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vlabodo [156]
3 years ago
9

Solve the simultaneous equations by substitution 3y - 2x = -7 5x + 3y = 28

Mathematics
2 answers:
iragen [17]3 years ago
4 0

Answer:

(5, 1 )

Step-by-step explanation:

Given the 2 equations

3y - 2x = - 7 → (1)

5x + 3y = 28 → (2)

Rearrange (1) expressing 3y in terms of x by adding 2x to both sides

3y = - 7 + 2x

Substitute 3y = - 7 + 2x into (2)

5x - 7 + 2x = 28

7x - 7 = 28 ( add 7 to both sides )

7x = 35 ( divide both sides by 7 )

x = 5

Substitute x = 5 into either of the 2 equations and solve for y

Substituting into (2)

5(5) + 3y = 28

25 + 3y = 28 ( subtract 25 from both sides )

3y = 3 ( divide both sides by 3 )

y = 1

solution is (5, 1 )

Otrada [13]3 years ago
4 0

Answer:

-(3y-2x= -7)

3y+5x= 28

-3y+2x= 7

3y+5x= 28

7x=35

7x/7= 35/7

x=5

5(5)+3y= 28

25+3y=28

3y= 28-25

3y=3

3y/3=3/3

y= 1

3y-2x= -7

3(1) -2x= -7

3-2x= -7

-2x= -7-3

-2x= -10

-2x/-2=-10/-2

x= 5

chekin

5x+3y=28

5(5)+3(1)=28

25+3=28

28=28

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Count the number of positive integers less than 100 that do not contain any perfect square factors greater than 1.

Possible perfect squares are the squares of integers 2-9.
In fact, only squares of primes need be considered, since for example, 6^2=36 actually contains factors 2^2 and 3^2.
Tabulate the number (in [ ])of integers containing factors of 
2^2=4: 4,8,12,16,...96 [24]
3^2=9: 9,18,....99  [11] 
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So the total number of integers from 1 to 99
N=24+11+3+2=40
=>
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3 years ago
Can someone answer this question please answer it correctly if it’s corect I will mark you brainliest
son4ous [18]

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3 years ago
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Use the method of undetermined coefficients to find the general solution to the de y′′−3y′ 2y=ex e2x e−x
djverab [1.8K]

I'll assume the ODE is

y'' - 3y' + 2y = e^x + e^{2x} + e^{-x}

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y'' - 3y' + 2y = 0

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For nonhomogeneous ODE (1),

y'' - 3y' + 2y = e^x

consider the ansatz particular solution

y = axe^x \implies y' = a(x+1) e^x \implies y'' = a(x+2) e^x

Substituting this into (1) gives

a(x+2) e^x - 3 a (x+1) e^x + 2ax e^x = e^x \implies a = -1

For the nonhomogeneous ODE (2),

y'' - 3y' + 2y = e^{2x}

take the ansatz

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Substitute (2) into the ODE to get

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y'' - 3y' + 2y = e^{-x}

take the ansatz

y = ce^{-x} \implies y' = -ce^{-x} \implies y'' = ce^{-x}

and solve for c.

ce^{-x} + 3ce^{-x} + 2ce^{-x} = e^{-x} \implies c = \dfrac16

Then the general solution to the ODE is

\boxed{y = C_1 e^x + C_2 e^{2x} - xe^x + xe^{2x} + \dfrac16 e^{-x}}

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balu736 [363]

Answer:

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