Solving word equations?

It's usually easier to change everything to cents so you don't need decimals. Choose your variable to stand for the more expensive coin (or if it's easier choose it to be the less expensive coin), then the total coins minus that variable stands for the other kind of coin. Or they might tell you how many more or less of the other coin there are.

Then just multiply each of these by the coin's value, add, and make equal the total value. Like on the first: d = number of dimes d + 2 = number of nickels $0.85 = 85 ¢ 10d + 5(d + 2) = 85 Or the last: q = number of quarters 110 - q = number of dimes $18.50 = 1850 ¢ 25q + 10(110 - q) = 1850.

All of these need two equations to solve them, so maybe that's why you're having trouble. In each case, name the variables: n for nickels, d for dimes, q for quarters. One equation will be of the form coin1 + coin2 = total number of coins The other equation will be in the form value*coin1 + value*coin2 = total value Aretha: 2 more nickels than dimes: n = d + 2 (1) Total value is 85, nickels are 5 and dimes are 10: 5n + 10d = 85 (2) divide equation (2) by 5, to make the calculation easier n + 2d = 17 substitute from equation (1) d + 2 + 2d 17 3d = 15 d = 5.So, n = 7 Check the solution by plugging both values into equation (2) 5*7 + 10*5 = 85 35 + 50 = 85 true, so you know the solution is correct: Aretha has 7 nickels and 5 dimes For the others I'll just give the equations, you should be able to solve them.

Box: n + d = 140 (1) 5n + 10d = 1115 (2) Jar: d + q = 110 (1) 10d + 25q = 1850 (2).

I cant really gove you an answer,but what I can give you is a way to a solution, that is you have to find the anglde that you relate to or peaks your interest. A good paper is one that people get drawn into because it reaches them ln some way.As for me WW11 to me, I think of the holocaust and the effect it had on the survivors, their families and those who stood by and did nothing until it was too late.