Matt and ming are selling fruit for a school fundraiser. customers can buy small boxes of oranges and large boxes of oranges. matt sold 3 small boxes of oranges and 14 large boxes of oranges for a total of $203. ming sold 11 small boxes of oranges and 11 large boxes of oranges for a total of $220. find the cost each of one small box of oranges and one large box of oranges.

a. The cost of each of one small box of oranges is equal to $7.

b. The cost of each of one large box of oranges is equal to $13.

Given the following data:

To find the cost of each of one small box of oranges and one large box of oranges:

In this exercise, we are required to translate the word problem into a mathematical (algebraic) expression and solve for the unknown variable (number):

Translating the word problem into an algebraic expression, we have;

   ....equation 1.

  ....equation 2.

Next, we would solve the system of equations by using substitution method:

From equation 1:

   ....equation 3.

Substituting eqn 3 into eqn 2:

L = $13

To find the value of S:

S = $7

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step-by-step explanation:

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3