Does the system shown always, sometimes, or never have a solution when a > b? ax + y = 8
bx + y = 5

always have a solution

Step-by-step explanation:

A linear equation in standard form is given by:

Ax + By = C, where A,B, C are real numbers and A,B is not equal to zero.

Two linear equations with equations Ax + By = C and Dx + Ey = F

Therefore the first equation becomes: (A/B)x + y = C/B and the second equation becomes (D/E)x + y = F/E ]

Both equations are said have a solution if A/B ≠ D/E, they have no solution if A/B = D/E and they have infinite solutions if A/B = D/E, C/B = F/E.

Given that:

when a > b

ax + y = 8

bx + y = 5

Since a ≠ b, but a > b, the system has a solution

when rotating a figure 180 degrees, imagine that you are able to take the figure and turn it clockwise or counterclockwise around a center point. to rotate a figure 180 degrees, you will need to apply the rule (x, y) → (-x, -y). start by using a coordinate grid with coordinates for each vertex of the figure. the center point of the coordinate grid is located at (0, 0), which is what you will rotate the figure around. write down the original coordinates of the shape you are going to rotate. then, apply the rule. for example, coordinate (1, 2) becomes (-1, -2). another example would be, coordinate (-4, -2) becoming (4, 2). after you change each original coordinate to the rotated coordinate, you will draw your new figure.

step-by-step explanation:

guess at question:

a vertical line intersects the graph of g(x,y) = 0 at two points. what conclusion follows about the function f(x) implicitly defined by g(x,f(x)) = 0?

the domain of f(x) cannot include the x coordinate of any such vertical line.

informally, f(x) is not a function, and

g(x,y)=0 cannot be solved for y uniquely.

the relation r = {(x,y) such that g(x,y) = 0} is a function if

for all (a,b) ∊ r and (c,d) ∊ r, a=c implies b=d.

step-by-step explanation: