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Course 1. 2-5. Addition Equations. Warm Up. Problem of the Day. Lesson Presentation. Course 1. Warm Up Determine whether each value is a solution. 1. 86 + x = 102 for x = 16 2. 18 + x = 26 for x = 4 3. x + 46 = 214 for x = 168 4. 9 + x = 35 for x = 26. yes. no. yes.

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  1. Course 1 2-5 Addition Equations Warm Up Problem of the Day Lesson Presentation Course 1

  2. Warm Up Determine whether each value is a solution. 1. 86 + x = 102 for x = 16 2.18 + x = 26 for x = 4 3.x + 46 = 214 for x = 168 4.9 + x = 35 for x = 26 yes no yes yes

  3. Problem of the Day After Renee used 40 m of string for her kite and gave 5 m to her sister for her wagon, she had 8 m of string left. How much string did she have to start with? 53 m

  4. Learn to solve whole-number addition equations.

  5. h + 14 82 h + 14 82 h h 68 ? The equation h + 14 = 82 can be represented as a balanced scale. To find the value of h, you need h by itself on one side of the scale. To get h by itself, first take away 14 from the left side of the scale. Now the scale is unbalanced. To rebalance the scale, take away 14 from the other side.

  6. Taking away 14 from both sides of the scale is the same as subtracting 14 from both sides of the equation. h + 14 = 82 –14 –14 h = 68 Subtraction is the inverse, or opposite, of addition. If an equation contains addition, solve it by subtracting from both sides to “undo” the addition.

  7. ? 65 + 87 = 152 ? 152 = 152 Additional Example 1A: Solving Addition Equations Solve the equation. Check your answer. x + 87 = 152 x + 87 = 152 87 is added to x. – 87 – 87 Subtract 87 from both sides to undo the addition. x = 65 Check x + 87 = 152 Substitute 65 for x in the equation.  65 is the solution.

  8. ? 72 = 18 + 54 ? 72 = 72 Additional Example 1B: Solving Addition Equations Solve the equation. Check your answer. 72 = 18 + y 72 = 18 + y 18 is added to y. –18 –18 Subtract 18 from both sides to undo the addition. 54 = y Check 72 = 18 + y Substitute 54 for y in the equation.  54 is the solution.

  9. ? 35 + 43 = 78 ? 78 = 78 Check It Out: Example 1A Solve the equation. Check your answer. u + 43 = 78 u + 43 = 78 43 is added to u. – 43 – 43 Subtract 43 from both sides to undo the addition. u = 35 Check u + 43 = 78 Substitute 35 for u in the equation.  35 is the solution.

  10. ? 68 = 24 + 44 ? 68 = 68 Check It Out: Example 1B Solve the equation. Check your answer. 68 = 24 + g 68 = 24 + g 24 is added to g. –24 –24 Subtract 24 from both sides to undo the addition. 44 = g Check 68 = 24 + g Substitute 44 for g in the equation.  44 is the solution.

  11. Additional Example 2: Social Studies Application Johnstown, Cooperstown, and Springfield are located in that order in a straight line along a highway. It is 12 miles from Johnstown to Cooperstown and 95 miles from Johnstown to Springfield. Find the distance d between Cooperstown and Springfield. distance between Johnstown and Springfield distance between Johnstown and Cooperstown distance between Cooperstown and Springfield = + 95 = 12 + d 95 = 12 + d 12 is added to d. Subtract 12 from both sides to undo the addition. –12 –12 83 = d It is 83 miles from Cooperstown to Springfield.

  12. Check It Out: Example 2 Patterson, Jacobsville, and East Valley are located in that order in a straight line along a highway. It is 17 miles from Patterson to Jacobsville and 35 miles from Patterson to East Valley. Find the distance d between Jacobsville and East Valley. distance between Patterson and East Valley distance between Patterson and Jacobsville distance between Jacobsville and East Valley = + 35 = 17 + d 35 = 17 + d 17 is added to d. Subtract 17 from both sides to undo the addition. –17 –17 18 = d It is 18 miles from Jacobsville to East Valley.

  13. Lesson Quiz Solve each equation. 1.x + 15 = 72 2. 81 = x + 24 3.x + 22 = 67 4. 93 = x+ 14 x = 57 x = 57 x = 45 x = 79 5. Kaitlin is 2 inches taller than Reba. Reba is 54 inches tall. How tall is Kaitlin? 56 inches

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