Ex: System of Equations Application – Coin Problem


– A WOMAN HAS 21 COINS
IN HER POCKET, ALL OF WHICH ARE DIMES
AND QUARTERS. IF THE TOTAL VALUE OF HER CHANGE
IS $3.90 HOW MANY DIMES AND HOW MANY
QUARTERS DOES SHE HAVE? SO FOR THIS PROBLEM WE’RE GOING
TO SET UP A SYSTEM OF EQUATIONS AND THEN SOLVE IT ALGEBRAICALLY. NOTICE THAT WE DON’T KNOW
THE NUMBER OF DIMES OR THE NUMBER OF QUARTERS, SO WE’LL ASSIGN VARIABLES
FOR THESE AMOUNTS. LET’S X=THE NUMBER OF DIMES AND WE’LL LET Y
=THE NUMBER OF QUARTERS. THE FIRST SENTENCE STATES THAT THE WOMAN HAS 21 COINS
IN HER POCKET, SO IF X IS THE NUMBER OF DIMES
AND Y IS THE NUMBER OF QUARTERS, THIS WOULD GIVE US THE EQUATION
X + Y MUST=21. NEXT WE’RE TOLD THE TOTAL VALUE
OF THE CHANGE IS $3.90. WELL, EVERY DIME IS WORTH
10 CENTS AND EVERY QUARTER IS WORTH
25 CENTS. SO IF WE WRITE THE VALUE
EQUATION IN CENTS, WE CAN AVOID DECIMALS. SO THE VALUE FROM THE DIMES
IS GOING TO BE 10 CENTS x X + THE VALUE OF THE QUARTERS
IS GOING TO BE 25 CENTS x Y, AND THIS MUST=$3.90,
WHICH WOULD BE 390 CENTS. AGAIN, WE’RE USING THE CENTS TO AVOID HAVING DECIMALS
IN OUR EQUATIONS. AND NOW TO SOLVE THIS SYSTEM
OF EQUATIONS ALGEBRAICALLY, WE CAN EITHER USE
THE SUBSTITUTION METHOD OR THE ELIMINATION METHOD. IF WE WANT TO USE
THE ELIMINATION METHOD, WE WANT TO ADD
THESE TWO EQUATIONS AND EITHER ELIMINATE THE X TERMS
OR THE Y TERMS. THE ONLY WAY
THAT’S GOING TO OCCUR IS IF THE X TERMS OR Y TERMS
ARE OPPOSITES. SO IF WE WANTED TO ELIMINATE
THE X TERMS, WE WOULD WANT THIS FIRST TERM
HERE TO BE -10X. SO WHAT WE’LL DO IS MULTIPLY
THE FIRST EQUATION BY -10 THEN LEAVE THE SECOND EQUATION
THE SAME. SO WE’RE GOING TO MULTIPLY THIS
ENTIRE EQUATION BY -10 THAT’S GOING TO GIVE US
-10X – 10Y=21 x -10 WOULD BE -210. WE’LL LEAVE THE SECOND EQUATION
THE SAME, SO WE HAVE 10X + 25Y=390. AND THEN WHEN WE ADD THESE
EQUATIONS TOGETHER, SINCE THE X TERMS ARE OPPOSITES,
THIS SUM WOULD BE 0. SO WE’RE LEFT WITH 15Y
MUST=-210 + 390, WHICH IS=TO 180. NOW WE’LL DIVIDE BOTH SIDES
BY 15, AND SO WE HAVE Y
=180 DIVIDED BY 15 IS=TO 12. AND SINCE Y IS=TO 12 THIS
MEANS WE HAVE 12 QUARTERS. AND NOW WE HAVE TO DETERMINE
HOW MANY DIMES WE HAVE. SO WE CAN USE THE EQUATION
X + Y=21 AND SUBSTITUTE 12 FOR Y. SO WE’D HAVE X + 12=21,
SUBTRACT ON BOTH SIDES, SO WE HAVE X=9. SO THIS TELLS US WE HAVE 9 DIMES
AND 12 QUARTERS. NOTICE HOW THIS GIVES US
21 COINS AND 9 X 10 CENTS + 12 x 25 CENTS
WOULD GIVE US A TOTAL OF $3.90. NOW LETS ALSO VERIFY THIS
GRAPHICALLY. REMEMBER IF WE GRAPH
THESE TWO LINEAR EQUATIONS ON THE SAME COORDINATE PLANE, THE POINT OF INTERSECTION
WOULD REPRESENT THE SOLUTION. SO HERE’S THE EQUATION
X + Y=21, AND HERE’S THE EQUATION
10X + 25Y=390, THEREFORE THIS POINT
OF INTERSECTION REPRESENTS OUR SOLUTION WHERE THE X COORDINATE
REPRESENTS THE NUMBER OF DIMES, WHICH IS 9, AND THE Y COORDINATE REPRESENTS
THE NUMBER OF QUARTERS, WHICH IS 12. OKAY, I HOPE YOU FOUND THIS
HELPFUL.  

4 thoughts on “Ex: System of Equations Application – Coin Problem

  1. Mr. Sousa, you are the man! I've passed ALL my college math classes (so far) with the use of your great videos-Pre-Calc, Calc I, II, III, Linear Alg, Diff Eq, and Probability/Data Analysis. You are a wonderful teacher, and I thank you humbly for your videos. I've developed such a fondness for mathematics that I've chosen it as a minor thanks to teachers like you. As a guy who has never liked, nor been any good at math historically, I'm extremely grateful there are people like you out there. Thank you, thank you, thank you!

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