Basic Mathematics for College Students

(Nandana) #1

Use the addition property of equality.


To solve an equationmeans to find all values of the
variable that make the equation true. We can
develop an understanding of how to solve equations
by referring to the scales shown on the right.
The first scale represents the equation


. The scale is in balance because the
weights on the left side and right side are equal. To
find , we must add 2 to the left side. To keep the
scale in balance, we must also add 2 to the right side.
After doing this, we see that is balanced by 5.
Therefore, must be 5. We say that we have solved
the equation and that the solution is 5.
In this example, we solved by
transforming it to a simpler equivalent equation,
.


Equivalent Equations

Equations with the same solutions are called equivalent equations.

The procedure that we used to solve x 2 3 illustrates the following property
of equality.


Addition Property of Equality

Adding the same number to both sides of an equation does not change its
solution.
For any numbers , , and ,
if , then

When we use this property of equality, the resulting equation is equivalent to the
original one.We will now show how it is used to solve x 2  3.


ab acbc

ab c

x 5


x 2  3

x 2  3

x

x

x

x 2  (^3) x – 2 = 3
Add
2
x = 5
11111
111
x
x − 2
Add
2
2
8.3 Solving Equations Using Properties of Equality 659
EXAMPLE (^2) Solve:
StrategyWe will use a property of equality to isolate the variable on one side of
the equation.
WHYTo solve the original equation, we want to find a simpler equivalent
equation of the form , whose solution is obvious.
Solution
We will use the addition property of equality to isolate on the left side of the
equation. We can undo the subtraction of 2 by adding 2 to both sides.
This is the equation to solve.
Add 2 to both sides.
On the left side, when 0 is added to a number, the result
x (^5) is the same number.
On the left side, the sum of a number and its opposite is zero:
x  0  (^5)  2  2 0. On the right side, add: 3 2 5.
x 2  2  3  2
x 2  3
x
xa number
x 2  3
Self Check 2
Solve:
Now TryProblem 37
n 16  33

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