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### Unit 4: Lesson 13

Lesson 16: Solving problems with systems of equations

# Systems of equations with substitution: coins

Sal solves a word problem about the number of nickels and quarters in a piggy bank by creating a system of equations and solving it. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

• why nickle is 0.05 and quarters are 0.25? In question it is not given how sal find it? • American coins are based on portions of a dollar, and the standards are as follows:

One dollar = 100 pennies. Each penny is \$0.01
One dollar = 20 nickles. Each nickle is \$0.05
One dollar = 10 dimes. Each dime is \$0.10
One dollar = 4 quarters. Each quarter is \$0.25
• Could you solve a coin problem with 3 variables? (Q+N+D)?

How would you do it (if it can be done)? • A quick question that came to my head..... How about if she had 17 coins or 19 coins, is it possible that the total price of the 19 coins still be worth 2.00 dollars, if she only had nickels and quarters. • If you tried to solve those you'd get a fraction as your answer, which although it would satisfy the equation, wouldn't be a real solution, since in the real world you can't have a fraction of a coin. Only some combinations of the number of coins and the total money will produce whole number solutions, and so not all combinations are possible. In fact, this is one way of checking if a combination is possible - solve the system, and see if the solution is whole numbers of coins.
• When substituting a negative number with a positive number with a variable, would the answer be negative? • How did u get value of n as 0.05 and q as 0.25 ?
(1 vote) • clear the decimals units and work with whole numbers • Oh, man. Can someone please help with one of these KA quiz questions? I got it right but don't understand how the equations can give 2 different answers. For instance, K + L = 450. And 3L = 190 + K. Both are true systems of equations that are provided. I added them together two different ways, still equal, but rearranged appropriately. Trial 1:

K + L = 450
K + 190 = 3L (I just reversed what was on each side of the equal sign)
2K + L + 190 = 450 + 3L
then subtract the L and 190 from both sides:
2K = 260 + 2L
Divide everything by 2:
K = 130 + L
The above turns out to be true, but not helpful on its own.

But then if you add them this way:
K + L = 450 (same as above)
Change:
3L = K + 90 (same as above)
To:
3L - K = 190 (same as second equation, just subtracting K from both sides and having the 3L on the on the left)

Add both equations up, the Ks cancel out and you're left with:
4L = 640

So L = 160 and K = 290

How is it possible that just rearranging the equations like that changes the end result? They are both correct, but only one gives direct answer leaving only one variable. Isn't that all we're doing when solving equations is rearranging anyway? I would have thought that as long as we don't mess up the equality, they both would provide the exact same result. So how does that lead us down 2 separate paths? And what do we do about it when solving future equations?

If anyone has the patience to read through and understand what I tried to explain, eternal thanks to you!
(1 vote) • It's not so much that you have different result as the first time you added the equations, you didn't finish the work. You never found the numeric values of L and K. Your second attempt is a correct approach.

Remember, to find numeric answers, you need to manipulate and add the equations in such a way to eliminate a variable. As long as you have 2 variables in the equation, you can't find the specific numeric values to solve the system.

In your 2nd attempt, you added and eliminated "k". This is the eliminate method because at the point your add the equations your goal is to eliminate a variable. You then have an equation with a single variable to find.

If you use substitution method, you solve one of the equations for a single variable. For example, change K+L=450 into K=450-L. You can then use the value of "k" to substitute into the other equation. The substitution forces "k" out of the equation leaving you with a single variable to find.
K+190=3L becomes 450-L+190=3L
If you solve this, you get the same result that you found of L=160.

Hope this helps.
• how would you graph this
(1 vote) • How do you embed things like times in the video and hyperlink them so someone can just click and see it?
(1 vote) • How do you solve x-y= 3 over 2x- 3y= -3 with substitution.
(1 vote) • Solve for x in the first equation:
x = y + 3
Now substitute your x into the second equation:
2 ( y + 3 ) - 3y = -3
Since we now have one equation with one variable, when can solve for y.
2y + 6 - 3y = -3 // -y + 6 = -3
-y = -9 // y = 9
Substitute y back into the 1st equation and solve for x.
x - 9 = 3 // x = -6
It doesn't matter which variable you solve for first, although you generally want to use the least complicated equation. The first equation had variables with coefficients of 1, so theat was the easiest.
(1 vote)

## Video transcript

As a birthday gift, Zoey gave her niece an electronic piggy bank that displays the total amount of money in the bank as well as the total number of coins. After depositing some number of nickels and quarters only-- so we only have nickels and quarters-- the display read money \$2.00, number of coins 16 How many nickels and quarters did Zoey put in the bank? So let's define some variables here. Let's let n equal the number of nickels. Maybe I'll write "let" here. Let's let q be equal to the number of quarters. So how many total coins do we have? We'', it's going to be the number of nickels plus the number of quarters. So we have the nickels plus the quarters need to be equal to-- well, it tells us we have 16 total coins. So if we add up the total number of nickels plus the number of quarters, we have 16 coins. So that's one equation right there. And then how much total money do we have? Well, however many nickels we have, we can multiply that times 0.05, and that'll tell us how much money we have in nickels. So 0.05 times the nickels plus the amount of money we have in quarters. Well, that'll just be \$0.25 per quarter, or 0.25 of \$1. So let me write 0.25 times the number of quarters. For example, if I had 4 quarters and no nickels, I'd have 4 times \$0.25 which is \$1. And no money due to nickels. So it's however may nickels times \$0.05 plus however many quarters times \$0.25. That's the total amount of money I have. And her piggy bank tells me that is \$2.00. That is equal to \$2.00. So we have two equations with two unknowns. We can solve for n and q. And let's do it by substitution. So the easiest thing that we could do here, let's solve for q over here. So if n plus q is equal to 16, we could subtract n from both sides of this equation. So if n plus q is equal to 16, if we subtract n from both sides, we get q is equal to 16 minus n. So all I did is I rewrote this first constraint right over there. So since this first constraint is telling us that q, the number of quarters, must be 16 minus the number of nickels, in the second constraint, every place that we see a q, every place we see quarters, we can replace it with 16 minus n. So let's do that. So the second constraint when we make the substitution becomes 0.05n plus 0.25. Instead of q, I'm going to write 16 minus n. That's what the first constraint tells us. q must be 16 minus n. That is going to be equal to \$2.00. We're solving this system by substitution. Now let's see if I can simplify this. So we have 0.05n plus-- let's distribute the 0.25 times the 16 and the 0.25 times the negative n. 0.25 times 16, that's the same thing as 1/4 times 16. That's just going to be 4. And then 0.25 times negative n is minus 0.25n. And that is going to be equal to \$2.00. Let me scroll down a little bit. I'll scroll down a little bit. See we have 0.05n minus 0.25n. So if I have 0.05 minus 0.25, let me combine these terms. So if I have 0.05 of something, and I'm going to subtract from that 0.25 of that something, that'll give me negative 0.20 of that something. If I combine these two terms, I get negative 0.20 or negative 0.2n. And then of course, I have the plus 4. Plus 4 is equal to \$2.00, or we could even just write 2 there. Now, we can isolate the n on the left-hand side by subtracting 4 from both sides. So let's subtract 4 from both sides. And we are left with, on the left-hand side, negative-- I could just write that is negative 0.20n is equal to 2 minus 4 is negative 2. And then we could divide both sides by negative 0.2. Or I could write negative 0.20, the same thing. We're not going to go too deep into the significance in all that. We're assuming that we have infinite precision on everything. So negative 2 divided by negative 0.2, these guys cancel out, and we are left with n is equal to-- the negatives cancel out. 2 divided by 0.2 is just going to be 10. n is equal to 10. And then we know that q is equal to 16 minus n from the first constraint. q is equal to 16 minus n, which is 10, which is going to be 6. So Zoey put in 10 nickels. I want to do that in a different color. She put in 10 nickels and 6 quarters in the bank. And we can verify it. So clearly she has 16 coins. So that part makes sense. 10 nickels, 6 quarters, that's 16 coins. That makes sense. And we could also verify that it's the right amount of money. 10 nickels are going to be \$0.50, 10 times \$0.05 each. So it's going to be \$0.50. And then 6 quarters is going to be \$1.50. So it's going to be \$1.50. So the total amount of money she has is \$0.50 plus \$1.50 which is \$2.00. So it all works out.