3x+(6-4x)=1

-x+6=1

-x=-5

x=5

## Related Questions

A system of linear inequalities is shown below. x + y _ 4 y

1) x + y ≥ 4 => y ≥ 4 – x

2) y < 2x – 3

B) Draw the lines for the following equalities:

1) y = 4 – x

2) y = 2x – 3

C) Shade the regions of solutions

1) The region that is over the line y = 4 – x

2) The region that is below the line y = 2x – 3

The solution is the intersection of both regions; this is the sector between both lines that is to the right of the intersection point, including the portion of the very line y = 4 – x and excluding the portion of the very line y = 2x – 3

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Firstly, we must calculate dx/dt ,

dx/dt = 4te^t + 4e^t=(4t + 4)e^t

dx/dt=0 implies (4t + 4)e^t=0 so 4t + 4 = 0 because e^t>0, and t= -1,

fort t = -1, x = 4 (-1)e^-1, and y=4(-1)e^1,

x= – 4 / e, and y= – 4 e

Then find the exact coordinates.

Highest point:- 4/e

Leftmost point: – 4e

Horizontal asymptote: lim 4te^_t

lim 4te^_t = lim 4/ e^t /t = 0, (x=o H.A), because e^t /t = infinity when t= infinity, so

t=infinity t=infinity

lim 4te^t =4 limte^t = 4 x o =0

t= – infinity t= – infinity

The curve has one horizontal asymptote and one vertical asymptote. Find them.

Horizontal asymptote: y= . 0

Vertical asymptote: x= 0

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