# System of Linear Equations

Mariel has 10 coins in her pocket. All of them are quarters and dimes, and their total value is \$1.60. How many of each type of coin does she have?

We can solve this by using a system of equations. We’re told that she has 10 coins in her pocket and that they’re all quarters and dimes.

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Question 1 of 5

Mini-test: System of Linear Equations

Where do  y = -x  and  y = 2 - x  intersect?

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D.
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Question 1 of 5

Question 2 of 5

Where do  y = 2  and  y = 6 - x  intersect?

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Question 2 of 5

Question 3 of 5

Find 'b' so that  y = 2x + b  and  y = b - 2x  intersect where y = -6.  Find x, too.

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Question 3 of 5

Question 4 of 5

When can  y = 2x + b  and  y = 2x - 6  intersect?

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E.

Question 4 of 5

Question 5 of 5

When can  y = -2x + b  and  y = 2x - 6  intersect?

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D.
E.

Question 5 of 5

Next lesson: Equations and Inequalities
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The following transcript is provided for your convenience.

So, if use D for the number of dimes, and Q for the number of quarters, then one of our equations could be her number of dimes plus her number of quarters is 10 total coins.

Next, we’re told that the total value is \$1.60. Well, dimes are worth 10 cents, so 10 times however many dimes she has, plus, quarters are worth 25 cents, so 25 times however many quarters she has, will give us a total of 160.

We’ve eliminated the decimal, so we just move the decimal over two times to the right for all of our numbers. So, instead of 10 cents, we have 10, instead of 25 cents, we have 25, and instead of a \$1.60, we have \$160, but you could leave the decimals in there if you wanted to.

Now, to solve this system, we’ve got three options. We could graph both of these and see where they intercept, we could use substitution, or elimination, and it’s your choice. For this particular problem, we’re going to use substitution.

So, we’re going to solve for D in our first equation, so we can substitute in our second equation. D is equal to negative Q plus 10. To solve for D,
subtract Q from both sides, and you get negative Q plus 10.

So, in our second equation, we’re going to use negative Q plus 10 to substitute for D. we have 10 times the quantity, instead of D, negative Q plus 10, plus 25Q, is equal to 160.

Now, we can solve for Q. To do that, first, we need to distribute. 10 times negative Q is negative 10Q, 10 times 10 is 100, so plus 100, plus 25Q, equals 160.

Now, to solve for Q, we need to combine like terms. Negative 10Q plus 25Q is 15Q, plus 100, is equal to 160.

Subtract 100 from both sides, and we get that 15Q is equal to 60. Now, we divide both sides by 15. 15 divided by 15 is 1, and 1 times Q is just Q, and we get that Q is 4, which means she has 4 quarters.  So, how many dimes does she have?

Well, we know that her number of dimes is equal to negative Q, so negative 4 plus 10. Therefore, her number of dimes is negative 4 plus 10, 6. So, we can say Mariel has 4 quarters and 6 dimes in her pocket.