# Problem Solving Using Systems Of Equations

An investor buys a total of 360 shares of two stocks.The price of one stock is per share, while the price of the other stock is per share. How many shares of each stock did the investor buy?

An investor buys a total of 360 shares of two stocks.The price of one stock is per share, while the price of the other stock is per share. How many shares of each stock did the investor buy?

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How many pounds of each kind of candy did he use in the mix?

The first and third columns give two equations: Multiply the first equation by 2 and subtract equations: Then He used 45 pounds of the \$2 candy and 5 pounds of the \$3.60 candy.

At the end of one interest period, the interest earned by the account exceeds the interest earned by the account by \$65. 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 \$2500 was invested at and \$3500 was invested at . He also invests \$1700 more than twice that amount at interest.

At the end of one interest period, the total interest earned was \$278. The last column gives an equation which can be solved for x: Then , so \$2100 was invested at and \$5900 was invested at . The first few involve mixtures of different things which cost different amounts per pound.

The investor bought 120 shares of the \$35 stock and 240 shares of the \$45 stock. I'll let x be the number of 32-cent stamps, let y be the number of 29-cent stamps, and let z be the number of 3-cent stamps. The last column says The number of 29-cent stamps is 10 less than the number of 32-cent stamps, so The number of 3-cent stamps is 5 less than the number of 29-cent stamps, so I want to get everything in terms of one variable, so I have to pick a variable to use.

Phoebe has some 32-cent stamps, some 29-cent stamps, and some 3-cent stamps. Since the last two equations both involve y, I'll do everything in terms of y. I'll solve for x in terms of y: Plug and into and solve for y: Then Phoebe has 20 32-cent stamps, 10 29-cent stamps, and 5 3-cent stamps. The setup will give two equations, but I don't need to solve them using the whole equation approach as I did in other problems.

The number of 29-cent stamps is 10 less than the number of 32-cent stamps, while the number of 3-cent stamps is 5 less than the number of 29-cent stamps. Since one variable is already solved for in the second equation, I can just substitute for it in the first equation. The larger number is 14 more than 3 times the smaller number. Let L be the larger number and let S be the smaller number. At the end of one interest period, the interest you earn is You now have dollars in your account.

The sum is 90: The larger number is 14 more than 3 times the smaller number: Plug into the first equation and solve: Then . Notice that you multiply the amount invested (the principal) by the interest rate (in percent) to get the amount of interest earned.

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.