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Transcript
2-5 AdditionEquations Equations 2-5 Addition Warm Up Problem of the Day Lesson Presentation Course 1 Course 11 Course 2-5 Addition Equations 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 Course 1 2-5 Addition Equations Learn to solve whole-number addition equations. Course 1 2-5 Addition Equations h + 14 82 h ? h + 14 h Course 1 82 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. 2-5 Addition Equations 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. Course 1 2-5 Addition Equations Additional Example 1A: Solving Addition Equations Solve the equation. Check your answer. x + 87 = 152 x + 87 = 152 – 87 – 87 87 is added to x. Subtract 87 from both sides to undo the addition. x = 65 Check x + 87 = 152 ? Substitute 65 for x in the equation. 65 + 87 = 152 ? 152 = 152 65 is the solution. Course 1 2-5 Addition Equations Additional Example 1B: Solving Addition Equations Solve the equation. Check your answer. 72 = 18 + y 72 = 18 + y –18 –18 54 = y Check 72 = 18 + y ? 72 = 18 + 54 ? 72 = 72 Course 1 18 is added to y. Subtract 18 from both sides to undo the addition. Substitute 54 for y in the equation. 54 is the solution. 2-5 Addition Equations Check It Out: Example 1A Solve the equation. Check your answer. u + 43 = 78 u + 43 = 78 – 43 – 43 u 43 is added to u. Subtract 43 from both sides to undo the addition. = 35 Check u + 43 = 78 ? 35 + 43 = 78 ? 78 = 78 Course 1 Substitute 35 for u in the equation. 35 is the solution. 2-5 Addition Equations Check It Out: Example 1B Solve the equation. Check your answer. 68 = 24 + g 68 = 24 + g –24 –24 44 = g Check 68 = 24 + g ? 68 = 24 + 44 ? 68 = 68 Course 1 24 is added to g. Subtract 24 from both sides to undo the addition. Substitute 44 for g in the equation. 44 is the solution. 2-5 Addition Equations 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 = 95 = 95 = 12 + d –12 –12 83 = d distance between Johnstown and Cooperstown 12 + distance between Cooperstown and Springfield + d 12 is added to d. Subtract 12 from both sides to undo the addition. It is 83 miles from Cooperstown to Springfield. Course 1 2-5 Addition Equations 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 = 35 = 35 = 17 + d –17 –17 18 = d distance between Patterson and Jacobsville 17 + distance between Jacobsville and East Valley + d 17 is added to d. Subtract 17 from both sides to undo the addition. It is 18 miles from Jacobsville to East Valley. Course 1