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Lesson Tutor : Algebra Lesson : Balancing Equations to Solve Variables (Finding ‘x’)

  /  Lesson Tutor : Algebra Lesson : Balancing Equations to Solve Variables (Finding ‘x’)

Basic Algebra – Lesson 6
Balancing Equations to Solve Variables
by Elaine Ernst Schneider

Objective(s): By the end of this lesson the student will be able to: 
balance equations to solve for variables (solve for ‘x’)


Pre Class Assignment:  Completion or review of Basic Algebra – Lesson 5


Resources/Equipment/Time Required: 


Outline:

In the last lesson, you learned to write expressions and to find the values of expressions.
When two expressions can be written to balance or equal one another, it is called an
equation. In other words, whatever is expressed on one side of the equal sign is calculated to be the exact same value as whatever is on the other side of the equal sign. IN other words, the left member and the right member of the equation “balance.”

Examples of equations:

5 + 7 – 2 = 2(5) (the answer on the left side is 10 and the answer on the right side is 10)

100/25 = 347 – 343 (the answer is 4 on the left of the equal sign and 4 on the right also)

Equations involving unknown variables are solved by balancing the left member and the right member. For example, in the equation y + 5 = 12, I know that the left member must equal 12 for it to balance with the right member. This means y would be 7 because 7 + 5 = 12. That one is easy enough to do the math without a formal procedure. But because algebraic equations can become much longer and more calculated, a system is necessary to solve for variable and balance equations.

Let’s take the above equation:

y + 5 = 12

The proper procedure in solving for y is to “isolate” y. This means we want y to stand by itself  on one side of the equal sign.

Now, keep in mind that we want every thing to “balance” on either side of the equal sign. This means whatever I do to the left member, I must do to the right member. So, to isolate y and solve the equation, I must “move” the 5 to the other side of the equal sign. To do this, I must make it zero on the left side of the equation by subtracting 5:

y + 5 = 12

y + 5 – 5 = 12 – 5

y + 0 = 12 – 5

y = 7

Try solving for the variable by keeping each side equal. Remember, the idea is to “balance” the equation, so what you do to one side, you must do to the other.

100/5 = y + 2

100/5 -2 = y +2 – 2

20 – 2 = y

18 = y  **

**The answer displayed prior to June 10/01 was incorrect and unreported. Thanks to the class of L. Crawford for the notice. JM

Now it’s time for you to try a few…


Assignment(s) including Answer key: 
EXERCISE:

1.   y – 10 = 17
2.   c + 4 = 29
3.   5y = 90
4.   1/3 t = 29
5.   6y = 72
6.   1/9 g = 58
7.   4y = 100
8.   12a = 132
9.   r + 9 = 48
10.  x + 79 = 422
Cut Here —————————————————–

ANSWER KEY:

1.  27
2.  25
3.  18
4.  87
5.  12
6.  522
7.  25
8.  11
9.  39
10. 343


Pre-Requisite To:  Basic Algebra Lesson 7

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