Posted by Kelsie on Tuesday, October 28, 2008 at 4:25pm.
Interesting-- some of your problems are similar to the ones that are stumping me! Maybe you would know some of mine. They are a few ones below yours, under Algebra by Celina. I'll try to help you.
#1, I can't help you with. For #2 and #3, I would choose one of the coins as a variable for your equation.
In 2, I would definitely use nickels as your variable. Here is an equation you could use to solve it:
x = nickels
4.75 = 2x + (x + 5)
I'm not positive that is the right equation or not, but it might be one you can try.
Hope I helped!!! :)
#1.
Let B = weight of the empty bucket.
Let x = weight of the water ===========================
B + x = 8 kg
Weight of water is 5 times the weight of the empty bucket; therefore, x = 5B.
Substitute 5B for x and solve for B, then x. You can go through the algebra but I have 1.333 kg for the weight of the empty bucket and 6.667 kg for the weight of the water. Check my thinking. If you have trouble, post your work and we can find the math error. I'll work on the other two problems shortly.
Let n = # nickels.
Let d = # dimes.
Let q = # quarters.
=====================
Helen has twice as many dimes as nickels; therefore, 2n=d
Helen as five more quarters than nickels; therefore, n+5 = q
Finally, the value is 4.75 if we add all of it together; so
0.05n + 0.10d + 0.25q = 4.75.
Substitute n+5 for q and substitute 2n for d and solve for n, then for d and q.
Check my thinking. Post your work if you get stuck and we can find your math error. I have # nickels = 7.
I don't get what you mean but substitute numbers for other numbers.
Ok. So what I got for my equation was
n+2n+(n+5)=4.75. Is that right
#3 is the same type of problem and its worked the same way. I'll leave that for you after you have finished with #2.
I am done with number 2. I kinda need the next equation