Examples

Example 1

 

 Solve 5x - 4 = 2x - 10.

	5x - 4	= 2x - 10
Solution:	-2x	-2x	(Step 3 above: subtract 2x from each side to get read of  the variable term from the right side)
	3x - 4	= -10
	+4	+4	(Step 4: adding four to both sides removes the constant term from the variable side.)
	3x	= -6
			(Step 5: divide both sides by 3 to get red of  the multiplier on the variable term)
	x = -2		(Simplify these fractions to obtain the answer

You can also check this solution by or to any equation by substituting your answer into the original equation.

5x - 4 = 2x - 10

5(-2) - 4 = 2(-2) - 10

-10 - 4 = -4 - 10

-14 = -14

Since both sides are the same when x = -2, you know now that you have correctly solved this equation.Congrats!

 

Example 2

  Anwer 7y - 2(3y - 5) = 3(3y + 1) - 9.

Solution: 7y - 6y + 10 = 9y + 3 - 9 (1) Multiply through the parentheses.

(2) Collect the like terms on each side.

At this point you should have at most.

1y + 10 = 9y - 6 two terms on each side know finish the rest of your problem. 

-9y -9y (3) Remove the variable from one side.

-8y + 10 = -6

-10 -10 (4) Remove the constant from the other

-8y = -16 side.

Divide off the variable coefficient.

This is the solution.

Example 3 An Identity

Solve 3(r + 1) - 5 = 3r -2.

3( r + 1) - 5 = 3r - 2  original equation

3r + 3 - 5 = 3r -2  Distributive Property

3r - 2=3r - 2        Reflexive Property of Equality

Since the expressions on each side of the equations are the same, this equation is an identity. the statement 3(r + 1) is true for the values of r in this.        

  Extra Chalenges

Solve these problems for extra practice.

Two grocery stores sell rice in bulk. The first charges $0.55 per pound. The second
charges $0.75 per pound for up to 3 pounds and $0.40 per pound for anything
over 3 pounds.
1. Write expressions for the cost of rice at each store in terms of the number of
pounds bought, assuming you buy more than 3 pounds.
2. Write and solve an equation that relates the two expressions from Exercise 1.
3. Interpret your result from Exercise 2.
4. Evaluate each expression from Exercise 1 for the value of x from Exercise 2.
Interpret the result.
5. Describe what happens for values of x less than the one found in Exercise 2
and for values greater than it. 

   

Additional Information  Steps For Solving Equations 1.        Use the Distributive Property so there is no more grouping symbols 2.      Simplify each side of the equation on both sides of the equals sign. 3.      Use the Additiaon and/or Subtraction Properties of Equality to get the variables on one side of the equals sign and the numbers without variables on the other side of the equals sign. 4.      Simply all that is possible on both sides of the equal sign. 5.      Use the Multiplication or Division Property of Equality to solve. If the solution is false, there is no solution. If the solution results in an identity, the solution is all numbers.

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