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Section 2.3 Solving Multi-Step Equations
Algebra 1
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Learning Targets Solve multi-step equations
Write and solve multi-step equations in a context Solve equations with variables on both sides Solve equations with special solutions (no solutions or many solutions) Write and solve an equation with variables on both sides
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Example 1: Solving equations Review
Solve for π₯: 11π₯β4=29 1. 11π₯=33 (add 4 to both sides) 2. π₯=3 (divide both sides by 11)
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Example 2: Solving equations Review
Solve for π₯: 4π₯+15=51 1. 4π₯=36 (subtract 15 to both sides) 2. π₯=9 (divide both sides by 4)
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Example 3: Solving equations Review
Find π: π+7 8 =5 1. π+7=40 (multiply both sides by 8) 2. π=33 (subtract 7 from both sides)
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Example 4: Solving equations Review
Find π: πβ25=115 π=140 (add 25 to both sides) 2. 2π=420 (multiply 3 to both sides) 3. π=210 (divide 2 to both sides) Alternative: π=210 multiply 3 2 , the reciprocal, to both sides. Combine steps 2&3 into one step.
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Example 5: Solving equations Review
Find 5π₯ 9 β11=β51 5π₯ 9 =β40 5π₯=β360 π₯=β72
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Example 6: Writing/Solving equations
If 15 less than 9 times a number is 30, what is the number? 1. 9π₯β15=30 2. 9π₯=45 (add 15 to both sides) 3. π₯=5 (divide both sides by 9)
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Example 7: Writing/Solving equations
If 12 more than 2 times a number is -10, what is the number? 1. 2π¦+12=β10 2. 2π¦=β22 (subtract 12 from both sides) 3. π¦=β11 (divide both sides by 2)
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Example 8: Writing/Solving equations
Find three consecutive integers with a sum of 21 Integer 1: π, Integer 2: π+1, Integer 3: π+2 π+ π+1 + π+2 =21 3π+3=21 3π=18 π=6 Integer 1: 6, Integer 2: 7, Integer 3: 8
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Example 9: Writing/Solving equations
The product of 9 and the sum of a number and 2 is 27. 1. 9 π₯+2 =27 2. 9π₯+18=27 (distribute the 9) 3. 9π₯=9 (subtract 18 from both sides) 4. π₯=1 (divide both sides by 9) Alternative: π₯+2=3 (divide by 9 after step 1), then continue to solve.
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Exit Ticket for Feedback
1. 10 more than the product of 7 and a number is 31, what is the number? 2. Find three consecutive integers with a sum of β36.
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