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lisov135 [29]
3 years ago
9

Solve for x: 2x-4/5=x+4

Mathematics
2 answers:
shepuryov [24]3 years ago
3 0

2x-4=5x+20

-3x=24

x=-8

ra1l [238]3 years ago
3 0

Answer:

2x -4/5 = x + 4

2x -4 /5× 5 = 5(x +4)

2x -4 = 5x +20

2x -4+4 = 5x +20+4

2x = 5x +24

2x -5x = 5x-5x+24

-3x = 24

-3x/-3 = 24/-3

x = -8

hope that will help you

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Answer:

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Step-by-step explanation:

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y=\frac{-3x-24}{4}

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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?
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First of all, the modular inverse of n modulo k can only exist if GCD(n, k) = 1.

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130 = 2 • 5 • 13

231 = 3 • 7 • 11

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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

6 = 1 • 5 + 1

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231 = 13 • 17 + 10

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