Jeans costs $15 each and shirts costs $9 each. you have at most $90 to spend clothing. you must purchase at least 3 pairs of jeans. let x be the amount of jeans purchased and y be the amount of shirts purchased. what is a possible solution?

The possible solutions are:

       6 jeans and 0 shirts

       5 jeans and 0 shirts

       5 jeans and 1 shirt

       4 jeans and 0 shirt

       4 jeans and 1 shirt

        4 jeans and 2 shirt

       4 jeans and 3 shirt

       3 jeans and 0 shirt

        3 jeans and 1 shirt

       3 jeans and 2 shirt

       3 jeans and 3 shirt

        3 jeans and 4 shirt

      3 jeans and 5 shirt

Cost of 1 jeans=$ 15

and cost of 1 shirt=$ 9

Let x be the amount of jeans purchased and

y be the amount of shirts purchased.

You must purchase at least 3 pairs of jeans.

This means that x ≥ 3---------(1)

The total amount to be spent in shopping is: Atmost $ 90

This means that:

  15x+9y ≤ 90------------(2)

From the graph we may see that the solution that lie in the feasible region are labeled in the graph.

( Since, the solution will be where we get the integer value of x and y in the feasible region)

30 units^2

step-by-step explanation:

area is l × w

the length is 3

and the width is 10

3 × 10= 30 units squared

that is the sum

step-by-step explanation:

there are no other answers without knowing the variables values