December 2014 / January 2015 Newsflash: Tips & Techniques

Upper Elementary and Middle School

Find the perimeter of the shape below. (Hint: there is enough information.)

Solution:

The length on the bottom of the rectangle is the same as the length on the top, although the top has been split into two pieces. Similarly the width on the left is 11 inches and the width on right is also 11 inches, again in two pieces. 

Now, the length of the side is given in inches and the length of the bottom is given in feet. Since we have to add to get the perimeter, the Law of SAMEness requires the units to be the same. Let’s change 2 feet into 24 inches (2 x 12). 

Note that it is not necessary to know the exact measurement of the length and width. All that matters is that we know the total length of each. 

With that, 11 in + 11 in + 24 in + 24 in = 70 inches.

Algebra 

If ax + 2a = c and x + 2 = 3, express a in terms of c. 

Solution: 

If ax + 2a = c, then a(x + 2) = c. Now, substitute “3” for “x + 2” and you get 3a = c, 

so a = c/3.

October/November 2014 Newsflash: Tips & Techniques

From time to time, we get lucky and the computation we have to do can be solved using a specific technique or trick that allows us to do effortless mental math.

Level 1:

12 x 125 = _______

The trick to this question is to see that “12” can be broken down into “3 x 4.”

            Think of “12 x 125” as “3 x (4 x 125).”

Now 4 x 125 = 500. (4 x 100 + 4 x 25)

Then, 3 x 500 = 1,500.

Try this (mentally): 8 x 175 = ________.

Level 2:

1/5 + 1/6 = __________

When you add two fractions, and the numerators of the fractions are both 1, the numerator of the answer is the sum of the denominators, and the denominator of the answer is the product of the denominators. In this problem, the numerator is   (5 + 6), and the denominator is (5 x 6).

So, 1/5 + 1/6 = 11/30. [Note: If the denominators are not relatively prime, the answer will have to be reduced.]

Try this (mentally): 1/12 + 1/13 = _____________.

Level 3:

2,000 ÷ 16 = ____________

The trick here is to recognize that 16 is a power of 2, and that powers of 2 make division into “half-of-a-half-of-a…”

In this case, since 16 is “2 to the fourth power,” we simply cut 2,000 in half 4 times.

So,

            Half of 2,000 is 1,000,

                        Half of 1,000 is 500,

                                    Half of 500 is 250, and finally,

                                                Half of 250 is 125.

So, 2,000 ÷ 16 = 125.

Try this (mentally): 3,000 ÷ 8 = ____________.

Note that there is more than one trick for each of these questions. See if you can find others!

Tips and Techniques: Transposing

Transposing is an important technique for students to master as they progress from Algebra I to Algebra II and beyond. It’s the process of moving quantities across the equal sign in an equation to make the problem solving process more efficient. Here’s how it works.

 Solving an equation means isolating the variable on one side of the equation, either the right side or the left. Generally it is better to isolate the variable on the left side of the equal sign because we read from left to right.

Beginning algebra students are taught the Law of Equations:  whatever you do on one side of an equation, you must do on the other side as well. 

Using the Law of Equations, to solve 2x – 3 = 13, we can add 3 to both sides, giving us

2x – 3 + 3 = 13 + 3 ===> 2x = 16

Then, we divide both sides by 2 to find that x = 8.

To solve 2x – 3 = 13 by transposing, we do the following:

Instead of adding 3 to both sides, we can transpose the 3 that’s already there by

moving it from the left side, over the equal sign, to the right side, and changing its sign from “–” to “+.”

Now we have  2x = 13 + 3, which becomes 2x = 16, the same answer we got using the Law of Equations. Again, we find that x = 8.

The beauty of this process is that it is visually easier to see how the parts of the equation change as we solve it. Instead of writing the transposed quantity twice (once on the left and once on the right), we literally pick up the quantity and move it to the other side. The price we have to pay for doing this is changing the sign of the quantity being moved. 

Let’s solve 3y + 2x – 3 = 7 for y.

Since we want to isolate y, we can transpose 2x and – 3.

This gives us y = –2x + 7 + 3. Simplifying, we get y = –2x + 10.

If we used the Law of Equations, we would have to write out

3y + 2x – 3 = 7 ===> 3y + 2x – 2x – 3 + 3 = 7 – 2x + 3

While simplifying the above does give us y = –2x + 10, it comes with a lot more work.

Note that most classroom teachers will not allow transposing until students learn to use the Law of Equations.

Try these:

1. Solve for x: 4x + 9 = 49
2. Solve for y: 2y + 7 – 3 = 24
3. Solve for y: 3x + 9y – 7 = 49
4. Solve for a: 3b – 4a – 6 + 4 = 49
5. Solve for a: 2b + 4a + 5 – 2 = 42 – 2a