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# help pls

0
233
3
+30

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

b+7/b+4 = c/9

Aug 2, 2022

#2
+2667
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
+30
+1

it is 8

Aug 2, 2022
#2
+2667
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.

BuilderBoi Aug 2, 2022
#3
+129690
+1

Very nice, BB!!!!

CPhill  Aug 2, 2022