Miguel and grace started collecting rare coins at the same time. back then, they had the same number of rare coins. miguel has been collecting 5 coins each week and he now has 38 coins. grace has been collecting 3 coins each week and she now has 24 coins. how many rare coins did they have altogether when they started collecting?
a) 3
b) 6
c) 7
d) 14

Option B is the correct answer

Step-by-step explanation:

Both collections follow arithmetic progression.

Back then, they had the same number of rare coins, which means their first term of AP is same. Let it be a.

Miguel's AP have a common difference 5 and Grace's AP have a common difference 3.

Let the number of weeks completed be n.

We have

           a + ( n-1 )x 5 = 38   eqn 1

           a + ( n-1 )x 3 = 24   eqn 2

eqn 1 - eqn 2

          a + ( n-1 )x 5 - (a + ( n-1 )x 3) = 38 - 24

          ( n-1 )x 2 = 14

          n-1 = 7

          n = 8

Substituting in eqn 1

           a + ( 8-1 )x 5 = 38

                     a = 38 - 35 =3

So at starting they both have 3 rare coins.

Total rare coins = 3 + 3 = 6

Option B is the correct answer

trapezoid

step-by-step explanation:

first i plotted all my points on a grid graph

i made sure all of them were in the correct quadrant

then i saw it was a trapezoid or a quadrilateral