Solve.EXAMPLE 1Solve: 4(2x − 3) + 7 = 3x + 5SolutionThere are no fractions, so we begin wi...

Solve.

EXAMPLE 1

Solve: 4(2x − 3) + 7 = 3x + 5

Solution

There are no fractions, so we begin with Step 2.

Step 2.

Step 3.

Step 4. Get all variable terms on the same side of the equation by subtracting 3x from both sides, then adding 5 to both sides.

Step 5. Use the multiplication property of equality to get x alone.

Step 6. Check.

The solution is 2 or the solution set is {2}.

EXAMPLE 2

Solve: 8(2 − t) = −5t

Solution

Step 2.

Step 4.

Step 5.

Step 6. Check.

The solution is .

EXAMPLE 3

Solve:

Solution

We begin by clearing fractions. To do this, we multiply both sides of the equation by the LCD of 2 and 3, which is 6.

Step 1.

Step 2.

There are no longer grouping symbols and no like terms on either side of the equation, so we continue with Step 4.

Step 4.

Step 5. The variable is now alone, so there is no need to apply the multiplication property of equality.

Step 6. Check.

The solution is 12.

EXAMPLE 4

Solve:

Solution

We clear the equation of fractions first.

Step 1.

Step 2. Next, we use the distributive property and remove parentheses.

Step 4.

Step 5.

Step 6. To check, replace a with 0 in the original equation. The solution is 0.

EXAMPLE 5

Solve: 0.25x + 0.10(x − 3) = 0.05(22)

Solution

First we clear this equation of decimals by multiplying both sides of the equation by 100. Recall that multiplying a decimal number by 100 has the effect of moving the decimal point 2 places to the right.

0.25x + 0.10(x − 3) = 0.05(22)

Step 1.

Step 2.

Step 3.

Step 4.

Step 5.

Step 6. To check, replace x with 4 in the original equation. The solution is 4.

EXAMPLE 6

Solve: −2(x − 5) + 10 = −3(x + 2) + x

Solution

The final equation contains no variable terms, and there is no value for x that makes 20 = −6 a true equation. We conclude that there is no solution to this equation. In set notation, we can indicate that there is no solution with the empty set, { }, or use the empty set or null set symbol, ➢. In this chapter, we will simply write no solution.

EXAMPLE 7

Solve: 3(x − 4) = 3x − 12

Solution

The left side of the equation is now identical to the right side. Every real number may be substituted for x and a true statement will result. We arrive at the same conclusion if we continue.

Again, one side of the equation is identical to the other side. Thus, 3(x − 4) = 3x − 12 is an identity and all real numbers are solutions. In set notation, this is {all real numbers}.

2(x + 3) − 5 = 5x − 3(1 + x)