IntroductionWhile it may not be efficient to write out the justification for each step when solving equations, it is important to remember that the properties of equality must always apply in order for an equation to remain balanced.

As equations become more complex, it may be helpful to refer to the properties of equality used in the previous lesson.

2.1.2: Solving Linear Equations

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Key Concepts• When solving equations, first take a look at the

expressions on either side of the equal sign.

• You may need to simplify one or both expressions before you can solve for the unknown. Sometimes you may need to combine like terms by using the associative, commutative, or distributive properties.

• Pay special attention if the same variable appears on either side of the equal sign.

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2.1.2: Solving Linear Equations

Key Concepts, continued• Just like with numbers, variables may be added or

subtracted from both sides of the equation without changing the equality of the statement or the solution to the problem.

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2.1.2: Solving Linear Equations

Solving Equations with the Variable in Both Expressions of the Equation

1. Choose which side of the equation you would like the variable to appear on.

2. Add or subtract the other variable from both sides of the equation using either the addition or subtraction property of equality.

3. Simplify both expressions.4. Continue to solve the equation.5. As with any equation, check that your answer is correct by substituting

the value into the original equation to ensure both expressions are equal.

Key Concepts, continued• Some equations may have no solution. This is the

case when, after you’ve completed all of the appropriate steps to solve an equation, the result is something impossible, like 2 = 6. The resulting equation is never true for any value of the variable.

• Some equations will be true for any value the variable is replaced with. This is the case when following all of the appropriate steps for solving an equation results in the same value on each side of the equal sign, such as 2x = 2x. The resulting equation is always true for any value of the variable.

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2.1.2: Solving Linear Equations

Key Concepts, continued• Other equations will only have one solution, where the

final step in solving results in the variable equal to a number, such as x = 5.

2.1.2: Solving Linear Equations

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Common Errors/Misconceptions• performing the wrong operation when isolating the

variable

• incorrectly combining terms

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2.1.2: Solving Linear Equations

Guided Practice

Example 1Solve the equation 5x + 9 = 2x – 36.

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2.1.2: Solving Linear Equations

Guided Practice: Example 1, continued

1. Move the variable to one side of the equation.Notice that the same variable, x, is on both sides of the equation: 5x is on the left of the equation and 2x is on the right. It makes no difference whether you choose to have the variables on the left or on the right; your solution will remain the same. It’s common to have the variable on the left, but not necessary.

It’s often easier to move the variable with the smallest coefficient to the opposite side of the equation. Here, 2x is smaller than 5x, so let’s move 2x.

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2.1.2: Solving Linear Equations

Guided Practice: Example 1, continued2x is positive, so to get it to the other side of the equal sign you will need to subtract it from both expressions in the equation.

It helps to line up what you are subtracting with the terms that are similar in order to stay organized. In this case, we are subtracting variables from variables.

2.1.2: Solving Linear Equations

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Guided Practice: Example 1, continuedWhen 2x is subtracted, it’s important not to forget the remaining terms of each expression. Look out for subtraction signs that now act as negative signs. Here, since 36 was originally being subtracted from 2x, the subtraction sign left behind makes 36 negative.

3x + 9 = –36

2.1.2: Solving Linear Equations

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Guided Practice: Example 1, continued

2. Continue to solve the equation 3x + 9 = –36.To isolate x, subtract 9 from both expressions in the equation.

Divide both expressions by the coefficient of x, 3.

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2.1.2: Solving Linear Equations

Guided Practice: Example 1, continued

3. The solution to the equation 5x + 9 = 2x – 36 is x = –15.A quick check will verify this. Substitute –15 for all instances of x in the original equation, and then evaluate each expression.

5x + 9 = 2x – 36

5(–15) + 9 = 2(–15) – 36

–75 + 9 = –30 – 36

–66 = –66

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2.1.2: Solving Linear Equations

Guided Practice: Example 1, continuedOur check verified that both sides of the equation are still equal; therefore, x = –15 is correct.

2.1.2: Solving Linear Equations

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Guided Practice: Example 1, continued

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2.1.2: Solving Linear Equations

Guided Practice

Example 5Solve the literal equation for b1.

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2.1.2: Solving Linear Equations

Guided Practice: Example 5, continued

1. Isolate b1.As we saw in Unit 1, to solve literal equations for a specific variable, we follow the same steps as solving equations.

In this equation, we could distribute over b1 + b2,

but this may cause more work for us. Instead, let’s

get rid of the fraction. Multiply both sides of the

equation by the inverse of , or 2.16

2.1.2: Solving Linear Equations

Guided Practice: Example 5, continued

2.1.2: Solving Linear Equations

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Guided Practice: Example 5, continued

2. Again, you could distribute h over b1 + b2, but it’s more efficient to divide both sides of the equation by h.

2.1.2: Solving Linear Equations

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Guided Practice: Example 5, continued

3. Finally, to solve the equation for b1, subtract b2 from both expressions of the equation.

2.1.2: Solving Linear Equations

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Guided Practice: Example 5, continued

4. The equation solved for b1

is .

The equation can be rewritten as using

the symmetric property of equality.

2.1.2: Solving Linear Equations

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Guided Practice: Example 5, continued

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2.1.2: Solving Linear Equations