Ex: System of Equations Application – Mixture Problem


– A CHEMIST NEEDS TO MAKE TWO
LITERS OF A 15% ACID SOLUTION FROM HER 10% ACID SOLUTION
AND A 35% ACID SOLUTION. SO WE’RE MIXING
THE 10% AND 35% SOLUTION TO CREATE TWO LITERS
OF 15% SOLUTION. WE WANT TO KNOW HOW MANY LITERS
OF EACH SHOULD SHE MIX TO GET THE DESIRED SOLUTION. SO WE’RE GOING TO NEED
TWO VARIABLES HERE AS WELL AS TWO EQUATIONS. LET’S LET X EQUAL THE NUMBER
OF LITERS OF THE 10% SOLUTION. AND WE’LL LET Y EQUAL THE NUMBER
OF NUMBER OF LITERS OF THE 35% SOLUTION. NOW SEE IF WE CAN WRITE
OUR TWO EQUATIONS. WE KNOW AFTER MIXING THESE
TWO AMOUNTS, WE’LL HAVE EXACTLY TWO LITERS
OF THE 15% SOLUTION. SO OUR QUANTITY EQUATION WILL BE
X + Y MUST EQUAL TWO LITERS. THE SECOND EQUATION IS GOING DEAL WITH THE
CONCENTRATIONS OF EACH SOLUTION. SINCE X IS EQUAL TO THE NUMBER
OF LITERS OF THE 10% SOLUTION AND Y IS EQUAL TO THE NUMBER
OF LITERS OF THE 35% SOLUTION, 10% x X, OR 0.10 x X + 35% x Y,
OR 0.35 x Y MUST EQUAL 15% OR 0.15 x THE TOTAL AMOUNT
OF THE SOLUTION, WHICH IS 2 LITERS. SO HERE’S OUR SYSTEM OF
EQUATIONS THAT WE HAVE TO SOLVE. THE FORM OF THE FIRST EQUATION
WILL BE GOOD TO WORK WITH, BUT LET’S TRANSFORM
THE SECOND EQUATION BEFORE WE SOLVE THE SYSTEM. LET’S START BY ELIMINATING
THE DECIMALS. IF WE MULTIPLY EVERYTHING
BY 100, THAT’S THE SAME AS MOVING
THE DECIMAL POINT TO THE RIGHT TWO PLACES. SO WE CAN WRITE THIS
AS 10X + 35Y EQUALS– AND THIS WOULD BE 15 x 2,
WHICH IS EQUAL TO 30. SO NOW WE’LL SOLVE THE SYSTEM
OF EQUATIONS USING THESE TWO EQUATIONS. SO AGAIN WE HAVE X + Y=2
AND THEN WE HAVE 10X + 35Y=30. NOW WE CAN SOLVE
THIS ALGEBRAICALLY USING ELIMINATION
OR SUBSTITUTION. IF WE CALL THIS EQUATION 1
AND THIS EQUATION 2, WE COULD VERY EASILY SOLVE
THE FIRST EQUATION FOR X OR Y. TO SOLVE THIS FOR X, WE WOULD
SUBTRACT Y ON BOTH SIDES. SO USING EQUATION 1,
WE WOULD HAVE X=2 – Y. AND THEN WE CAN PERFORM
SUBSTITUTION INTO EQUATION 2. IF X=2 – Y,
WE CAN REPLACE X WITH 2 – Y. NOW EQUATION 2
AFTER SUBSTITUTION, WOULD BE 10 x 2 – Y,
INSTEAD OF TIMES X, + 35Y=30. AND NOW WE HAVE ONE EQUATION
WITH ONE VARIABLE, SO NOW WE’LL SOLVE FOR Y. WE HAVE 20 – 10Y + 35Y=30. WELL HERE WE HAVE 2Y TERMS SO
WE CAN COMBINE THESE TWO TERMS. -10Y + 35Y=25Y,
SO WE’LL HAVE 20 + 25Y=30. SUBTRACT 20 ON BOTH SIDES WOULD
GIVE US 25Y=10. DIVIDE BOTH SIDES BY 25. AND WE HAVE Y=10/25
WHICH DOES SIMPLIFY. THERE’S A COMMON FACTOR OF 5
HERE. SO WE HAVE Y=2/5. REMEMBER Y IS THE NUMBER
OF LITERS OF THE 35% SOLUTION. SO AS AN ORDERED PAIR WE KNOW
THAT Y=2/5 WE STILL HAVE TO GO BACK
AND DETERMINE X, BUT WE DO KNOW FROM BEFORE
X=(2 – Y). SO WE’D HAVE 2 – 2/5,
WHICH IS 1 3/5. SO WE HAVE 1 3/5 LITERS
OF THE 10% SOLUTION. WE MAY WANT TO CONVERT
THESE TO DECIMALS. 3/5 IS THE SAME AS 0.6. SO THE MIXTURE CONTAINS 1.6
LITERS OF THE 10% SOLUTION AND 2/5 AS A DECIMAL WOULD BE
0.4 LITERS OF THE 35% SOLUTION. OKAY. I HOPE YOU FOUND
THIS HELPFUL.

6 thoughts on “Ex: System of Equations Application – Mixture Problem

  1. thanks this really helped. please check our videos @ https://www.youtube.com/channel/UCKfUqyk2gJHeKuAdy_8vBtA

  2. I've been trying to find a video that went into depth on how to do these kind of problems, and they always skipped over something, or cheated, and since I have to show work, I couldn't do that. This helped me learn how to do these kinds of problems. Thanks so much for creating this, you just got a like and a sub! <3

  3. What if it's like this? "You need a 50% alcohol solution. On hand, you have a 45 mL of a 40% alcohol mixture. You also have 65% alcohol mixture. How much of the 65% mixture will you need to add to obtain the desired solution?"

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