Solve using system by elimination:
-3x + 4y = -18
x = -2y - 4

Step-by-step explanation:

equation 1 : -3x + 4y = -18

equation 2: x = -2y -4

transformed equation 2: x + 2y = -4

-3x + 4y = -18

x + 2y = -4

By elimination means we have to make one variable equal to zero or we have to take one variable out of the whole equation.

Let's eliminate the variable x. To do so, let's multiple the second equation to +3. Therefore,

-3x + 4y = -18

(3)x + (3)(2y) = (3)(-4)

-3x + 4y = -18

3x + 6y = -12

Now that both equations have the same value of x but different signs, we can now eliminate x by adding the two equations.

-3x + 4y = -18

3x + 6y = -12

0 + 10y = -30

Dividing both sides by 10, we have y = -3.

Now that we have the value of y, we can just substitute the value of y into either one of the equation OR we can do the process again but this time we need to eliminate y to find the value of x.

-3x + 4y = -18

x + 2y = -4

To eliminate y, we need to multiply the second equation by -2.

-3x + 4y = -18

(-2)x + (-2)2y = -4(-2)

-3x + 4y = -18

-2x - 4y = 8

Now that we have the same values but different signs of y, we can now add the two equations again.

-3x + 4y = -18

-2x - 4y = 8

-5x + 0 = -10

Dividing both sides by -5, we have x = 2.

CHECKING:

Let's take the second original equation,

x = -2y -4

Let's substitute the values we have found into the equation:

2 = -2(-3) - 4

2= 6 - 4

2 = 2

Both sides of the equation produced the same values. Therefore, the values are correct.

step-by-step explanation:

from the graph attached below:

domain of graph is the value of x for which the given function profit function p(x) is defined.

(a) the domain represents the number of years for which the outlet will earn a profit.

range of profit function defines the lowest to highest value of profit earned.

(b) the range of this function is (0 , 160 ). for this function, the range represents the outlet's profit earned.

"image"

step-by-step explanation:

in most of these problems, the notation is that the point with a "prime" is the image of the point without. that is, b' is the image of b.

here, there is no dilation or reflection. there is simply translation (with no direction shown). which figure you call the "preimage" and which one you call the "image" is up to you. as a matter of convention, the naming is as described above.