if p+q+r=16, b+c=4 and x,y,z are the solution of linear equations



(x/r)+(y/r-b)+(z/r-c)=1 then x+y+z is

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How do i solve equations like these find no. of ordeted pairs for 10y+10x=xy-1?
You will get infinite number of ordered pairs for this equation. 10y + 10 x = xy –1 y = – (1 + 10x)/(10 – x) & x = – (1 + 10y)/(10 – y) This is not defined for x = 10 & y = 10
the gcd of two positive integer is 81 and their lcm is 5103 , find the number
i want answer from you
anand kapare one month ago
729 and 567 GCD=81 Therefore numbers are 81a and 81b for a and b are co-prime. LCM=81*a*b = 5103=81*63 therefore ab=63 a,b=9,7 so numbers are 81*7=567 and 81*9 =729 GCD of 15,20,35 is 5...
Karan Yadav one month ago
find gcd of 15,20,35 . also find x, y, z such that 15x+20y+35z= gcd
anand kapare one month ago
let (1^4+2^4+3^4+....n^4)=f(n).then the sum of ∑ (2r-1)^4=?(where r=1.....n) f(2n)-16f(n ) f(2n)-f(n) f(2n)-4f(n) noe of these
use the summation terms given in this equation (1^4+2^4+3^4+....n^4)=f(n) and the given summation to find can be seen as the sum of odd terms of (2n-1) terms.First try to write f(2n) an...
Sourabh Singh one month ago
Hello, I would like to ask you that do you only entertain Indians or also other nationalities? Thanks.
this forum is for all over the world student, most of our student are NRIs , so its for all of you.
Sher Mohammad 3 months ago
2.f(x) is a differentiable funcn. such that f(x)f(y) +2 = f(x) + f(y) +f(xy); f’(0)=0, f’(1)=2, then find f(x)
Given, f(x) f(y) + 2 = f(x) + f(y) + f(xy) f’(0) = 0, f’(1)= 2 Differentiatiing above expression with respect to y, f(x) f’(y) = 1 + f’(xy) x For y=1, f(x)...
Shobhit Varshney one month ago
thank u sir
Arpit Dhankar one month ago
prove that four straight lines represented by the equation 12x 2 +7xy -12y 2 =0 and 12x 2 +7xy -12y 2 -x +7y -10 form a square
Hello Student, 12x2+7xy -12y2=0 12x2+7xy -12y2-x +7y -10 with first x=(-7y +- yroot(49+144*4))/24 you will get 2 lines from here with second do the same thing x=(-7y+1 +- root((7y-1)^2...
Arun Kumar 4 months ago
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