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Algebra 14 Online
OpenStudy (anonymous):

Please help I need to finish this today or I'm grounded and I will not graduate! :()

OpenStudy (anonymous):

OpenStudy (anonymous):

hero (hero):

Standard form of a linear equation is Ax + By = C

hero (hero):

You have \(x + \dfrac{4y}{7} + 2 = 0\) Start by doing this: 1. Subtract 2 from both sides 2. Multiply both sides by 7

OpenStudy (anonymous):

\[x+4y=14\]

OpenStudy (anonymous):

\[7x+4y=14\]

OpenStudy (anonymous):

?

OpenStudy (anonymous):

@Hero which one would it be?

OpenStudy (anonymous):

@suyogsomavanshi

OpenStudy (anonymous):

General form of linear equation is Ax + by = c We are having x +( 4y /7) + 2 = 0 So we have to first get rid of 2 from left hand side can you try for it ?

OpenStudy (anonymous):

\[x+(4/7)=-2\]

OpenStudy (anonymous):

Correct but there is y with (4/7) right ??? So we get x + 4y / 7 = -2 Now we need to get rid of 7 from bottom of 4y Any idea ??

hero (hero):

If you multiplied both sides by 7 you should have gotten 7x + 4y = -14

OpenStudy (anonymous):

ok so time both sides by 7

OpenStudy (anonymous):

correct getting this ????

hero (hero):

You had it, but you forgot the negative in front of the 14

OpenStudy (anonymous):

so 7x+4y=-14 ok I didn't have the -2 before thanks and what about the first attachment

OpenStudy (anonymous):

So this would be in our desired form like Ax+By= c 7x+4y=-14 Is it clear to you ?

OpenStudy (anonymous):

yes it is thanks!

OpenStudy (anonymous):

OpenStudy (anonymous):

Any doubt ???

OpenStudy (anonymous):

I don't even know where to begin on the question I just uploaded?

OpenStudy (anonymous):

@acxbox22 @mathmale

OpenStudy (anonymous):

@suyogsomavanshi

OpenStudy (anonymous):

\[\left[\begin{matrix}4 & 7 \\ 1 & 2\end{matrix}\right]*\left(\begin{matrix}24 \\ 7\end{matrix}\right)\] i also need help with this

hero (hero):

@Erob, post it as a separate question. Close this question and post a new one.

OpenStudy (anonymous):

First we would try to fix first one

OpenStudy (anonymous):

We have to consider two odd numbers as 2n + 1 and 2n + 3 getting this ?

OpenStudy (anonymous):

Hello ... ???

OpenStudy (anonymous):

I'll post the third one seperae and k

OpenStudy (anonymous):

2n+1 and 2n+3 got it

OpenStudy (anonymous):

@suyogsomavanshi

OpenStudy (anonymous):

Now sum of them is = 2n + 1 + 2n + 3 = ---- ?

OpenStudy (anonymous):

4n+4

OpenStudy (anonymous):

perfect Now it says there sum is increased by 15 so how you could write it ??

OpenStudy (anonymous):

(4n+4)+15

OpenStudy (anonymous):

or just 4n+19

hero (hero):

\[\begin{matrix} \times & \left[ \begin{matrix} 24 \\ 7 \end{matrix} \right] \\ \begin{bmatrix} 4 & 7 \\ 1 & 2 \end{bmatrix} & \left[ \begin{matrix} \square \\ \square \end{matrix} \right] \end{matrix}\]

hero (hero):

(4)(24) + 7(7) (1)(24) + 2(7)

OpenStudy (anonymous):

thanks!

hero (hero):

That's how you set it up if you want to figure out the size of the solution matrix.

hero (hero):

It also helps if you want to figure out whether or not the matrices can even be multiplied

OpenStudy (anonymous):

We have to wind up first question you are right for 4n + 19 Now it says it is equal to 47 ...... So we get 4n + 19 = 47 getting this ???

OpenStudy (anonymous):

yes so 4n=28?

OpenStudy (anonymous):

and n=7?

OpenStudy (anonymous):

Perfect

OpenStudy (anonymous):

Thanks!

OpenStudy (anonymous):

So as we assume our two numbers 2n+1 and 2n+3 Now just plug n=7 into this and find out numbers can you try ?

OpenStudy (anonymous):

15 and 17?

OpenStudy (anonymous):

Perfect

OpenStudy (anonymous):

Is it clear to you ?

OpenStudy (anonymous):

Yes it is thanks!

OpenStudy (anonymous):

You are most welcome :)

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