Which of the points are solution to the inequality? check all that apply. y < 2x + 3 a. (-3,3) b. (-2,-2) c. (-1,1) d. (0,1) e. (2,5)

Which of the points are solution to the inequality? Check all that apply. 
y < 2x + 3 
A. (-3,3)
B. (-2,-2)
C. (-1,1)
D. (0,1)
E. (2,5)

We can sketch each point on the graph and find out if each point lies in the shaded zone.  If so, it is a solution to the inequality, otherwise, it's not.
We see from the graph attached that 
A is outside of the shaded area, so it is NOT a solution.
B is inside the shaded area, it is a solution.
C is on the boundary line.  However, the boundary line is dotted, meaning only a strict inequality qualifies (i.e. equality does not count), so it is NOT a solution.
D is within the shaded area, so it is a solution.
E is within the shaded area, so it is a solution.

Alternate solution: 
We can substitute the values of (x,y) into the inequality and see if the inequality is satisfied.  If the evaluated values satisfy the inequality, then (x,y) used is a solution.
A. (-3,3) => 3<2(-3)+3=-3  since 3>-3, so (-3,3) is NOT a solution.
B. (-2,-2) => -2<2(-2)+3=-1 since -2<-1, so it (-2,-2) is a solution.
C. (-1,1) => 1<2(-1)+3=1 since 1=1 (and not <1), so (-1,1) is NOT a solution
D. (0,1) => 1<2(0)+3=3 since 1<3, (0,1) is a solution
E. (2,5) => 5<2(2)+3=7 since 5<7, (2,5) is a solution.

So we come to the same conclusions at the graphical method.

Ifound this on google “that means that the graph of a function with direct variation has these features: it is a straight line: because the slope is the ratio y/x = k. it passes through the origin (0,0): because y = kx = k (0) = 0.” i hope this you for what your looking for