Problem Solving Using Systems Of Equations

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Let x be the number of shares of the $35 stock and let y be the number of shares of the $45 stock.

The first and third columns give Multiply the first equation by 45, then subtract the second equation: Since , I have .

The first and third columns give the equations Multiply the second equation by 10 to clear decimals: Solve the equations by multiplying the first equation by 25 and subtracting it from the second: Then , so .

Thus, 42 of the $2.50 seats and 36 of the $10.50 seats were sold. A total of 300 tickets are sold, and the total receipts were $4140. The first and third columns give the equations Multiply the first equation by 15 and subtract equations: Then There were 120 tickets sold for $12 each and 180 tickets sold for $15 each.

To review how this works, in the system above, I could multiply the first equation by 2 to get the y-numbers to match, then add the resulting equations: If I plug into , I can solve for y: In some cases, the whole equation method isn't necessary, because you can just do a substitution. The first few problems will involve items (coins, stamps, tickets) with different prices.

If I have 6 tickets which cost each, the total cost is If I have 8 dimes, the total value is This is common sense, and is probably familiar to you from your experience with coins and buying things.The first and third columns give the equations Multiply the second equation by 10 to clear the decimals: Multiply the first equation by 6 and subtract equations: Then Use 15 gallons of the solution and 35 gallons of the solution.Amounts of a alcohol solution and a alcohol solution are to be mixed to produce 24 gallons of a alcohol solution.At the end of one interest period, the interest earned by the account exceeds the interest earned by the account by . I have Rewrite the equations: Clear the decimals by multiplying the second equation by 100: Multiply the first equation by 3 and subtract equations to solve for y: Then 00 was invested at and 00 was invested at . He also invests 00 more than twice that amount at interest.At the end of one interest period, the total interest earned was 8. The last column gives an equation which can be solved for x: Then , so 00 was invested at and 00 was invested at . The first few involve mixtures of different things which cost different amounts per pound.You'll see that the same idea is used to set up the tables for all of these examples: Figure out what you'd do in a particular case, and the equation will say how to do this in general. If there are twice as many nickels as pennies, how many pennies does Calvin have? In this kind of problem, it's good to do everything in cents to avoid having to work with decimals. If the words seem too abstract to grasp, try some examples: If you have 3 nickels, they're worth cents. With this arrangement: There are many correct ways of doing math problems, and you don't have to use tables to do these problems.But they are convenient for organizing information --- and they give you a pattern to get started with problems of a given kind (e.g.This gives Solve the equations by multiplying the first equation by 160 and subtracting it from the second: Hence, and . Then the number of pounds of (pure) silver in the 50 pounds is That is, the 50 pounds of alloy consists of 10 pounds of pure silver and pounds of other metals. Suppose you have 80 gallons of a solution which is acid.She needs 8 pounds of raisins and 9 pounds of nuts. Notice that you multiply the number of pounds of alloy by the percentage of silver to get the number of pounds of (pure) silver. Then the number of gallons of (pure) acid in the solution is So you can think of the 80 gallons of solution as being made of 16 gallons of pure acid and gallons of pure water.By the way --- How does "percent" fit the pattern of the earlier problems, where I had things like "dollars per ticket" or "cents per nickel"?In fact, "percent" is short for "per centum", and 00 is divided between two accounts, one paying interest and the other paying interest.

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