Equation Solving. Grade 7 Math: Solving Simple Linear Equations Subtopic Navigator Introduction: What is an Equation? One-Step Equations (Addition/Subtraction) One-Step Equations (Multiplication/Division) Two-Step Equations Equations with Variables on Both Sides Checking Your Solutions Translating Words to Equations Word Problems - Number Problems Word Problems - Real World Applications Practice Exercises Conclusion Learning Objectives Understand what an equation is and what it means to "solve" one Solve one-step equations using inverse operations Solve two-step equations systematically Check solutions by substitution Translate word phrases into algebraic equations Solve word problems by creating and solving equations Apply equation-solving skills to real-world situations Introduction: What is an Equation? An equation is a mathematical statement that shows two expressions are equal. It contains an equals sign (=) and often includes variables (letters that represent unknown numbers). Key Terms: Equation: A statement that two expressions are equal (e.g., $x + 5 = 12$) Variable: A letter that represents an unknown number (usually $x$, $n$, etc.) Solution: The value of the variable that makes the equation true Solving an equation: Finding the value(s) of the variable that make the equation true ⚖️ Think of equations like balance scales! ⚖️ When we solve equations, we want to keep the scale balanced. Whatever we do to one side, we must do to the other side to maintain balance. Addition & Subtraction. One-step equations require only one operation to solve. For addition/subtraction equations, we use the inverse operation (opposite operation) to isolate the variable. Inverse Operations: Addition ↔ Subtraction (they undo each other) If $x + a = b$, then $x = b - a$ (subtract a from both sides) If $x - a = b$, then $x = b + a$ (add a to both sides) Example 1: Solving $x + 7 = 15$ Step-by-Step Solution: Equation: $x + 7 = 15$ Goal: Get $x$ alone on one side Since 7 is added to $x$, we subtract 7 from both sides: $x + 7 - 7 = 15 - 7$ Simplify: $x = 8$ Check: $8 + 7 = 15$ ✓ Correct! Example 2: Solving $y - 12 = 20$ Step-by-Step Solution: Equation: $y - 12 = 20$ Since 12 is subtracted from $y$, we add 12 to both sides: $y - 12 + 12 = 20 + 12$ Simplify: $y = 32$ Check: $32 - 12 = 20$ ✓ Correct! Common Mistake: Incorrect: $x + 5 = 10$ → $x = 10 + 5$ (Wrong! Should subtract, not add) Correct: $x + 5 = 10$ → $x = 10 - 5$ Remember: Do the opposite operation! Practice Questions Solve: $n + 8 = 15$ Solve: $x - 3 = 12$ Solve: $15 = a + 9$ Solve: $b - 7 = 20$ Explain in your own words why we use inverse operations to solve equations. Multiplication/Division For multiplication/division equations, we also use inverse operations. Remember: Multiplication and division undo each other. Inverse Operations (Multiplication/Division): Multiplication ↔ Division (they undo each other) If $ax = b$, then $x = frac{b}{a}$ (divide both sides by a) If $frac{x}{a} = b$, then $x = ab$ (multiply both sides by a) Note: In algebra, $ax$ means $a times x$ Example 1: Solving $3x = 18$ Step-by-Step Solution: Equation: $3x = 18$ (means $3 times x = 18$) Since $x$ is multiplied by 3, we divide both sides by 3: $frac{3x}{3} = frac{18}{3}$ Simplify: $x = 6$ Check: $3 times 6 = 18$ ✓ Correct! Example 2: Solving $frac{x}{4} = 5$ Step-by-Step Solution: Equation: $frac{x}{4} = 5$ (means $x ÷ 4 = 5$) Since $x$ is divided by 4, we multiply both sides by 4: $frac{x}{4} times 4 = 5 times 4$ Simplify: $x = 20$ Check: $20 ÷ 4 = 5$ ✓ Correct! Equation Type Inverse Operation Example Solution $x + a = b$ Subtract a $x + 5 = 12$ $x = 7$ $x - a = b$ Add a $x - 8 = 15$ $x = 23$ $ax = b$ Divide by a $4x = 28$ $x = 7$ $frac{x}{a} = b$ Multiply by a $frac{x}{6} = 3$ $x = 18$ Principles Practice Solve: $5x = 45$ Solve: $frac{m}{7} = 4$ Solve: $12 = 3y$ Solve: $frac{n}{8} = 6$ What operation would you use to solve $frac{x}{9} = 11$? Why? Two-Step Equations Two-step equations require two operations to solve. We need to undo the operations in reverse order of the "order of operations" (PEMDAS/BODMAS). Solving Two-Step Equations: 1. First, undo addition/subtraction (farthest from the variable) 2. Then, undo multiplication/division (closest to the variable) Remember: Always do the same operation to both sides! Example 1: Solving $2x + 5 = 13$ Step-by-Step Solution: Equation: $2x + 5 = 13$ Step 1: Undo addition (subtract 5 from both sides): $2x + 5 - 5 = 13 - 5$ $2x = 8$ Step 2: Undo multiplication (divide both sides by 2): $frac{2x}{2} = frac{8}{2}$ $x = 4$ Check: $2(4) + 5 = 8 + 5 = 13$ ✓ Correct! Example 2: Solving $frac{n}{3} - 4 = 8$ Step-by-Step Solution: Equation: $frac{n}{3} - 4 = 8$ Step 1: Undo subtraction (add 4 to both sides): $frac{n}{3} - 4 + 4 = 8 + 4$ $frac{n}{3} = 12$ Step 2: Undo division (multiply both sides by 3): $frac{n}{3} times 3 = 12 times 3$ $n = 36$ Check: $frac{36}{3} - 4 = 12 - 4 = 8$ ✓ Correct! Common Mistake - Wrong Order: For $2x + 5 = 13$ Incorrect: Divide by 2 first → $x + 5 = 6.5$ (Wrong! Can't divide just the $2x$) Correct: Subtract 5 first, THEN divide by 2 Remember: Always undo addition/subtraction BEFORE multiplication/division! Application Practice Solve: $3x + 7 = 22$ Solve: $frac{a}{5} - 2 = 3$ Solve: $4y - 10 = 26$ Solve: $frac{m}{4} + 6 = 11$ Explain why we undo addition/subtraction before multiplication/division in two-step equations. Equations with Variables on Both Sides Sometimes variables appear on both sides of the equation. Our goal is to get all variables on one side and constants on the other. Strategy for Variables on Both Sides: 1. Move variables to one side (add/subtract variable terms) 2. Move constants to the other side (add/subtract numbers) 3. Solve the resulting one- or two-step equation Tip: Try to move the smaller variable term to avoid negative coefficients Example 1: Solving $3x + 2 = 2x + 9$ Step-by-Step Solution: Equation: $3x + 2 = 2x + 9$ Step 1: Move variables to left (subtract $2x$ from both sides): $3x - 2x + 2 = 2x - 2x + 9$ $x + 2 = 9$ Step 2: Move constants to right (subtract 2 from both sides): $x + 2 - 2 = 9 - 2$ $x = 7$ Check: Left: $3(7) + 2 = 21 + 2 = 23$, Right: $2(7) + 9 = 14 + 9 = 23$ ✓ Example 2: Solving $5n - 12 = 3n + 4$ Step-by-Step Solution: Equation: $5n - 12 = 3n + 4$ Step 1: Move variables (subtract $3n$ from both sides): $5n - 3n - 12 = 3n - 3n + 4$ $2n - 12 = 4$ Step 2: Move constants (add 12 to both sides): $2n - 12 + 12 = 4 + 12$ $2n = 16$ Step 3: Solve (divide both sides by 2): $frac{2n}{2} = frac{16}{2}$ $n = 8$ Check: Left: $5(8) - 12 = 40 - 12 = 28$, Right: $3(8) + 4 = 24 + 4 = 28$ ✓ Technique Practice Solve: $4x + 3 = 2x + 11$ Solve: $7y - 5 = 2y + 15$ Solve: $8n + 1 = 5n + 16$ Solve: $2a - 7 = a + 3$ When solving equations with variables on both sides, why do we move variable terms first? Checking Your Solutions Always check your solutions by substituting them back into the original equation. This verifies your work and helps catch mistakes. The Verification Process: 1. Write the original equation 2. Substitute your solution for the variable 3. Simplify both sides using order of operations 4. Check if both sides are equal 5. If equal, your solution is correct. If not, re-check your work. Example 1: Checking $x = 5$ for $3x - 4 = 11$ Verification: Original equation: $3x - 4 = 11$ Substitute $x = 5$: $3(5) - 4 = 11$ Simplify: $15 - 4 = 11$ $11 = 11$ ✓ Both sides are equal! Conclusion: $x = 5$ is the correct solution. Example 2: Finding an Error Suppose someone solved $2x + 8 = 20$ and got $x = 8$ Check their work: Original: $2x + 8 = 20$ Substitute $x = 8$: $2(8) + 8 = 20$ Simplify: $16 + 8 = 20$ $24 = 20$ ✗ This is false! Their solution $x = 8$ is incorrect. They should re-solve. Common Verification Errors: 1. Substituting into the wrong equation 2. Making calculation errors when checking 3. Forgetting order of operations when simplifying 4. Not writing the check clearly Tip: Always check in a separate space from your solving work Method Practice Check if $x = 4$ is a solution to $5x - 3 = 17$ Check if $n = 6$ is a solution to $2n + 7 = 19$ For the equation $frac{x}{3} + 5 = 9$, someone got $x = 12$. Check their answer. Explain why checking solutions is important. Create your own equation and check a correct and incorrect solution. Translating Words to Equations To solve word problems, we first need to translate English phrases into algebraic expressions and equations. Keywords tell us which operations to use. Word/Phrase Mathematical Meaning Example sum, plus, increased by, more than Addition (+) "5 more than x" → $x + 5$ difference, minus, decreased by, less than Subtraction (−) "8 less than y" → $y - 8$ product, times, multiplied by, of Multiplication (×) "twice a number" → $2n$ quotient, divided by, ratio of Division (÷) "a number divided by 4" → $frac{x}{4}$ is, equals, gives, results in Equals (=) "x plus 3 is 10" → $x + 3 = 10$ Example 1: Translating "Five more than twice a number is seventeen" Translation: Step 1: "a number" → let it be $x$ Step 2: "twice a number" → $2x$ Step 3: "five more than twice a number" → $2x + 5$ Step 4: "is seventeen" → $= 17$ Equation: $2x + 5 = 17$ Example 2: Translating "The quotient of a number and 3, decreased by 4, is 8" Translation: Step 1: "a number" → let it be $n$ Step 2: "quotient of a number and 3" → $frac{n}{3}$ Step 3: "decreased by 4" → $frac{n}{3} - 4$ Step 4: "is 8" → $= 8$ Equation: $frac{n}{3} - 4 = 8$ Skills Practice Translate: "Three times a number increased by 7 is 19" Translate: "Eight less than five times a number equals 22" Translate: "The sum of a number and 9, divided by 2, is 11" Translate: "Twice a number decreased by 5 equals the number plus 7" Create your own word phrase that translates to $4x - 3 = 25$ Word Problems - Number Problems Number problems involve finding unknown numbers based on relationships between them. These are often solved by letting a variable represent the unknown number. Steps for Solving Word Problems: 1. Read the problem carefully 2. Define a variable (let x = ...) 3. Write an equation based on the problem 4. Solve the equation 5. Check your answer in the original problem 6. Answer the question with a complete sentence Example 1: Consecutive Numbers The sum of three consecutive numbers is 54. What are the numbers? Solution: Step 1: Let $x$ = first number Step 2: Then $x+1$ = second number, $x+2$ = third number Step 3: Write equation: $x + (x+1) + (x+2) = 54$ Step 4: Solve: $3x + 3 = 54$ $3x = 51$ $x = 17$ Step 5: Check: $17 + 18 + 19 = 54$ ✓ Step 6: Answer: The numbers are 17, 18, and 19. Example 2: Number Relationships If 5 times a number is decreased by 12, the result is 43. Find the number. Solution: Step 1: Let $n$ = the number Step 2: "5 times a number" → $5n$ Step 3: "decreased by 12" → $5n - 12$ Step 4: "the result is 43" → $5n - 12 = 43$ Step 5: Solve: $5n = 55$ → $n = 11$ Step 6: Check: $5(11) - 12 = 55 - 12 = 43$ ✓ Step 7: Answer: The number is 11. Verification Practice The sum of a number and 15 is 32. Find the number. Twice a number decreased by 8 equals 20. What is the number? If 7 is added to 3 times a number, the result is 31. Find the number. The sum of two consecutive even numbers is 46. Find the numbers. Explain why defining a variable is an important first step in solving word problems. Real World Applications Equations help us solve practical problems in everyday life: shopping, measurements, age problems, and more. Example 1: Shopping Problem Sarah bought 3 notebooks and a $2 pen. She spent $14 total. How much did each notebook cost? Solution: Step 1: Let $n$ = cost of one notebook (in dollars) Step 2: Cost of 3 notebooks = $3n$ Step 3: Total cost = cost of notebooks + cost of pen Step 4: Equation: $3n + 2 = 14$ Step 5: Solve: $3n = 12$ → $n = 4$ Step 6: Check: $3(4) + 2 = 12 + 2 = 14$ ✓ Step 7: Answer: Each notebook costs $4. Example 2: Age Problem Mike is 5 years older than Lisa. In 3 years, the sum of their ages will be 31. How old is Lisa now? Solution: Step 1: Let $L$ = Lisa's age now Step 2: Mike's age now = $L + 5$ Step 3: In 3 years: Lisa will be $L + 3$, Mike will be $(L+5) + 3 = L + 8$ Step 4: Sum in 3 years: $(L+3) + (L+8) = 31$ Step 5: Solve: $2L + 11 = 31$ $2L = 20$ $L = 10$ Step 6: Check: Now: Lisa 10, Mike 15. In 3 years: Lisa 13, Mike 18. Sum: 13+18=31 ✓ Step 7: Answer: Lisa is 10 years old now. Example 3: Measurement Problem The length of a rectangle is 3 cm more than twice its width. The perimeter is 36 cm. Find the dimensions. Solution: Step 1: Let $w$ = width (in cm) Step 2: Length = $2w + 3$ Step 3: Perimeter formula: $P = 2l + 2w$ Step 4: Equation: $2(2w + 3) + 2w = 36$ Step 5: Solve: $4w + 6 + 2w = 36$ $6w + 6 = 36$ $6w = 30$ $w = 5$ Step 6: Length = $2(5) + 3 = 13$ Step 7: Check: Perimeter = $2(13) + 2(5) = 26 + 10 = 36$ ✓ Step 8: Answer: Width = 5 cm, Length = 13 cm. Application Practice 6 movie tickets cost $54. How much does one ticket cost? Tom is twice as old as his sister. In 5 years, the sum of their ages will be 34. How old is Tom now? A carpenter cuts a 72-inch board into two pieces. One piece is twice as long as the other. How long is each piece? After spending $15 on lunch, Maria has $27 left. How much money did she have originally? Create your own real-world word problem that can be solved with an equation. Practice Exercises Solve: $x + 9 = 17$ Solve: $5n = 65$ Solve: $frac{y}{4} = 7$ Solve: $3x - 8 = 16$ Solve: $frac{m}{5} + 2 = 9$ Solve: $4x + 3 = 2x + 15$ Translate and solve: "Seven less than four times a number is 21." Check if $x = 6$ is a solution to $2x + 5 = 19$ The sum of two consecutive integers is 47. Find the numbers. A bike rental costs $12 plus $3 per hour. If you paid $27, how many hours did you rent the bike? Solve: $7 - 2x = 15$ Solve: $5(x - 3) = 20$ The width of a rectangle is 8 cm less than its length. The perimeter is 64 cm. Find the dimensions. Explain the difference between an expression and an equation. Give an example of each. Create and solve a two-step equation word problem about saving money. Show/Hide Answers Exercise 1: $x + 9 = 17$ → $x = 17 - 9$ → $x = 8$ Exercise 2: $5n = 65$ → $n = 65 ÷ 5$ → $n = 13$ Exercise 3: $frac{y}{4} = 7$ → $y = 7 × 4$ → $y = 28$ Exercise 4: $3x - 8 = 16$ → $3x = 24$ → $x = 8$ Exercise 5: $frac{m}{5} + 2 = 9$ → $frac{m}{5} = 7$ → $m = 35$ Exercise 6: $4x + 3 = 2x + 15$ → $2x = 12$ → $x = 6$ Exercise 7: Let $n$ = the number. Equation: $4n - 7 = 21$ → $4n = 28$ → $n = 7$ Exercise 8: Check: $2(6) + 5 = 12 + 5 = 17$, but $17 ≠ 19$, so $x = 6$ is NOT a solution. Exercise 9: Let $x$ = first integer, $x+1$ = second integer. Equation: $x + (x+1) = 47$ → $2x + 1 = 47$ → $2x = 46$ → $x = 23$. Numbers: 23 and 24. Exercise 10: Let $h$ = hours rented. Equation: $12 + 3h = 27$ → $3h = 15$ → $h = 5$. Rented for 5 hours. Exercise 11: $7 - 2x = 15$ → $-2x = 8$ → $x = -4$ Exercise 12: $5(x - 3) = 20$ → $5x - 15 = 20$ → $5x = 35$ → $x = 7$ Exercise 13: Let $l$ = length, then $w = l - 8$. Perimeter: $2l + 2(l-8) = 64$ → $2l + 2l - 16 = 64$ → $4l = 80$ → $l = 20$. Width = $20 - 8 = 12$. Dimensions: 20 cm by 12 cm. Exercise 14: Expression: A mathematical phrase without an equals sign (e.g., $3x + 5$). Equation: A mathematical statement with an equals sign showing two expressions are equal (e.g., $3x + 5 = 17$). Exercise 15: Sample problem: Sarah saves $5 each week and already has $30 saved. How many weeks until she has $75? Solution: Let $w$ = weeks. Equation: $30 + 5w = 75$ → $5w = 45$ → $w = 9$. It will take 9 weeks. Conclusion/Recap Great work! You've learned how to solve various types of linear equations and apply them to word problems. Remember these key points: Equation Solving Strategies: 1. One-step equations: Use inverse operations (opposites) 2. Two-step equations: Undo addition/subtraction first, then multiplication/division 3. Variables on both sides: Move variables to one side first 4. Always check solutions by substitution 5. Word problems: Read carefully, define variables, write equations, solve, check, answer Equations are like puzzles—each step brings you closer to finding the missing piece (the variable's value). With practice, solving equations becomes faster and more intuitive. Look for equations in real life—they're everywhere! Clip It! Share your ANSWER in the Chat. Indicate TITLE e.g Linear Equation 1. .....2. e.t.c
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