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Applied Linear Equations: Mixture Problem  Concept
Carl Horowitz
Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upperlevel math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
In order to understand graphing inverse functions, students should review the definition of inverse functions, how to find the inverse algebraically and how to prove inverse functions. The graphs of inverse functions and invertible functions have unique characteristics that involve domain and range. Techniques for graphing inverse functions can make it easier to graph certain functions by hand.
In this example we are going to be using linear equations to solve a mixture problem. And so for this particular word problem it's all about caffeine. We're in need of a little caffeine boost, we go to a local coffee shop and buy a drink that is 10 percent caffeine. You're still dragging a little bit after that so you go back buy another drink that is 10 percent, 10 ounces larger and 40 percent caffeine. So a lot more caffeine than that one, in total we just consumed 28 percent of caffeine. How large was our second drink? Okay, so there is our set up. And for these types of problems I find it most useful to first sort of make a diagram as to what is going on.
Okay, so we have a first cup, have a second cup and we have a sort of ending result cup, your stomach if you will or some sort of collaborate collection of all that caffeine. So our first drink is 10 percent caffeine and it never tells us how large it is, so if we don't know how large something is, we can assign it a variable x. Okay, our second drink is 40 percent caffeine and we don't know how big that is either but we do know it is 10 ounces larger than the first. So if the first is 5 ounces, the second would be 15 we're just adding 10 onto that first number. So this just turns out to be x+10. And our sort of a collection is 20 percent caffeine and the total amount of liquid is these two previous cups added together. So we had our x, our x+10 throw those together, 2x+10.
Okay. So we now have a diagram which we then need to turn into a equation for us to solve. Say this first cup was 10 percent, sorry, 10 ounces of liquid. It's 10 ounces of liquid and 10 percent caffeine, you actually ingested 10 percent of 10, 1 ounce. So we really take the percent of caffeine times the quantity figure out how much caffeine is actually in a glass. So this one turns out to be 0.1 times x. Remember when you're turning a percentage into a decimal, move your decimal place over two spots. Same idea for this one, 40 percent turns into 0.4 times a quantity, x+10. And the last one over here, 28 percent 0.28, times an amount of liquid, 2x+10.
Okay, so, you could just solve this out right here through here, dealing with decimals. In general, I suggest getting rid of your decimals just because, decimals and fractions still sometimes have this sphere of and kind of freeze up. So multiply everything by a number to get rid of all our decimals. Here we have a tenth, here we have a tenth and here we have hundredths. So if you multiply everything by a hundred, all the decimals go away. By, by 100, 100 times a tenth is 10. 100 times fourth, 40 and 100 times 0.28 is 28, 2x+10.
Okay, we now have a nice linear equation. No decimals, solve it out as we would any other linear equation. So distribute our terms to, so distribute that 40 and that 28. 10x stay the same, that gives us 40x, 40 times 10 is 400. 28 times 2, 56 and 28 times 10 is 280. Okay, combine like terms on each side. So 10x, 40x gives us 50x+400=56x+280. Okay, so now we have x's and other terms, combining all our x's. I try to keep my x's positive so I will subtract this 50x over to the other side. Get our 280 over to the other side as well, so subtract 280 and then simplify. So we've got rid of this 50x, 400280=120, 56x50x=6x and that 280 disappears. So we're left with 120=6x, to solve this out divide by six, x=20.
Okay, now we need to always make sure we're actually solving for what the question asked for. The question asked for how large was our second drink. We found x here and that is actually the amount of our first drink. So if our first drink is 20 ounces that makes our second, twenty plus 10, 30 ounces. So our answer then is 30 ounces for our second drink. So given a word problem, sort of made a diagram, that get out sort of how to organize information. Change it into a linear equation which we then solved out and made sure we gave the proper answer.
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Carl Horowitz
B.S. in Mathematics University of Michigan
He knows how to make difficult math concepts easy for everyone to understand. He speaks at a steady pace and his stepbystep explanations are easy to follow.
Sample Problems (2)
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Applied Linear Equations: Mixture Problem
Problem 1 4,387 viewsA nut farmer mixes some peanuts that sell for $2.50/lb with almonds that sell for $5/lb to make a 12pound mixture worth $3/lb.
How many pounds of peanuts were in the mixture? 
Applied Linear Equations: Mixture Problem
Problem 2 905 views
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