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How many quarters and nickels are in 5.20 if there are only 36 coins?
Q + N = 36;.25Q + .05N = 5.20;.25(36-N) +.05N = 5.20;9- .20N = 5.20;3.8 = .20N;N = 19 and Q = 17
How do you make 2.65 out of dimes and nickels only using thirty-one coins?
This is a simple algebra question:d + n = 31.1d + .05n = 2.65Subsituting in for "d" using "n" we have:d = 31 - n3.1 - .1n + .05n = 2.65-0.05n = -0.45Solve for n:n = 9d + 9 = 31d = 22Thus, there are 9 nickels and 22 dimes.
How do you solve 25q plus 05n equals 10.50?
The "solution" is a pair of numbers that make the statement true when youplug in the numbers in place of 'q' and 'n'.So far, there are an infinite number of them. In fact, if you graph this equation, thegraph is a straight line, and every point on the line is a solution to the equation.If you expect to get one single number for 'q' and one single number for 'n', thatcan't happen until you have another equation to go along with this one. You alwaysneed the same number of equations as the number of 'unknowns' in them.(Here you have two 'unknowns' ... 'q' and 'n' .)
You save 42 nickels and dimes that total 3.35 how many nickels and dimes do you save?
This problem can be solved using a system of equations.If we call the number of nickles n and the number of dimes d, we can write the following two equations:n + d = 42 (The number of nickels and dimes is 42).05n + .1d = 3.35 (5 cents for every nickel plus 10 cents for every dime is $3.35)Then:n + d = 425n + 10d = 335 (Switch to cents instead of dollars, gets rid of decimals)n = 42 - dWe have now just defined n in terms of d. Now every time you see n in the second equation, you can replace it. This is called substitution.5n + 10d = 335 (from above)5(42 - d) + 10d = 335 (substituting for n)Follow through with the algebra, and d (# of dimes) = 25, then there must be 17 nickels.25 dimes, 17 nickels
