x=1 and x=9 both are valid
|3x-7|=|3*1-7|=|-4|=4(modulus of any value is +ve)
2x+2=2*1+2=4.
Hence valid
[color=#ed1c24] Bunuel, I have a question. Usually while solving modulus questions we take two cases 1) x>0 2) x<0. According to the first statement when x>0 we get x=9 which is valid, but when x<0 we get x=1 (which is not valid). Now, in some of the earlier questions when x<0 and if we got a positive value for it we neglected it and considered that x had only one valid
...
|3x-7|=|3*1-7|=|-4|=4(modulus of any value is +ve)
2x+2=2*1+2=4.
Hence valid
ArunpriyanJ wrote:
[color=#ed1c24] Bunuel, I have a question. Usually while solving modulus questions we take two cases 1) x>0 2) x<0. According to the first statement when x>0 we get x=9 which is valid, but when x<0 we get x=1 (which is not valid). Now, in some of the earlier questions when x<0 and if we got a positive value for it we neglected it and considered that x had only one valid
...
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