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Solving Linear |
Read them in order
Stick Diagram Solutions: Fifth Example
(brought to you by the number a, unknown on both sides
with fractional coefficients)
See Animated Example: (2/3)x + 5¼ = (4/3)x+1¼ below.
Now in the equation (¾)a + 16 = (¼)a+ 24 we imagine that a represents an unknown length. In the stick below, the top stick has length (¾)a+16 while the bottom stick has length (¼)a+ 24. The equation say both sticks have the same length as shown.
| (¾)a
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16
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| (¼)a
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24
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Cutting off or subtracting 16 from both sticks (adding -16) gives
| (¾)a
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| (¼)a
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8
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a stick of length (¾)a on top and a stick of length (¼)a+ 8 on bottom. This second stick diagram represents the equation (¾)a = (¼)a+ 8
Cutting off or subtracting (¼)a from both sticks (adding -(¼)a) gives
| (½)a
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| 8
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or
| (½)a
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| 8
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That gives (½)a = 8. Let replicate twice to get the stick diagrams
| (½)a
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(½)a
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| 8
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8
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and
| a
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| 16
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So a = 16.
A Solution Table for (¾)a + 16 = (¼)a+ 24 follows.
If I was solving (¾)a + 16 = (¼)a+ 24 before a 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 eaplain the operations.
Remember what we do one stick in a pair, we must do do the other, to keep the lengths after the operation, the same. If two stick have the same length, we cut 5 from and 8 from the other, the resulting pair of sticks will not the same the length.
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Solution Table for (¾)a + 16 = (¼)a+ 24 |
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| Stick Diagram | Operation | Equation | |||||||||||||||||||||
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Initial Equation Given |
(¾)a + 16 = (¼)a+ 24 | |||||||||||||||||||||
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Subtract 16 (add -16) |
(¾)a = (¼)a + 8 | |||||||||||||||||||||
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subtract (¼)a | (½)a = 8 | |||||||||||||||||||||
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Double (a.k.a. times or multiply by 2) |
(½)a + (½)a = 8 + 8 | |||||||||||||||||||||
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Simplify |
a = 16 | |||||||||||||||||||||
Check: Does a = 16 satisfy (¾)a + 16 = (¼)a+ 24
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Left Hand Side |
=(¾)a + 16 =(¾)16 + 16 =12 +16 = 28 |
Right Hand Side |
= (¼)a+ 24 =(¼)16+ 24 =4+24 =28 |
Conclusion: a = 16 satisfies (¾)a + 16 = (¼)a+ 24
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.
Animated Example: (2/3)x + 5¼ = (4/3)x+1¼
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