Question

solve the equation
n/3+1=4-n
the right answer is 2 whole 1 upon 4
but please explain this​

Answer 1

Answer:

n/3+1=4-n

taking constants and variable in their own sides,

n/3+n=4-1

taking LCM of the denominators,

n/3+3n/3=3

4n/3=3

Cross multiplication,

n=9/4

Answer 2

Answer:

Given the series [−4+(−1)+2+⋅⋅⋅⋅x=437] in A.P.

Then,

a=−4

d=−1−(−4)=3

S

n

=437

Then, using sum formula A.P.

S

n

=

2

n

[2a+(n−1)d]

⇒437=

2

n

[2(−4)+(n−1)3]

⇒874=n[−8+3n−3]

⇒874=3n

2

−11n

⇒3n

2

−11n=874

⇒3n

2

−11n−874=0

⇒3n

2

−(57−46)n−874=0

⇒3n

2

−57n−46n−874=0

⇒3n(n−19)−46(n−19)=0

⇒(n−19)(3n−19)=0

⇒n−19=0,3n−19=0

⇒n=19,n=

3

19

Hence, n=19

So,

x=a+(n−1)d

x=−4+(19−1)3

x=−4+54

x=50

Hence, this is the answer

Step-by-step explanation:

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