Question

x_4-x_2=3.1_2-x_3 solution​

Answer 1

Answer:

all of xi where i=1,2,3,4,5 are non negative

and

0≤x1≤3

1≤x2<4

and

x3≥15

Attempt

first used 0≤x1≤3 and

1≤x2<4 and then find the number of solutions

violating 0≤x1≤3 ,and1≤x2<4

will give x1>3,andx2>3

Now number of solutions are there to the equation x1+x2+x3+x4+x5=21,

=(

21+5−1

21

)=12,650

Now number of solutions are there to the equation x1+x2+x3+x4+x5=21, such that 0≤x1≤3 and 1≤x2<4

=12650−number of solutions are there to the equation x1>3,andx2>3

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

solving number of equation for x1>3,andx2>3

let x1=x

′

1

+3

x2=x

′

2

+3

our equation becomes

x

′

1

+3+x

′

2

+3+x3+x4+x5=21

⇒x

′

1

+x

′

2

+x3+x4+x5=15

∴ number of equation=

(

15+5−1

15

)=3876

Now,

Now number of solutions are there to the equation x1+x2+x3+x4+x5=21, such that 0≤x1≤3 and 1≤x2<4

=12650−3876=8,774

now among 8,774 we have to find x3≥15

for x3≥15,

let x3=x

′

3

+15

our equation becomes

x1+x2+x

′

3

+15+x4+x5=21,

x1+x2+x3+x4+x5=6

∴ number of equation =(

6+5−1

6

)=210

so final answer=8,774−210=8,564