Solving with inverse operations in N, Z and Q, checking, and word problems
Mathematics Prep 1 Term 1, Unit 2, Lesson 2
GoalSolve one- and two-step equations and equations with brackets or x on both sides, then check.
New words
equivalent equations · equations with the same solution setinverse operation · the operation that undoes another, like − for +both sides · the left and right of the = signcheck · put the answer into the original equation
Explanation
An equation is like a balance scale. If we do the same thing to both sides, it stays balanced and the new equation is equivalent. We may add the same number to both sides, subtract it, multiply both sides by it, or divide both sides by it (not by 0). Undo the operations in the opposite order: for 3x + 5 = 20, subtract 5 first, then divide by 3. With brackets, divide by the number outside or expand. With x on both sides, move the x terms to one side first. At the end, check in the original equation. If the answer is not in the substitution set, the solution set is ∅.
Worked examples
Solve 3x + 5 = 20. Subtract 5 from both sides: 3x = 15. Divide both sides by 3: x = 5. Check: 3 × 5 + 5 = 20, which is true.
Brackets and x on both sides. Solve 4(x − 2) = 20: divide by 4 to get x − 2 = 5, then add 2: x = 7. Solve 5x − 3 = 2x + 9: subtract 2x, 3x − 3 = 9; add 3, 3x = 12; divide, x = 4. Check: 5 × 4 − 3 = 17 and 2 × 4 + 9 = 17.
Word problem. A ribbon 72 cm long goes round a card whose length is 8 cm more than its width x. 2(x + x + 8) = 72, so x + x + 8 = 36, 2x = 28, x = 14. The sides are 14 cm and 22 cm. Check: 2 × (14 + 22) = 72. Also 3x = 7 has the solution set ∅ in Z but {7/3} in Q.
Class activity
Balance play. With a grown-up, use a hand balance or a see-saw drawing. Put 3 equal bags and a 5 g weight on one side and a 20 g weight on the other. What does each bag weigh? Write the equation and solve it.
Practice
1
Solve 2x + 3 = 15.
2x = 12, so x = 6. Check: 12 + 3 = 15.
2
Solve 7x − 2 = 3x + 10.
4x − 2 = 10, so 4x = 12 and x = 3. Check: 21 − 2 = 19 and 9 + 10 = 19.
3
A taxi charges 6 LE to start and 4 LE per km. A ride costs 30 LE. How many km?
4x + 6 = 30, 4x = 24, x = 6 km.
4
Solve 5x − 4 = 21 in N.
5x = 25, x = 5, and 5 ∈ N. The solution set is {5}.
5
Omar solves 2x + 6 = 14 and writes x + 6 = 7. What is wrong? Solve it.
He divided only one term. Subtract 6 first: 2x = 8, so x = 4. Check: 8 + 6 = 14.