Click to see equation, show work pls!

x = 6, 9

Step 1:

To solve this equation, we need to subtract 3 from both sides like this:

Step 2:

We square both sides (multiply using FOIL) to get rid of the radical.

If you did not understand how these were done, here is an example:

(x + 1)(x + 2)

First terms: x * x = x^2

Outer terms: x * 2 = 2x

Inner terms: 1 * x = x

2x - x = x

Last terms: 1 * 2 = 2

x^2 + x + 2

Step 3:

Solve for x:

9x - 45 = x^2 - 6x + 9

x^2 - 6x + 9 (+45) = 9x - 45 (+45)

x^2 - 6x + 54 (-9x) = 9x (-9x)

x^2 - 15x + 54 = 0

Factor:

(x^2 - 6x) + (- 9x + 54)

x(x - 6)        - 9(x - 6)

Factor out (x - 6):

(x - 6)(x - 9) = 0

x - 6 = 0; 6 - 6 = 0

x - 9 = 0; 9 - 9 = 0

Our anwer: x = 6, 9

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.