Answer:
The graph will be a solid line with a y-intercept of negative four and a slope of seven. The graph will be shaded above the line.
Step-by-step explanation:
In IR2 we can write any line in this form :
[tex]y=ax+b[/tex]
Where ''a'' is the slope and when [tex]x=0[/tex] ⇒
[tex]y=(a).(0)+b\\y=b[/tex] ⇒ [tex](0,b)[/tex] is the point in which the line intersects the y-axis and ''b'' is the y-intercept.
To solve the linear equality [tex]y\geq7x-4[/tex] we must first work with the expression [tex]y=7x-4[/tex]
This line has a slope of 7 given that 7 is multiplying the variable x and a y-intercept of - 4
This line will divide the plane into two regions :
The region above the line and the region below the line
To decide which region we need to shade, we can put any arbitrary point in the expression [tex]y\geq 7x-4[/tex] and see what happen.
For example the point [tex](0,0)[/tex] ⇒
[tex]y\geq 7x-4\\0\geq (7).(0) - 4\\0\geq -4[/tex]
This final expression is true, therefore the point [tex](0,0)[/tex] belongs to the region [tex]y\geq 7x-4[/tex] and we conclude that we need to shade above the line.
Given that in the expression [tex]y\geq 7x-4[/tex] we have the symbol [tex]\geq[/tex] and not > , we must include the line in the region. The line will be a solid line.
The correct description is '' The graph will be a solid line with a y-intercept of negative four and a slope of seven. The graph will be shaded above the line ''
I add a graph of the linear equality.
