Love of food
There are three families in a certain restaurant. The first family bought 2 Spaghetti's, 4 French fries and 6 glasses of Juice and it cost them 110 coins.
The second family bought 3 Spaghetti's, 8 French fries and 3 glasses of Juice and it cost them 155 coins.
The third family bought 5 Spaghetti's, 2 French fries and 10 glasses of Juice and it cost them 170 coins.
My question is, if I will buy a spaghetti, a french fry and a glass of juice, how much would it cost me?
The answer is 35.
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WE WILL SOLVE IT USING SYSTEMS OF LINEAR EQUATIONS, WE WILL LET THE PRICE OF THE SPAGHETTI AS X,THE FRIES AS Y AND THE JUICE AS Z. SO WE HAVE THE EQUATIONS:
2X+4Y+6Z=110 3X+8Y+3Z=155 AND 5X+2Y+10Z=170
AND BY SOLVING...
2(2X+4Y+6Z=110)-->4X+8Y+12Z=220 AND BY PAIRING IT WITH 3X+8Y+3Z=155 WE CAN USE ELIMINATION TO ELIMINATE 8Y SO WE HAVE X+9Z=65
AND ANOTHER IS 2(5X+2Y+10Z=170)-->10X+4Y+20Z=340 AND BY PAIRING IT WITH 2X+4Y+6Z=110 WE CAN USE ELIMINATION TO ELIMINATE 4Y SO WE HAVE 8X+14Z=230 AND BY PAIRING IT WITH X+9Z=65 WE GOT X=65-9Z AND SUSTITUTE IT TO 8X+14Z=230 8(65-9Z)+14Z=230 520-72Z+14Z=230 -58Z=-290 Z=5 WHICH IS THE PRICE FOR OUR JUICE
AND WE WILL SUSTITUTE THIS VALUE TO GET THE VALUE OF OUR X X+9Z=65 X+9(5)=65 X+45=65 X=65-45 X=20 WHICH IS THE PRICE FOR OUR SPAGHETTI
AND LASTLY,WE WILL GET THE PRICE OF OUR FRENCH FRIES 2X+4Y+6Z=110 2(20)+4Y+6(5)=110 40+4Y+30=110 4Y+70=110 4Y=110-70 4Y=40 Y=10
SO IF I WILL BUY A SPAGHETTI,A FRENCH FRY AND A GLASS OF JUICE, IT WOULD COST ME 20+10+5 =35 COINS ALL IN ALL