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goldfiish [28.3K]
3 years ago
14

1. 6x+2y=- 4 3x+4y=-17

Mathematics
1 answer:
zmey [24]3 years ago
8 0

Answer:

x = 1

y = -5

Step-by-step explanation:

In this question, we need to find what the values of "x" and "y" are.

To do this, we must solve the equations.

Solve:

6x + 2y = -4

3x + 4y = -17

Lets solve for "y" first:

6x + 2y = -4

-2(3x + 4y = -17) = -6x - 8y = 34

Now subtract the two equations from each other.

6x + 2y = -4

-6x - 8y = 34

-6y = 30

Divide both sides by -6

y = -5

Now, we know that the "y" variable's value is -5.

Lets find the value of "x" by plugging -5 to "y" in one of the equations:

6x + 2(-5) = -4

6x - 10 = -4

Add 10 to both sides.

6x = 6

Divide both sides by 6.

x = 1

The value of "x" is 1.

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What is the smallest integer $n$, greater than $1$, such that $n^{-1}\pmod{130}$ and $n^{-1}\pmod{231}$ are both defined?
olasank [31]

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

\begin{cases} 17x \equiv 1 \pmod{130} \\ 17y \equiv 1 \pmod{231} \end{cases}

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

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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)

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

X * 5% = $1.50

X = $1.50 / 5%

Or by using simple rule of three:

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Net + tax = Total Charge

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