a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5, find the value of a + b + c + d. a +1 = b + 2 b + 2 = c+3 c + 3 = d+4 d+4=a+b+c+d+5 Ok start from the first equation, A-B=1 -> = A=1+b B-C=1 -> = B = 1+c C-D=1 -> = C = 1+d -1=a+b+c d+4=a+b+c+d+5 d+4=(1+b)+(1+c)(1+d)+d+5 d+4=8 +b+c+d+d 0= 4 + b+c+d -1=(1+B) + (1+C) + (1+D) -1 = 3+B+C+D -4 = B+C+D Not too sure from there, i think thats the answer but the A is missing, but i think thats because the +d-d cancelled out earlier. Hope that helps xD If not then we get hmm -4 = B+C+D we needa Add A to that, but we need a value for. -4=B+C+D + (1+b) -5 = 2b+C+D IF B = A then the answer is -5 let me try find B=A Wait, b can't equal A so thats wrong. meh ignore this If its any use enjoy
a+b+c+d+5 = a+1 0a + 1b + 1c + 1d = -4 a+b+c+d+5 = b+2 1a + 0b + 1c + 1d = -3 a+b+c+d+5 = c+3 1a + 1b + 0c + 1d = -2 a+b+c+d+5 = d+4 1a + 1b + 1c + 0d = -1 So now you have 4 linear equations, and 4 variables to solve 0a + 1b + 1c + 1d = -4 1a + 0b + 1c + 1d = -3 1a + 1b + 0c + 1d = -2 1a + 1b + 1c + 0d = -1 Now this problem becomes solving the system of linear equations. So if you're lazy, just plug it into a matrix to solve. Otherwise you'll have to do a heck of a lot of substitution. Check this article out on how to solve a system of linear equations: http://en.wikipedia.org/wiki/System_of_linear_equations
a +1 = b + 2 b + 2 = c+3 c + 3 = d+4 d+4=a+b+c+d+5 a=B+1 b=c+1 c=d+1 a=b+1 a-1=c+1 a-2=d+1 a+a+a-1+a-2 4a-3 = ? 4a = 3 A=3/4 B=-1/4 C= -5/4 D= -9/4 -12/4 = 3 a+b+c+d = 3 GOT IT! and tested
Ummmmmmmmm, SoC, I did it my method and I got a different answer... I tested it too... A = 2/3 B = -1/3 C = -4/3 D = -7/3 A+B+C+D = -10/3
aha ur correct I didnt use the 4th equation 4+d=a+b+c+d+5 Meaning using my method It could be a/n, a+2n/n, a+3n/n and a+4n/n But, A is limited due to the 4th equation, meaning your answers are correct zero. Thought i missed something
Hmmm well you have four variables (a,b,c,d) and several equations to solve them. A+1=B+2 B+2=C+3 C+3=D+4 D+4=A+B+C+D+5 The key equation here is A+B+C+D+5 because it contains all the variables. If we can isolate one variable, the equation can be solved. We do by solving for A in terms of B, C, and D. A+1=B+2 B=A-1 A+1=C+3 C=A-2 A+1=D+4 D=A-3 Now we insert these back into the "master" equation. A+(A-1)+(A-2)+(A-3)+5=A+1 A+A-1+A-2+A-3+5=A+1 4A-1=A+1 4A=A+2 3A=2 A=2/3 Now that we know A, we can find the rest of the variables using the original series of equations. B=(2/3)-1 B=-1/3 C=(2/3)-2 C=-4/3 D=(2/3)-3 D=-7/3 Finally, just plug those into A+B+C+D. (2/3)+(-1/3)+(-4/3)+(-7/3)=-10/3 I hope I'm right! Although it looks as if SoC and Zer0 have beaten me to the punch.