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Unit 2 Honors Review 2015-2016 Objective: to review for unit 2 Quest tomorrow Warm-up: Complete handout in the front of the classroom **Note: if you have not completed your compound inequalities project it needs to be completed by Friday (2 point classwork grade) Remember: QUEST TOMORROW! Solving Multi-Step Equations • Make sure to distribute everything first Be careful: -(3+4) = -3 - 4 • Combine like terms • Solve for your variable! Common Mistake: 4x = 5 your solution should be 5 x= 4 Clearing with Fractions: 1. Find your LCD: 𝑞 15 1 5 − = 3 5 LCD = 15 2. Multiply each term by your LCD: 15 𝑞 15 1 15 3 ∙ − ∙ = ∙ 1 15 1 5 1 5 3. Cross Simplify: q-3=9 4. Solve: q= 12 Clearing Decimals: 1. Determine what the most amount of decimal places you have to move in each term to get each term to be non decimal. **Note: when we move a decimal place we’re moving by 10, 100, 1000, … 2. Move that many decimal places for ALL TERMS (don’t forget about variables or constants) 3. Solve Example: 1.5 = 1.2y-5.7 (1 decimal place) 15 = 12y – 57 (15+57) = 12y 72 =12y y=6 **Note if a question has both fractions and decimals convert all of them to one or the other, if one is a repeating decimal convert all to fractions! Solving Equations with Variables on Both Sides Steps: 1. Distribute everything on both sides of your equal sign • Example: 2. Combine like terms on each side of the equal sign 3x + 15-9 = 2(x+2) 3. Move your SMALLEST variable to the opposite side using inverse operations 3x + 6 = 2x + 4 4. Move your constant to the opposite side **Variables should be on one side, constants should be on the other** 5. SOLVE for your variable 3x + 15 – 9 = 2x + 4 -2x -2x x+6=4 -6 -6 x = -2 Solving Equations with Variables on Both Sides Special Cases • Identity/ All Real Solutions x + 4 -6x = 6-5x-2 -5x + 4 = -5x + 4 + 5x +5x 4=4 **This is a true statement therefore, it is ALL REAL SOLUTIONS • No Solution -8x + 6 + 9x = -17 + x x + 6 = x -17 -x -x 6 = -17 **This will never be a true statement, therefore, it is NO SOLUTION Solving Absolute Value Equations They should have 2 solutions! Think back to |x| = 4, how many different values can x be? (4 or -4) 2 solutions… Steps: 1. Isolate your variable (get it all by itself by using inverse operations) **Note: If you have 7|x+2| DO NOT DISTRIBUTE!! 2. Remove your absolute value signs and create 2 equations a) Original equation b) Negate the solution (make it negative) 3. Solve each equation 4. Check to make sure the solutions work in the original equation Example: 1-|10-2k|=-25 -|10-2k| = -26 |10-2k| = 26 Case1: 10-2k = 26 k = -8 Case 2: 10-2k= -26 k = 18 Check: plug values into original equation make sure it works Your solution: {-8,18} Solving Absolute Value Equations (Special Cases) One Case = 1 solution No Solution: |x+3|+4 = 4 2 - |2x-5|=7 |x+3| = 0 -|2x-5| = 5 **you can’t have a negative 0 so there’s only one case** Case 1: x+ 3 = 0 x = -3 Solution: {-3} |2x-5| = -5 **STOP!! Can you ever have an absolute value = negative number?!! NO, therefore your solution is: NO SOLUTION INEQUALITIES!! • Reminders: **When you divide or multiply by a negative switch the inequality sign **If you have your variable on the left, that is the direction that you can shade on your graph ** If you’re not shading above your graph, make sure its really really dark so I can see it and don’t mark it wrong **Open circles = < or > **Closed Circle = < or > **To rewrite an inequality, you flip the sign and the order Example: x < 4 is equivalent to 4 > x And graphed as Solving Inequalities with Variables on Both Sides • Just as with equations: 1. distribute everything 2. combine like terms 3. then move the “SMALLEST” variable to the other side of the inequality and the constants to the other side 4. Then graph (moving the variable to the left of the inequality makes this a lot easier!) Example: 5(4+x) < 3(2+x) 20 + 5x < 6 + 3x 20 + 2x < 6 2x < -14 x < -7 Solving Inequalities with Variables on both sides Special Cases • All Real Numbers: • No Solution **True Statement, with no variables** • **False statement with no variables** Example: Example: x+5> x+3 -x -x 5 > 3 (always true) ALL REAL SOLUTIONS **you can conclude this from the original inequality, how? 2 (x+3) < 5 + 2x 2x + 6 < 2x + 5 -2x -2x 6< 5 This will never be true therefore, NO SOLUTION! What is the difference between and and or? AND means intersection A B A B -what do the two items have in common? OR means union -if it is in one item, it is in the solution Compound Inequalities “AND” “OR” Solutions that they BOTH have collaboratively Solutions they the Each have + what they have collaboratively **Should be written as one inequality.. **CAN NOT BE WRITTEN AS ONE INEQUALITY** Smallest value < variable < large value Example: Example: 4 < x+2 < 8 -2 -2 -2 2< x< 6 **If this confuses you solving simultaneously split them into two separate inequalities, and solve** Remember your solution should only be the values that are “SHARED” 2x < 6 or x < 3 OR 3x > 12 x> 4 7) 2x < -6 and 3x ≥ 12 1. Solve each inequality for x 2. Graph each inequality 3. Combine the graphs 4. Where do they intersect? 5. They do not! x cannot be greater than or equal to 4 and less than -3 Solution!! No 2 x 6 2 2 x 3 o -3 3x 12 3 3 x4 -6 1 o ● 4 0 7 Graph x ≥ -1 or x ≤ 3 -5 0 5 The whole line is shaded!! Graph: x > 4 or x > -2 Solving Absolute Value Inequalities 1. Isolate the absolute value 2. Create two inequalities 1. 2. Remove the absolute value signs, keep absolute value expression, keep the inequality sign, and keep the value Remove the absolute value sign, keep absolute value expression, change inequality sign and change value sign Example: |x-8| + 5 > 11 |x-8| > 6 Case 1: x-8 > 6 x > 14 Case 2: x-8 < -6 x< 2 This was a greater than inequality therefore it will have an OR graph! Follow the same steps for an AND graph RECALL: GreatOR and LessthAND Solving Absolute Value Inequalities (Special Cases) All Real Numbers No Solution **True Statement** **False Statement** Example: Example: |x-6| +7 > 2 |x+12| -5 < -6 |x-6| > -5 |x+12| < -1 True, absolute values will always be a positive number, therefore any positive number will always be greater than -5, and therefore it is true for any value of x False, this is saying that an absolute value is going to be less than a negative number, will this ever be true??? No because absolute values are always positive