Solving Equations with Multiple Variables

  • Thread starter Leaping antalope
  • Start date
In summary, the conversation discusses solving a system of equations with four variables. The suggested method is to create equations in the form of d=... and use elimination to obtain two equations with two variables. Then, using the fifth equation, one can create another equation with two variables and use elimination again to solve for the remaining variables. This process is repeated until all four variables are obtained.
  • #1
Leaping antalope
44
0
Could someone solve these equations for me? It seems complex, but I believe there is a easy way to find out a, b, c, and d...

(1) 8a+4b+2c+d=11
(2) 27a+9b+3c+d=44
(3) 64a+16b+4c+d=110
(4) 125a+25b+5c+d=220
(5) 216a+36b+6c+d=385
Find a, b, c, and d.

Thanks~
 
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  • #2
At first glance it appears you have too many equations.

Other than that, why don't you just try elimination?
 
  • #3
Also, some of them can be divided into smaller numbers to.

Some are also proportional and for that reason you can get ride of one equation.

Honestly take it nice and slow, so you make know mistakes, and you'll get it.

Note: Soon you'll learn about matrices and thank "godmath" for it.
 
  • #4
Keep multiplying the equations by whole numbers to cancel out variables by subtraction like 2-1=A 3-2=B 4-3=C
then result B-A=X C-B=Y
Y-X
that should leave you with one varible equals a number
 
  • #5
Solution

(1) 8a+4b+2c+d=11
(2) 27a+9b+3c+d=44
(3) 64a+16b+4c+d=110
(4) 125a+25b+5c+d=220
(5) 216a+36b+6c+d=385

(2)-(1)= (A)
(3)-(2)= (B)
(4)-(3)= (C)

(A)= 19a + 5b + c = 33
(B)= 37a + 7b + c = 66
(C)= 61a + 9b+ c = 110

(B)-(A)=(X)
(C)-(B)=(Y)

(X)= 18a + 2b = 33
(Y)= 24a + 2b = 44

(Z) = (Y)-(X)

(Z)= 6a=11
a=11/6

therefore
by Equation (X)
(X)= 18a + 2b = 33
18(11/6)+ 2b = 33
b=0

therefore
by equation (A)
19a + 5b + c = 33
19(11/6) + 5(0) + c = 33
c=-11/6

thefore by Equation 1
8a+4b+2c+d=11

8(11/6)+4(0)+2(-11/6)+d=11
d=0


Summary
[tex] a=11/6 [/tex]
[tex] b=0 [/tex]
[tex]c= -11/6[/tex]
[tex]d= 0[/tex]
 
  • #6
Tide said:
At first glance it appears you have too many equations.

Other than that, why don't you just try elimination?

indeed you need x number of equations to solve for x number of variables

in this case i needed 4 equations since there were four variables being solved for
a,b,c,d
 
  • #7
Leaping antalope said:
Could someone solve these equations for me? It seems complex, but I believe there is a easy way to find out a, b, c, and d...
Take all of the equations and make them of the form d= ...
Then, you can put the two non d sides of the equation together to create 2 pairs in which d is elimintated entirely.

You can repeat with these 2 equations to eliminate one of the other variables. This will leave you with 2 variables.

You can use the 5th equation to start over with one of the other 4, to obtain another formula using 2 variables. Then, add them up to eliminate one of the variables. Once you have the value of one of the variables, you can fill it in the others, and repeat to discover the others.
 

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