Stick Diagram Solutions: Sixth Example
(brought to you by the number q, unknown on one side,
simple fraction for a coefficient)
Now in the equation (¾)q + 17 = 32 we imagine that q represents an unknown
length. In the stick below, the top stick has length (¾)q + 17 = 32 while the bottom stick
has length 32. The equation say both sticks have the same length, here 32.
Cutting off or subtracting 17 from both sticks (adding -17) gives
a stick of length (¾)q on top and a stick of length 15 = 32-17 on bottom. This
second stick diagram represents the equation (¾)q =15.
But (¾)q = (¼)q + (¼)q + (¼)q. So a third of the length of (¾)q is +
(¼)q. To find (¼)q, we will take cut
each stick into thirds:
The stick diagram suggests that (¼)q =5
Replicating the above diagram (¼)q four times (multiplying by 4 the
reciprocal of ¼) gives
(¼)q
|
(¼)q
|
(¼)q
|
(¼)q
|
5
|
5
|
5
|
5
|
and simplifying suggest q = 20
We could have gone directly to the latter conclusion by multiplying the
diagram
for equation (¾)q =15, or equivalently
by the reciprocal (4/3) of ¾ to find
| q = |
4
3 |
( |
3
4 |
q) |
= |
4
3 |
(15) |
= |
4
3 |
(3 x 5) |
= |
4 x5 |
= |
20 |
|
A Solution Table for (¾)q + 17 = 32 follows.
If I was solving (¾)q + 17 = 32 in class, I would just fill in
the table and skip the work before it. Each table consists of a diagram in
the left column, a description of what is done or given in the middle column,
and the equivalent equations in the rightmost column. At the moment, you
are required to draw the stick diagram in the solution of the equation. That is
a crutch. Later on, only the equation column is required with a few words
to explain the operations.
|
Solution Table for (¾)q + 17 = 32 |
| Stick Diagram |
Operation |
Equation |
|
|
Initial Situation
Given |
(¾)q + 17 = 32 |
|
|
Subtract 10
(a.k.a Add -10) |
(¾)q =15 |
|
|
A third of (¾)q is (¼)q. So q is also a third of 15. (Multiply by
to go from 2nd row to 3rd.)
|
(¼)q = 5
|
(¼)q
|
(¼)q
|
(¼)q
|
(¼)q
|
5
|
5
|
5
|
5
|
| optional diagram - keep if it helps.
|
|
Replicate four times or multiply by 4
to go from 3rd row to 4th and then simplify
|
(¼)q +
(¼)q+(¼)q+(¼)q
= 5 +5 +5 +5 = 4 * 5
Or q = 20
|
Conclusion:
|
Check if q = 20 satisfies (¾)q + 17 = 32?
|
Left Hand Side |
=(¾)q + 17
=(¾)20 + 17
= 15 + 17
= 32
|
Right Hand Side |
= 32 |
Note: We can always check whether a number is a solution of an
equation or not by computing the left and right sides of an equation for or at
that number. If the two sides differ, the number is not a solution.
- If you are asked to show that a number satisfies an equation you do the
check.
- If you are asked to find a solution to an equation algebraically you
should show some work (besides trial and error) that leads to the solution.
Then you should check the solution.
Solutions of equations can always be checked. So before you hand-in an
answer, you can always check whether it is correct or not. And if it is not
correct, you should say so if you do not have time to find the correct
answer. Instructors want to see how you obtain the solution. If your
arithmetic without a calculator is usually good, the odd error in your work is
not as important as you showing that you have master an algebraic method for
solving problems.