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## Algebra 1

### Course: Algebra 1 > Unit 6

Lesson 6: Systems of equations word problems- Age word problem: Imran
- Age word problem: Ben & William
- Age word problem: Arman & Diya
- Age word problems
- System of equations word problem: walk & ride
- Systems of equations word problems
- System of equations word problem: no solution
- System of equations word problem: infinite solutions
- Systems of equations word problems (with zero and infinite solutions)
- Systems of equations with elimination: TV & DVD
- Systems of equations with elimination: apples and oranges
- Systems of equations with substitution: coins
- Systems of equations with elimination: coffee and croissants
- Systems of equations: FAQ

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# Systems of equations with substitution: coins

CCSS.Math: , , , ,

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?(8 votes)
- 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(24 votes)

- Could you solve a coin problem with 3 variables? (Q+N+D)?

How would you do it (if it can be done)?(9 votes)- You can have as many variables as you want, as long as you have the same number of equations as variables. You have to subtract or add Q and N, N and D, and Q and D. Then you solve it similarly to the 2 variable ones.(18 votes)

- 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.

Explain please!(6 votes)- 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.(7 votes)

- 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)

By adding together, we get:

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!(3 votes)- 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.(6 votes)

- How did u get value of n as 0.05 and q as 0.25 ?(2 votes)
- The problem is dealing with nickels and quarters. A nickel is worth 5 cents or $0.05. A quarter is worth 25 cents or $0.25.

Note: n and q are the numbers of each type of coins. To get the value of all the nickels, Sal needs to multiply "n" with the value of nickel = $0.05n. Similarly, the value of all the quarters = $0.25q.

Hope this helps.(6 votes)

- When substituting a negative number with a positive number with a variable, would the answer be negative?(3 votes)
- clear the decimals units and work with whole numbers(3 votes)
- how would you graph this(2 votes)
- If you really want to graph it, you would have to solve for one of the variables in both equations, and then you would have a independent and a dependent variable, graph with y intercept and slope, but the numbers might not be whole numbers which make graphing more accurate.(2 votes)

- How do you solve x-y= 3 over 2x- 3y= -3 with substitution.(2 votes)
- 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.(2 votes)

- How do you embed things like times in the video and hyperlink them so someone can just click and see it?(2 votes)

## 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.