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# 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.

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• 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.
• 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! • 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 did u get value of n as 0.05 and q as 0.25 ? • When substituting a negative number with a positive number with a variable, would the answer be negative? • clear the decimals units and work with whole numbers • how would you graph this • How do you solve x-y= 3 over 2x- 3y= -3 with substitution. • 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. 