There's no need for a 3rd equation. We aren't solving matrices here.

You should have gotten the two equations:

a = 1.2b - 5 ①
c = (b - 3)/0.95 ②

The common term is b so we need to find out the range of b.

Since you are given a,b,c < 70, substitute ① and ② into the inequality.

a < 70

Sub ①,

1.2b - 5 < 70
1.2b < 75
b < 75/1.2
b < 62.5
b ≤ 62 (since prices to nearest dollar)


c < 70

Sub ②,

(b - 3)/0.95 < 70
b - 3 < 66.5
b < 69.5
b ≤ 68 (since prices to nearest dollar)


Overall, b ≤ 62

Next , we look at ① again.

a + 5 = 1.2b
a + 5 = 6b/5

Since a is to the nearest dollar, it is a positive integer. Then (a + 5) is also a positive integer. So 6b/5 must also be a positive integer in order to equal (a + 5).

Using the concept of prime factorisation,

Now 6 = 2¹ × 3¹ ,but 5 = 5¹. 6 cannot be divided exactly by 5

So 6b/5 = (2¹×3¹)b/5¹ and for it to be an integer, b must be a multiple of 5


We can say b = 5k , k ∈ Z+ (for all positive integers k)

Since b ≤ 62, then b can only be 5,10,15,...55,60


Then, we look at ② again

0.95c + 3 = b
b - 3 = 19c/20


Since b is to the nearest dollar, it is a positive integer. Then (b - 3) is also a positive integer. So 19c/20 must also be a positive integer in order to equal (b - 3).

Now 19 is a prime number = 19¹ × 1 , but 20 = 2² × 5¹. 19 cannot be divided exactly by 20.

So 19c/20 = 19c/(2²×5¹) and for it to be an integer, c must be a multiple of 20

We can say c = 20k , k ∈ Z+ (for all positive integers k)


Since c < 70, c can only be 20 or 40 or 60.

So b can only be :

19(20)/20 + 3 = 19 + 3 = 22
Or
19(40)/20 + 3 = 38 + 3 = 41
Or
19(60)/20 + 3 = 57 + 3 = 60

Only the last one is a multiple of 5.

So b = 60, c = 60

a = 6(60)/5 - 5 = 72 - 5 = 67

Some other things we can observe :


6b/5 = a + 5
6b/5 - 5 = a


a + 5 = 6b/5
a + 5 = 6(5k/5)
a + 5 = 6k


Since b is a multiple of 5, then 6b/5 is a multiple of 6 and also a multiple of 3 and 2.

This tells us that 6b/5 is even. Subtracting 5 gives us an odd number so a must be odd.

How to derive the general integer solution to the question :

Since 19c/20 = b - 3 , and we let c = 20k, k ∈ Z+,

Then, 19(20k)/20 = b - 3
19k = b - 3
b = 19k + 3


Now 19k + 3 must be divisible by 5 since b is an integer that is a multiple of 5.

Neither 19 nor 3 are divisible by 5, and we don't know which values of k would result in b being a multiple of 5 and which wouldn't.


We rewrite k as something else. This ensures that no matter what value of the unknown is substituted, b is multiple of 5.


Further breaking it down,

19k + 3
= 20k - k + 5 - 2
= 20k + 5 - k - 2
= 5(4k + 1) - (k + 2)


We just need to redefine k + 2 as a multiple of 5 and we rewrite everything else.

The simplest we can let k be is k = 5n + 3 where n is a integer, n ≥ 0. This makes k + 2 = 5n + 5 , which is a multiple of 5.



So,

b = 19(5n + 3) + 3 = 95n + 60
c = 20(5n + 3) = 100n + 60

a = 6(95n + 60)/5 - 5 = 114n + 72 - 5
= 114n + 67


This links all three variables to a single unknown.


First few possible sets of a,b,c (if their prices are not restricted to < 70)

67,60,60
181,155,160
295,250,260
405,345,360
523,440,460