Solving Linear Inequalities Included in this presentation:  Solving Linear Inequalities  Solving Compound Inequalities  Linear Inequalities Applications

Steps to Solving Linear Inequalities 1.Simplify both sides of the inequality. (Get rid of parenthesis and put “like terms” together. 2.Isolate the variable. When isolating the variable in an inequality you can do the following to both sides of the inequality to keep equivalent statements. Add or subtract a number Multiply or divide by a positive number. Multiply or divide by a negative number AND reverse the direction of the inequality sign. The only difference in solving for a variable in a linear inequality and solving for a variable in a linear equation is…. WHEN YOU DIVIDE OR MULTIPLY BY A NEGATIVE NUMBER IN AN INEQUALITY, YOU HAVE TO CHANGE THE INEQUALITY SIGN.

Graphing Solutions When you are graphing solutions to inequalities if the sign is then you put an open circle at the endpoints. If the sign is then you put a closed circle at the endpoints.

Example 1: Variables on 1 side Step 1: Distribute the 2. Step 2: Combine 3x and 2x. Step 3: Add the 10 to the right side. Step 4: Divide by 5 to isolate the variable. Solve the inequality for x, then graph the solution. The graph is on the next slide.

Example 1 continued Solve the inequality for x, then graph the solution. This means that any number substituted in for x in the original statement, that is larger than or equal to 14/5, will make a true statement.

Example 2: Variables on 1 side Solve the inequality for x, then graph the solution. Step 1: Distribute the -7. Step 2: Combine “like” terms Step 3: Add 2 to both sides. Step 4: Divide by -14 (WHEN YOU DIVIDE BY A NEGATIVE NUMBER, YOU MUST CHANGE THE INEQUALITY SIGN. The graph is on the next slide.

Example 2 continued Solve the inequality for x, then graph the solution. This means that any number substituted in for x in the original statement, that is larger than 0, will make a true statement.

Example 3: Variables on both sides Solve the inequality for x, then graph the solution. Step 1: Distribute the 5. Step 2: Combine “like” terms Step 3: Subtract 10x from both sides. Step 4: Add 15 to both sides Step 5: Divide by -3. (WHEN YOU DIVIDE BY A NEGATIVE NUMBER, YOU MUST CHANGE THE INEQUALITY SIGN. The graph is on the next slide.

Example 3: Variables on both sides Solve the inequality for x, then graph the solution. This means that any number substituted in for x in the original statement, that is larger than or equal to -6, will make a true statement.

Example 4: Variables on both sides Solve the inequality for x, then graph the solution. Step 1: Distribute the -2. Step 2: Multiply by 3 to get rid of the fraction. Step 3: Add 18x to both sides. Step 4: Subtract 3 from both sides. Step 5: Divide by 13. The graph is on the next slide.

Example 4: Variables on both sides Solve the inequality for x, then graph the solution. This means that any number substituted in for x in the original statement, that is smaller than 21/13 will make a true statement.

Compound Inequalities Compound inequalities are inequalities that are paired together with “and” or “or.” The solutions of an “ And ” compound inequality are bounded by endpoints and include all the numbers in between the endpoints. This is an “and” compound inequality This means that the values of x have to be greater than -3 AND less than 5. The graph of the solutions looks like…..

Solving “AND” Inequalities Solve the inequality for x, then graph the solution. When solving AND inequalities, you need to isolate the variable in the middle by adding, subtracting, multiplying, and dividing numbers to each section. Remember when you divide by a negative or multiply by a negative number you will still need to change the inequality sign. This means that the values of x have to between 2 and 8 or equal to 2 and 8 to make the “AND” inequality true.

Example 5: “AND” Compound Inequality Solve the inequality for x, then graph the solution. Step 1: Subtract 4 from each section. Step 2: Divide each section by 3. This means that the values of x have to between -14/3 and 2/3 to make the “AND” inequality true.

Example 6: “AND” Compound Inequality Solve the inequality for x, then graph the solution. Step 1: Distribute the -5 through the middle section. Step 2: Add 40 to each section. Step 3: Divide each section by -5. Remember when you divide by a negative number you change the inequality signs. Its not necessary to rewrite the inequality but it may help interpret the answer. This means that the values of x have to between -11 and -10 or equal to - 11 and -10 to make the “AND” inequality true.

Compound Inequalities Compound inequalities are inequalities that are paired together with “and” or “or.” The solutions of an “ OR ” compound inequality are not bounded by endpoints and the solutions go in opposite ways. This is an “or” compound inequality This means that the values of x have to be less than or equal to -4 or greater than or equal to 1. The graph of the solutions looks like…..

Solving “OR” Inequalities Solve the inequality for x, then graph the solution. Step 1: Subtract 9 Step 2: Divide by 2 Step 1: Subtract 3 Step 2: Divide by 5 When solving “OR” compound inequalities solve each inequality separately and then graph the solutions on the same number line. This means that the values of x have to be less than 1 or greater than 7 to make the “OR” inequality true.

Example 7: Solving “OR” Inequality Solve the inequality for x, then graph the solution. Steps: 1.Distribute Combine like terms. 3.Subtract 6 4.Divide by -2 Flip sign of inequality when divide by a negative number Steps: 1.Distribute Combine like terms. 3.Subtract 10 4.Divide by -3 Flip sign of inequality when divide by a negative number This means that the values of x have to be greater than 4 or less than -2 to make the “OR” inequality true.

Linear Inequalities Applications When given word problems that can be represented with linear inequalities, follow the following steps. 1. Define the variable. 2. Write the inequality the represents the situation. 3. Solve the inequality to state the answer.

Example 8: Linear Inequalities Applications The product of 5 and a number, x, decreased by 9 is at most 7. What are the possible solutions of x? Let “x” represent the number we don’t know. Product means multiply, decreased by 9 means subtract 9 and at most means that it can be equal to 7 or smaller than 7….so The possible x values that will be solutions of this inequality are any numbers that are smaller than 16/5 or equal to 16/5.

Example 9: Linear Inequalities Applications The product of -2 and the quantity of the sum of a number, x, and 4 is either less than -9 or more than 5. What are the possible solutions of x? Let “x” represent the number we don’t know. Product means multiply, quantity means that I have to have the sum of x and 4 and then multiply that to the -2.….so The possible x values that will be solutions of this inequality are any numbers that are smaller than -13/2 or greater than ½.

Example 10: Linear Inequalities Applications The quantity of twice a number, x, increased by 7 is between 2 and 28. What are the possible solutions of x? Let “x” represent the number we don’t know. Twice a number means multiply by 2, increased by 7 means add 7, between means the quantity is in between those 2 numbers but can’t equal those numbers….so The possible x values that will be solutions of this inequality are any numbers that are larger than -5/2 AND less than 21/2.

Example 10: Linear Inequalities Applications You have $130 and go shopping. You pay $7.25 for lunch and the rest of the money you have to buy shirts. If shirts cost $15 a piece, what is the inequality that would represent the number of shirts you could buy? How many shirts could you buy? Let “x” represent the number of shirts that we buy. 15 times the quantity of shirts you buy has to be less than or equal to the amount of money you have after lunch. So it says to spend less than or equal to the amount of money you have left you would have to buy shirts….which means you can buy 8 shirts or less. You can not buy part of a shirt so your solutions would be limited to 8, 7, 6, 5, 4, 3, 2, 1, or 0 shirts.