4. Subtract 6 from the sum of x and 2.
5. What should be added to x to get 0?
Answers
Answer:
sorry I didn't understand that.....
Answer:
hi
Step-by-step explanation:
[12/11, 6:15 PM] Saisurya Tamizhiniyan: Question :- if x+3/2 +3x= 5(x-3)+ x+23/5,
then what is the value of x ?
Solution :-
+x + (3/2) + 3x= 5(x-3) + x + 23/5
x will be cancel from both sides,
+ (3/2) + 3x = 5(x-3) + 23/5
+ (3/2) + 3x = 5x - 15 + 23/5
+ 5x - 3x = (3/2) + 15 - (23/5)
Taking LCM of RHS denominator now,
+ 2x = (3*5 + 15*10 - 23*2)/10
+ 2 x = (15 + 150 - 46) / 10 !!
- 2x = (165 - 46) / 10
+ 2x = 119/10
Taking 2 RHS side, it will be in denominator,
- x = (119) / (10 *2)
X = (119/20)
= 5
[12/11, 6:15 PM] Saisurya Tamizhiniyan: Let the unknown number be 'x'
Let the sum of terms be 'S'
Let the no.of terms be 'n'
Let the average be 'A'
Average=Sum of terms/No.of terms
Sum of terms=One third of x+Quarter of
x+One twelfth of x
S=(x/3)+(x/4)+(x/12)
S=(4x+3x+x)/12
S=8x/12
S=2x/3
n=3
A=(2x/3)×(1/3)
A=2x/9
As per the problem, the unknown number is 56 more than average
i.e x=56+(2x/9)
=> x-(2x/9)=56
=> (9x-2x)/9=56
=> 7x/9=56
=> 7x=504
=> x=72
Therefore the required number is 72