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valentina_108 [34]
4 years ago
5

Solve the following simultaneous linear congruences.

Mathematics
1 answer:
Anastaziya [24]4 years ago
5 0

a. The moduli are coprime, so you can apply the Chinese remainder theorem directly. Let

x=4\cdot5+3\cdot5+3\cdot4

  • Taken mod 3, the last two terms vanish, and 20\equiv2\pmod3 so we need to multiply by the inverse of 2 modulo 3 to end up with a remainder of 1. Since 2\cdot2\equiv4\equiv1\pmod3, we multiply the first term by 2.

x=4\cdot5\cdot2+3\cdot5+3\cdot4

  • Taken mod 4, the first and last terms vanish, and 15\equiv3\pmod4. Multiply by the inverse of 3 modulo 4 (which is 3 because 3\cdot3\equiv9\equiv1\pmod4), then by 2 to ensure the proper remainder is left.

x=4\cdot5\cdot2+3\cdot5\cdot3\cdot2+3\cdot4

  • Taken mod 5, the first two terms vanish, and 12\equiv2\pmod5. Multiply by the inverse of 2 modulo 5 (3, since 3\cdot2\equiv6\equiv1\pmod5) and again by 3.

x=4\cdot5\cdot2+3\cdot5\cdot3\cdot2+3\cdot4\cdot3\cdot3

\implies x=238

By the CRT, we have

x\equiv238\pmod{3\cdot4\cdot5}\implies x\equiv-2\pmod{60}\implies\boxed{x\equiv58\pmod{60}}

i.e. any number 58+60n (where n is an integer) satisifes the system.

b. The moduli are not coprime, so we need to check for possible contradictions. If x\equiv a\pmod m and x\equiv b\pmod n, then we need to have a\equiv b\pmod{\mathrm{gcd}(m,n)}. This basically amounts to checking that if x\equiv a\pmod m, then we should also have x\equiv a\pmod{\text{any divisor of }m}.

x\equiv4\pmod{10}\implies\begin{cases}x\equiv4\equiv0\pmod2\\x\equiv4\pmod5\end{cases}

x\equiv8\pmod{12}\implies\begin{cases}x\equiv0\pmod2\\x\equiv2\pmod3\end{cases}

x\equiv6\pmod{18}\implies\begin{cases}x\equiv0\pmod2\\x\equiv0\pmod3\end{cases}

The last congruence conflicts with the previous one modulo 3, so there is no solution to this system.

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Katrina buys a 42-ft roll of fencing to make a rectangular play area for her dogs.
vekshin1

Answer:

(a) l = -w + 21

(b) Domain: 0 <em>(See attachment for graph)</em>

(c) f(w) = -w + 21

Step-by-step explanation:

Given

2(l + w) = 42

l = length

w = width

Solving (a): A function; l in terms of w

All we need to do is make l the subject in 2(l + w) = 42

Divide through by 2

l + w = 21

Subtract w from both sides

l + w - w = 21 - w

l  = 21 - w

Reorder

l = -w + 21

Solving (b): The graph

In (a), we have:

l = -w + 21

Since l and w are the dimensions of the fence, they can't be less than 1

So, the domain of the function can be 0

--------------------------------------------------------------------------------------------------

To check this

When w = 1

l = -1 + 21

l = 20

(w,l) = (1,20)

When w = 20

l = -20 + 21

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(w,l)= (20,1)

--------------------------------------------------------------------------------------------------

<em>See attachment for graph</em>

<em></em>

Solving (c): Write l as a function f(w)

In (a), we have:

l = -w + 21

Writing l as a function, we have:

l = f(w)

Substitute f(w) for l in l = -w + 21

l = -w + 21 becomes

f(w) = -w + 21

5 0
3 years ago
The answers to this question cousin
podryga [215]
If R is between Q and S then QR+RS=QS

given
QR=x+4
RS=3x-1
and QS=27

x+4+3x-1=27
combine like terms
4x+3=27
minus 3 both sides
4x=24
divide both sides by 4
x=6

sub back
QR=x+4
QR=6+4
QR=10

RS=3x-1
RS=3(6)-1
RS=18-1
RS=17

x=6
QR=10
RS=17
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3 years ago
Elisa ​says, "If you subtract 14 from my number and multiply the difference by -3 ​, the result is -33 ​." What is Elise​'s ​num
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Answer:

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