Solving System Of Equation Word Problems

Calvin mixes candy that sells for .00 per pound with candy that costs .60 per pound to make 50 pounds of candy selling for .16 per pound.How many pounds of each kind of candy did he use in the mix?How many pounds of raisins and how many pounds of nuts should she use?Suppose she uses x pounds of raisins and y pounds of dried fruit.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.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"?

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.

If 36 is added to the number, the digits interchange their place. Solution: Let the digit in the units place is x And the digit in the tens place be y.

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.

The first and third columns give the equations Multiply the second equation by 100 to clear the decimals. Suppose you have 50 pounds of an alloy which is silver.

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.

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