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Stick Diagram Solutions: Third Example
(brought to you by the number a, unknown on both sides 
with whole number coefficients)  

see animated solution for 3a+ 2 = 5a+4  (slightly different from text problem)

Now in the equation 5a + 16 = 3a+ 24 we imagine that a represents an unknown length. In the stick below, the top stick has length 5a+16 while the bottom stick has length 3a+24. The equation say both sticks have the same length, here 32.

5a
16
3a
24

Cutting off or subtracting 16 from both sticks (adding -16) gives

5a
3a
8

a stick of length 5a on top and a stick of length 3a+ 8 on bottom. This second stick diagram represents the equation  5a = 3a + 8 

Cutting off or subtracting 3a from both sticks (adding -3a) gives

2a
8

But 2a = a + a. So half of the length of 2a = 8 is a.  Therefore the stick diagrams

a
a
8
8

and 

a
8

imply a = 4.

But 3a = a + a + a. One third of the length of 3a is a. To find a, we will take cut each stick into two equal pieces.

A Solution Table for 5a + 16 = 3a+ 24 follows.  

If  I was solving 5a + 16 = 3a+ 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 explain 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.  

Solution Table for 5a+16 = 3a+ 24

Stick Diagram Operation Equation
5a
16
3a
24
Initial Equation
Given
5a+16= 3a + 24
5a
3a
8
Subtract 16
(a.k.a add -16)
5a = 3a + 8
2a
8
Subtract 3a
(a.k.a add -3a)
2a = 8
a
4
Take 
1
2
 
a = 4

Check: Does a = 4 satisfy 5a+16 = 3a + 24

Left Hand Side 

=5a+16
=5(4)+16
=20 +16
= 36

 Right Hand Side  

= 3a+ 24
=3(4)+24
=12+24
=36

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: 3a +4 = 5a+2

 

Solving Linear
Equations
|(Feb 14, 2005)
with & then without stick diagrams plus testing solutions -do not hand-in untested solutions;  solving word problems; and solving systems: -  essentially one unknown, essentially triangular & general

Skill in arithmetic with fractions is a must for algebra. 

Folder Chapters -
lesson groups

Proper Use of Equal Sign
A. Letters and Lengths
B.. Solving Linear Eq'ns - WS
C. Solving Linear Eq'ns - No Sticks
D. Almost One
E: 2D Systems - Sub Method.
E:  Continued
E: Still More
F. Larger Systems


Area Intro
(i) x + 20 = 29 WS
(ii) 2x + 5 = 20 WS
(iii) 3x + 10 = 32 WS
(iv) 5a + 16 = 3a+ 24 WS
(v)  (½)x + 8 = 24½ WS
(vi)  (¾)a + 16 = (¼)a+ 24 WS
(vii) (¾)q + 17 = 32 WS
(viii) 13 =[2/3]x +7 twice WS
(ix) Animated Examples WS
(a) Integral Coefficients (A)
(b) Integral Coefficients (B)
(c) Fractional Coefficients
(d) With parameters



 



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