# 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.

How did you get 10 and 25?

More money more problems.

You saved me on this one thanks man

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!

Now, if only I could speel….