A variable by itself doesn't have a co efficient
First, "boxes of two sizes" means we can assign variables: Let x = number of large boxes y = number of small boxes "There are 115 boxes in all" means x + y = 115 [eq1] Now, the pounds for each kind of box is: (pounds per box)*(number of boxes) So, pounds for large boxes + pounds for small boxes = 4125 pounds "the truck is carrying a total of 4125 pounds in boxes" (50)*(x) + (25)*(y) = 4125 [eq2] It is important to find two equations so we can solve for two variables. Solve for one of the variables in eq1 then replace (substitute) the expression for that variable in eq2. Let's solve for x: x = 115 - y [from eq1] 50(115-y) + 25y = 4125 [from eq2] 5750 - 50y + 25y = 4125 [distribute] 5750 - 25y = 4125 -25y = -1625 y = 65 [divide both sides by (-25)] There are 65 small boxes. Put that value into either equation (now, which is easier?) to solve for x: x = 115 - y x = 115 - 65 x = 50 There are 50 large boxes.
Answer:13/34
Step-by-step explanation:
5r-4 and r=-5 so it would be 5(-5)=-20, then -4 , so is is -24
i would pick number C) since i think it would be the correct answer
