Word problem please help?

A = B + 5 and AB = 84 substitute into the second equatioon

(B + 5)B = 84

B^2 + 5B = 84

B^2 + 5B - 84 = 0 factor

(B - 7)(B + 12) = -

B = 7 or B = -12.. since you can't have a negative age, then B is 7

AB = 84

A(7) = 84

A = 12 So B = 7 and A = 12

Consecutive Integers. Product is 63? Well the only ones it could be are 7 and 9

Mathematically you have this.

First odd integer is = x

the next odd integer is x + 2

So multiply and equal to 63

(x)(x + 2) = 63

x^2 + 2x = 63

x^2 + 2x - 63 = -

factor

(x + 9)(x - 7) = 0

x = -9 or x = 7 chose the 7

Therefore 7 and 9 are the numbers.

Coin problem

dimes = 0.10 and quarters are 0.25

total coins = 12

d + q = 12

0.10d + 0.25q = 1.65 take the first equation and rearrange it to become d = 12 - q, now sub that into the second equation.

0.10(12 - q) + 0.25q = 1.65

1.2 - 0.10q + 0.25q = 1.65

1.2 + 0.15q = 1.65

0.15q = 1.65 - 1.2

0.15q = 0.45

q = 0.45 / 0.15

q = 3 So you have 3 quarters.

Can you do the rest now?