Last question, could you please help me. Graphing the system of inequalities.

to graph y >= 2X-1

you have to convert it to function like this y = 2X-1 and the get any number for the variable of x and calculate what value of y will be ex. if X=0 then y= -1 and if x= 1/2 then y= 0 and if x=1 then y = 3 then we now have three ordered pairs for x and y which are (0,-1) (1/2,0) (1,3) no2 we will draw this point and since it is linear function it will be drawn as straight line after we draw the straight line we will ask ourselves a question is the sign given in the question is

1- if it is (less then or equal to)(<=) or (more then or equal to)(>=) we draw the straight line solid one to include all the points on it as in the first inequality

2- if the sign is less than(>) or more than(<) then we draw the line dashed

since the the first inequality we drawn (y >= 2X-1) has equal then we will draw it solid

then we will assume that X=0 and y=0 as the origin point we ignored the equal sign here be we need to know the area that describe the > sign we already finished drawing if it was equal only and calculate it 0 > 2×0 -1 here we put 0 instead of x and y in the same equation so is 0 > 2×0-1 is true (is this statement is true) well 2×0-1 = -1 then the right hand side equal -1 and left hand side = 0 then is 0 > -1 yes this statement is true the the area this inequality apply is towards the origin point then we will draw a solid straight line cuz if the equal sign and the ordered pairs we got em as if it were a function by ignoring the > sign and the area we will shade that apply this inequality is towards the origin point

the 2 other inequality are drawn the same way and then u take the intersection area this is will be the answer

i think it might be a but you might want a second opinion