Al
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?