Fiona solved the equation shown. – StartFraction 5 Over 3 EndFraction v plus 4 equals 8 minus StartFraction 1 Over 3 EndFraction v.(6x – 3) = –
A table titled Fiona's Solution with 3 columns and 5 rows. The first row is, blank, Steps, Resulting equation. The second row has the entries, 1, Use the distributive property to simplify. StartFraction one-half minus 2 x plus 1 equals negative StartFraction 13 Over 2 EndFraction. The third row has the entries, 2, blank, StartFraction 3 Over 2 minus 2 x equals negative StartFraction 13 Over 2 EndFraction. The fourth row has the entries, 3, Isolate the variable expression by using the addition property of equality, negative 2 x equals negative 8. The fifth row has the entries, 4, Isolate the variable by using the division property of equality, x equals 4.
What is the missing step of her solution?

Simplify by combining like terms.
Simplify by using addition property of equality.
Simplify by using the multiplication property of equality.
Simplify by using the division property of equality.

55/i dont know i do it for the points

step-by-step explanation:

b. x + 1

step-by-step explanation:

i'am smart

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'.

idk

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