Introduction

1. Introduction While 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

2. 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. 2.1.2: Solving Linear Equations

3. 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. 2.1.2: Solving Linear Equations

4. 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. 2.1.2: Solving Linear Equations

5. 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

6. Common Errors/Misconceptions performing the wrong operation when isolating the variable incorrectly combining terms 2.1.2: Solving Linear Equations

7. Guided Practice Example 1 Solve the equation 5x + 9 = 2x – 36. 2.1.2: Solving Linear Equations

8. Guided Practice: Example 1, continued 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. 2.1.2: Solving Linear Equations

9. Guided Practice: Example 1, continued 2x 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

10. Guided Practice: Example 1, continued When 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

11. Guided Practice: Example 1, continued 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. 2.1.2: Solving Linear Equations

12. Guided Practice: Example 1, continued 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 2.1.2: Solving Linear Equations

13. Guided Practice:Example 1, continued Our check verified that both sides of the equation are still equal; therefore, x = –15 is correct. ✔ 2.1.2: Solving Linear Equations

14. Guided Practice: Example 1, continued 2.1.2: Solving Linear Equations

15. Guided Practice Example 5 Solve the literal equation for b1. 2.1.2: Solving Linear Equations

16. Guided Practice:Example 5, continued 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. 2.1.2: Solving Linear Equations

17. Guided Practice: Example 5, continued 2.1.2: Solving Linear Equations

18. Guided Practice: Example 5, continued 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

19. Guided Practice: Example 5, continued Finally, to solve the equation for b1, subtract b2 from both expressions of the equation. 2.1.2: Solving Linear Equations

20. Guided Practice: Example 5, continued The equation solved for b1 is . The equation can be rewritten as using the symmetric property of equality. ✔ 2.1.2: Solving Linear Equations

21. Guided Practice: Example 5, continued 2.1.2: Solving Linear Equations