Question

4. Subtract 6 from the sum of x and 2.
5. What should be added to x to get 0?
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Answer 1

Answer:

sorry I didn't understand that.....

Answer 2

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