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# a+b=4ab b+c=5bc c+a=3ac

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a+b=4ab b+c=5bc c+a=3ac
Mar 1, 2015

#2
+27310
+5

Another approach to this is to divide through by the product on the right-hand side of each equation, so that we get:

1/b + 1/a = 4

1/c + 1/b = 5

1/a + 1/c = 3

Now let A = 1/a, B = 1/b and C = 1/c, so the equations become

B + A = 4            (1)

C + B = 5            (2)

A + C = 3            (3)

Subtract (3) from (2) to get B - A = 2

Add this to (1) to get 2B = 6 so B = 3, so b = 1/3

Put B back into (1) to get A = 1, so a = 1

Put B back into (2) to get C = 2, so c  = 1/2

.

Mar 2, 2015

#1
+94235
+5

a+b=4ab   b+c=5bc   c+a=3ac

a= b(4a - 1)     a = c(3a - 1)

b = a/(4a -1)  c = a/(3a -1)

b + c = 5bc

a/(4a -1) + a/(3a -1)  = 5a^2/[(4a -1)(3a -1)]

[a(3a -1) + a(4a -1)]/ [(4a -1)(3a -1)]  = 5a^2/[(4a -1)(3a -1)]

3a^2 - a + 4a^2 - a  = 5a^2

2a^2 - 2a = 0

a^2 - a = 0

So

a(a -1) = 0     0 is trivial....so a = 1

So  b = a/[4(a) -1] = 1 / [4(1) -1] = 1/3

And  c = a/[3(a) -1] = 1 /[3(1)-1] = 1/2

Check

a + b = 1 + 1/3 = 4/3 = 4ab =4(1)(1/3) = 4/3

b + c = 1/2 + 1/3  = 5/6 =5bc = 5(1/3)(1/2) = 5/6

a + c = 1 + 1/2 = 3/2 = 3ac = 3(1)(1/2) = 3/2

Mar 1, 2015
#2
+27310
+5

Another approach to this is to divide through by the product on the right-hand side of each equation, so that we get:

1/b + 1/a = 4

1/c + 1/b = 5

1/a + 1/c = 3

Now let A = 1/a, B = 1/b and C = 1/c, so the equations become

B + A = 4            (1)

C + B = 5            (2)

A + C = 3            (3)

Subtract (3) from (2) to get B - A = 2

Add this to (1) to get 2B = 6 so B = 3, so b = 1/3

Put B back into (1) to get A = 1, so a = 1

Put B back into (2) to get C = 2, so c  = 1/2

.

Alan Mar 2, 2015