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avatar+28 

How many pairs of integers $(b,c)$ satisfy the equation

 

b+7/b+4 = c/9

 Aug 2, 2022

Best Answer 

 #2
avatar+2455 
0

We have: \({b+7 \over b + 4} = {c \over 9}\)

 

Cross multiply: \(9(b+7) = c(b + 4)\)

 

Distribute both sides: \(9b + 63 = bc + 4c \)

 

Subtract 9b from both sides: \(bc + 4c - 9b = 63\)

 

Factor out b: \(b(c-9) + 4c = 63\)

 

Subtract 36 from both sides (Simon's Favorite Factoring Trick): \(b(c-9) + 4c - 36 = 27\)

 

Factor out c: \(b(c-9) + 4(c-9) = 27\)

 

Simplify the left-hand side: \((b+4)(c-9) = 27\)

 

The (positive and negative) factor pairs of 27 are (1,27), (-1,-27), (-3,-9) and (3,9). 

 

Each of those cases has 2 pairs of integers that work, which makes for \(4 \times 2 = \color{brown}\boxed{8}\) integers. 

 Aug 2, 2022
 #1
avatar+28 
+1

it is 8

 Aug 2, 2022
 #2
avatar+2455 
0
Best Answer

We have: \({b+7 \over b + 4} = {c \over 9}\)

 

Cross multiply: \(9(b+7) = c(b + 4)\)

 

Distribute both sides: \(9b + 63 = bc + 4c \)

 

Subtract 9b from both sides: \(bc + 4c - 9b = 63\)

 

Factor out b: \(b(c-9) + 4c = 63\)

 

Subtract 36 from both sides (Simon's Favorite Factoring Trick): \(b(c-9) + 4c - 36 = 27\)

 

Factor out c: \(b(c-9) + 4(c-9) = 27\)

 

Simplify the left-hand side: \((b+4)(c-9) = 27\)

 

The (positive and negative) factor pairs of 27 are (1,27), (-1,-27), (-3,-9) and (3,9). 

 

Each of those cases has 2 pairs of integers that work, which makes for \(4 \times 2 = \color{brown}\boxed{8}\) integers. 

BuilderBoi Aug 2, 2022
 #3
avatar+124597 
+1

Very nice, BB!!!!

 

 

 

cool cool cool

CPhill  Aug 2, 2022

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