Amy,Brian, and Carla had some basketball cards. Amy and Brian had 1250 cards. Amy and Carla had 830 cards. Brian had 4 times as many cards as Carla. How many cards did any have?

Question

I need help please

Answer 1

In order to solve this problem, let's represent it with couple of equations we can work our way through. For simplicity, we can represent Amy with A, Brian with B, and Carla with C.

Since we know "Amy and Brian had 1250 cards", our first equation will be:

A + B = 1250

We also know that "Amy and Carla had 830 cards", which we can represent with

A + C = 830

Finally, we are told "Brian had 4 times as many cards as Carla", which we can represent with

B = 4C.

By multiplying our second equation (A + C = 830) by four, we can work to remove C from the picture, making this solvable. Doing so gets us

4A + 4C = 3320

Then, by substituting B for 4C, we get

4A + B = 3320

Now, we can simply subtract our first equation (A + B = 1250) from the one we just created we get

3A = 2070

Simply dividing both sides of the equation by 3 allows us to solve for A,

A = 690

Then, we can use this to help find other solutions, starting with our second equation, meaning A + C = 830 can now be understood as 690 + C = 830. Simply by subtracting 690 from both sides of the equation, we find that C = 140.

Finally, since we already know B = 4C, we simply can multiply 140 by 4, ultimately finding that B = 560.

So, in summary, Amy has 690 cards, Brian has 560, and Carla has 140.