First of all, the modular inverse of n modulo k can only exist if GCD(n, k) = 1.
We have
130 = 2 • 5 • 13
231 = 3 • 7 • 11
so n must be free of 2, 3, 5, 7, 11, and 13, which are the first six primes. It follows that n = 17 must the least integer that satisfies the conditions.
To verify the claim, we try to solve the system of congruences

Use the Euclidean algorithm to express 1 as a linear combination of 130 and 17:
130 = 7 • 17 + 11
17 = 1 • 11 + 6
11 = 1 • 6 + 5
6 = 1 • 5 + 1
⇒ 1 = 23 • 17 - 3 • 130
Then
23 • 17 - 3 • 130 ≡ 23 • 17 ≡ 1 (mod 130)
so that x = 23.
Repeat for 231 and 17:
231 = 13 • 17 + 10
17 = 1 • 10 + 7
10 = 1 • 7 + 3
7 = 2 • 3 + 1
⇒ 1 = 68 • 17 - 5 • 231
Then
68 • 17 - 5 • 231 ≡ = 68 • 17 ≡ 1 (mod 231)
so that y = 68.
(3t^2 + 2 + (-2)) + (t^2 + 3t - 4) - (4 + (-5))
Apply the distributive property to remove parentheses.
3t^2 + 2 - 2 + t^2 + 3t - 4 - 4 + 5
Combine like terms.
<h3>4t^2 + 3t - 3 is the simplified form of the given expression.</h3>
get the equation in slope intercept form
2x-3y=9
subtract 2x from each side
-3y = -2x +9
divide by -3
y = 2/3 x -3
slope = 2/3
y intercept = -3
x intercept set y=0 and solve
0 = 2/3 x -3
add 3 to each side
3 = 2/3 x
multiply each side by 3/2
9/2 = x
the x intercept is 9/2 or 4 1/2
m=miles
cost for 1 mile ride= $2.20
c=cost
rule c=0.20m+2.00
c=0.20x1+2
c=0.20+2
c+2.20