s=-35 and 2r-5s=-30?
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we have to subtract these two equation and 2nd equation's sign will be changed in subtraction
3r - 5s = - 35 .....(1)
2r - 5s = - 30 ......(2)
- ....+ ........+
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r = - 5
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put this value of r in (1)
3(-5) - 5s = -35
=> -15 - 5s = -35
=> 15+5s = 35
=> 5s. = 35-15
5s = 20
s = 20/5 = 4
3r - 5s = - 35 .....(1)
2r - 5s = - 30 ......(2)
- ....+ ........+
----------------------------
r = - 5
----------------------------
put this value of r in (1)
3(-5) - 5s = -35
=> -15 - 5s = -35
=> 15+5s = 35
=> 5s. = 35-15
5s = 20
s = 20/5 = 4
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