Nickels and Dimes = some amount, how many nickels?

[LearninDaMath]

1. The problem statement, all variables and given/known data

Bob has 50 coins, all nickels and dimes, worth a total of 4.85. How many nickels does he possess?

2. Relevant equations

n + d = 50

5n + 10d = 485

Multiplying the first equation by 10, we obtain 10n+10d=500

Subtracting the second equation from the third equation, we get 5n=15, so n = 3

3. The attempt at a solution

I have the solution. My question is:

Why do you multiply the first equation by 10 to begin with? Also, why do you then subtract one equation from the other? What is the rationale to it?

Thanks in advance

[Mark44]

1. The problem statement, all variables and given/known data

Bob has 50 coins, all nickels and dimes, worth a total of 4.85. How many nickels does he possess?

2. Relevant equations

n + d = 50

5n + 10d = 485

Multiplying the first equation by 10, we obtain 10n+10d=500

Subtracting the second equation from the third equation, we get 5n=15, so n = 3

3. The attempt at a solution

I have the solution. My question is:

Why do you multiply the first equation by 10 to begin with? Also, why do you then subtract one equation from the other? What is the rationale to it?

Thanks in advance

If you multiply both sides of an equation by the same number, you get an equivalent equation (same solution set). Multiplying the first equation by 10 results in 10n + 10d = 500.

If you subtract the same number from both sides of an equation, you also get an equivalent equation. In this case, you subtracted 5n + 10d from 10n + 10d, and 485 from 500, resulting in 5n = 15, or n = 3. Since 5n + 10d = 485, you are actually subtracting the same number from both sides of the equation 10n + 10d = 500.

Another way to do this problem is to solve for either n or d in the first equation, and then substitute into the second equation.

[LearninDaMath]

ah i get it, it's like doing a "systems of equations" problem. Thanks! And yea, as soon as I posted the question, my sister recognized that it can also be solved using substition as you mentioned.