In any case, what, exactly, do you mean by "intercepts". You could mean the x and y -intercepts but those have y-coordinate 0 or x-coordinate 0, respectively. You could mean the point of intersection (not "intercept") of the two lines but that is only a single point. The main reason for confusion is that you haven't stated a problem! You give two inqualities but you haven't said what you are to do with them.
"Graph" the points (x,y) that satisfy both? Okay, if that is the problem then the first thing you should do is graph the equalities . That is, first graph 4x+ 10y= 40 and -6x- 8y= 48. When x= 0, 4(0)+ 10y= 40 so y= 4. (0, 4) is the y-intercept for the first line. When y= 0, 4x+10(0)= 40 so x= 10. (10, 0) is the x-intercept. Ah, now I see what you meant before. I think of the "intercepts" as points but you were just giving the non-zero component. Now, since a line is determined by two points, you can just draw the line through (0, 4) and (10, 0). Similarly, (-8, 0) and (0, -6) are the x and y intercepts are the line -6x -8y= 48. Draw the line through those two point.
That gives you two intersecting straight lines. You will particularly want to note the coordinates of the point where the two lines intersect. You can do that by solving the two equations 4x+ 10y= 40 and 6x- 8y= 48. A good start would be to multiply the first equation by 3 and the second equation by 2 and then adding.
Those two lines are the boundaries of the region you want. They divide the plane into four regions. Choose one point in each region to determine if points in that region satisfy the two inequalites or not. Note that one of the inequalities has an "=". Mark that line darker to indicate that points on that line are included in the regions satisfying the inequality.