Math CentralQuandaries & Queries


Question from Mukulu, a student:

Solve the equation simultaneously X/5=(Y+2)/2= (Z-1)/4 ……………….eqt 1
3X+4Y+2Z-25=0 ………………eqt 2

Hi Mukulu,

If we know that X/5=(Y+2)/2= (Z-1)/4 we can use substitution to find each of the values. First we need to find the value of one of the variable, x is the easiest to solve for:

X/5=(Y+2)/2= (Z-1)/4

X=5(Y+2)/2= 5(Z-1)/4 now we can use either 5(Y+2)/2 or 5(Z-1)/4 in the place of X, I will choose 5(Y+2)/2



15(Y+2)/2 + 4Y+2Z-25=0

23Y/2 + 2Z=10

Since we know (Y+2)/2= (Z-1)/4 then (Y+2)/2- (Z-1)/4 =0 or Y/2 -Z/4=-5/4

Now using simple linear algebra, you can solve for Y & Z using

23Y/2 + 2Z=10

Y/2 -Z/4=-5/4

Once you know Y & Z, you can use X=5(Y+2)/2= 5(Z-1)/4 to find X

Hope this helps,


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