Is this correct or no?

Mathematics

2 answers:

Vikentia [17]3 years ago

8 0

Answer: The steps are not correct in several places. The correct result is "x > 2"

Explanation:

First row: they multiply both sides by 2, but on the right hand side the multiplication should act on the entire term "(3-2x)" so 2(3-2x)=6-4x

Second row should be now: 3x-8 > 6-4x | now adding 8 to both sides:

Third row should look like this : 3x > 14 - 4x

Fourth row: we want to bring the term 4x to the left so it should be something like this:

3x > 14 - 4x | add "4x" to both sides

which results in

3x + 4x > 14 - 4x + 4x

7 x > 14 | divide both sides by 7

x > 14/7

x > 2

Let me know if you have questions

Tanya [424]3 years ago

4 0

Yes.??? dont really Know sorry

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Find the solution of the given initial value problem:<br><br> y''- y = 0, y(0) = 2, y'(0) = -1/2

igor_vitrenko [27]

Answer: The required solution of the given IVP is

$y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.$

Step-by-step explanation: We are given to find the solution of the following initial value problem :

$y^{\prime\prime}-y=0,~~~y(0)=2,~~y^\prime(0)=-\dfrac{1}{2}.$

Let $y=e^{mx}$ be an auxiliary solution of the given differential equation.

Then, we have

$y^\prime=me^{mx},~~~~~y^{\prime\prime}=m^2e^{mx}.$

Substituting these values in the given differential equation, we have

$m^2e^{mx}-e^{mx}=0\\\\\Rightarrow (m^2-1)e^{mx}=0\\\\\Rightarrow m^2-1=0~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mx}\neq0]\\\\\Rightarrow m^2=1\\\\\Rightarrow m=\pm1.$