  Home Q&As More Thanksgiving Word Problems (Challenging) Halloween Math Costumes What Do You Think of the Jodi Arias Trial? Christmas Balance Scale Math Thanksgiving Balance Scale Problem See all... Ask a Question Products About Mitch Adler Contact Us Question Dear Mitch, Now that New Year's Eve has come and gone, could you tell the answer to the New Year's Eve balance question you had with the 3 scales and the hats and noisemakers? Thanks, and Happy New Year to you! Larry K. (and no, I am NOT Larry King!!) Answer Dear Larry K.,  Of course! The question was: How many of the plastic decorative cakes would you need to place on the empty side of the third scale, scale C, in order to form a balance with the other side of that scale, which held 3 party hats and 3 noisemakers?  And the ANSWER is: 6. You would need to put SIX 'cakes' on the empty side of Scale C in order to get that scale balanced.  Why?First, let's review what the illustrations depicted: The first scale, Scale A, had 1 cake and 3 noisemakers on the left, and that collection was balanced by 2 hats on the right.  The second scale, Scale B, had 4 noisemakers on the left, and they were balanced by just 1 hat and 1 'cake' on the right. So, using a very basic and 'old-fashioned' algebraic approach, we represent each object with a letter.  We'll represent each hat with an h. We'll represent each noisemaker with an n. We'll represent each cake with a c. And we'll represent each point of balance with an equal sign (which, of course, is the definition of "equal", i.e., 3 + 4 is BALANCED with 7. So, with these simple and straightforward algebraic substitutions of letters for objects, we get:  Scale A:      c + 3n = 2h Scale B:           4n  =  h + c Scale C:   3n + 3 h = ? c As I mentioned in an explanation to another recent balance scale question, when it comes to these 'systems of equations' (which simply means more than one equation that we are trying to solve in conjunction with each other), the 'golden rule' is this: You need to have AT LEAST the same number of equations as variables.  So, since we have 3 variables (h, n, and c), and only two equations (because scale c is not yet a completed equation, and therefore cannot be included in the tallying of equations), we need to come up with a third equation. BUT DON'T PANIC, because this is an easier thing to do than it sounds.  All we have to do is add the first equation to the second equation, and the easiest way to do that is vertically. So: The equation from scale A told us:   c + 3n = 2h And, the equation from scale B told us:           4n  =   h + c                         adding them, we get:   c + 7n  = 3h + c Just as we could remove a pound (or ounce or ton or whatever we wish) from one side of a balance scale if we remove the exact same amount from the other side of that scale, we can subtract one c from each side of the equation above, which gives us: 7n = 3h  Now, to get n by itself (so we can identify and work with its value or "weight"), we divide both sides by 7 And we get: n = 3/7 (h)  So now, using our new name for n ("3/7 h") we can go back to the equation depicted on scale A and plug our new version of n into every place we come across an n. And we get: C + 3(3/7h) = 2h  Why?  Because even though it may look more complicated and 'uglier' than it was before we did that, in math it is MUCH easier to solve an equation with fewer variables than more, despite how much 'uglier' the newer version of the equation might look.  Sometimes, equations that are more complicated-looking are actually simpler mathematically! STEP 2:  We want to find out how many 'cakes' each of the other items is worth (weighs). So...  C + 9/7 h = 2h Next, to get C by itself, we subtract everything else on its side of the equation (and, of course, to keep things balanced, we do the same to the other side.)  So: C = 2h  – 9/7(h) C = 14/7 (h) - 9/7(h)  (WHY 14/7??  Where'd THAT come from? We need a 'common denominator' to add or subtract fractions, so 14/7 is the most helpful name here for 2!)  So: c =  5/7(h) But, since we want to find out the value (weight) of h, we multiply both sides of the equation by the reciprocal (a fraction turned upside down), and get: h =  7/5(c) STEP 4:  Go back to scale B: 4n = 7/5c + c 4n = 7/5c + 5/5(c)  (remember common denominators... anything over itself is equal to one, so 5/5 is perfect...)  4n = 12/5(c)   n =  12/5(c)             4       n = 3/5(c) STEP 5: Back to scale C: 3n + 3h = ? c 3  x  3/5(c) + 3  x  7/5(c) =  ? c 9/5 (c) + 21/5(c)  =  ? c  30/5(c) =    6 (c) And, in words, to answer the original question, we substitute the original words for the letters, and get: SIX CAKES YEAH! I hope that helps!  Happy New Year! Mitch © 2019. Mitch Adler. All rights reserved.