How to Express Ages Algebraically?

Understanding how to express ages algebraically is a useful skill, especially when solving word problems in math. Let’s break it down step by step.

Basic Concepts

Variables

In algebra, we use variables to represent unknown values. For example, if we don’t know someone’s age, we can use a variable like $x$ to represent it.

Relationships Between Ages

Often, problems will describe relationships between different ages. For example, if John is 5 years older than Mary, and Mary’s age is represented by $m$, then John’s age can be expressed as $m + 5$

Example Problem

Let’s say you are given a problem:
“Alice is twice as old as Bob. In 5 years, Alice will be 3 times as old as Bob. How old are they now?”

Step-by-Step Solution

  1. Define the Variables: Let $a$ represent Alice’s current age and $b$ represent Bob’s current age.
  2. Set Up Equations Based on the Problem:
    • Alice is twice as old as Bob: $a = 2b$
    • In 5 years, Alice will be 3 times as old as Bob: $a + 5 = 3(b + 5)$
  3. Solve the Equations:
    • Substitute $a = 2b$ into the second equation: $2b + 5 = 3(b + 5)$
    • Distribute and simplify: $2b + 5 = 3b + 15$
    • Solve for $b$: $2b + 5 – 3b = 15$
    • $-b + 5 = 15$
    • $-b = 10$
    • $b = -10$
    • Since age can’t be negative, we need to re-evaluate our setup. In this case, there’s an error in the problem statement or our interpretation.

Checking for Errors

To avoid confusion, always recheck your problem setup and ensure your equations are correct. Sometimes, problems might have typos or missing information.

Practice Problem

“Sarah is 4 years younger than Tom. In 2 years, Tom will be twice as old as Sarah. How old are they now?”

Solution

  1. Define Variables: Let $t$ represent Tom’s age and $s$ represent Sarah’s age.
  2. Set Up Equations:
    • Sarah is 4 years younger than Tom: $s = t – 4$
    • In 2 years, Tom will be twice as old as Sarah: $t + 2 = 2(s + 2)$
  3. Solve the Equations:
    • Substitute $s = t – 4$: $t + 2 = 2((t – 4) + 2)$
    • Simplify: $t + 2 = 2(t – 2)$
    • Distribute: $t + 2 = 2t – 4$
    • Solve for $t$: $t + 2 – 2t = -4$
    • $-t + 2 = -4$
    • $-t = -6$
    • $t = 6$
    • Find $s$: $s = t – 4 = 6 – 4 = 2$

So, Tom is 6 years old and Sarah is 2 years old.

Conclusion

Expressing ages algebraically involves defining variables and setting up equations based on the relationships described in the problem. Practice is key to mastering this skill, so try solving different types of age-related problems to get comfortable with the process.

Citations

  1. 1. Purplemath – Solving Age Problems
  2. 2. Khan Academy – Algebraic Expressions
  3. 3. Math is Fun – Algebra Basics

Related

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H + HO2 → O2 + H2 k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O2 k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) H + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-5 s^-1) φ

Table 1 Reactions, rate constants and activation energies used in the model* No. Reaction kopt (M⁻¹ s⁻¹) 1 OH + H₂ → H + H₂O 3.74 x 10⁷ 2 OH + HO₂ → HO₂ + OH⁻ 5 x 10⁹ 3 OH + H₂O₂ → HO₂ + H₂O 3.8 x 10⁷ 4 OH + O₂ → O₂ + OH 9.96 x 10⁹ 5 OH + HO₂ → O₂ + H₂O 7.1 x 10⁹ 6 OH + OH → H₂O₂ 5.3 x 10⁹ 7 OH + e⁻aq → OH⁻ 3 x 10¹⁰ 8 H + O₂ → HO₂ 2.0 x 10¹⁰ 9 H + HO₂ → H₂O₂ 2.0 x 10¹⁰ 10 H + H₂O₂ → OH + H₂O 3.44 x 10⁷ 11 H + OH → H₂O 1.4 x 10¹⁰ 12 H + H → H₂ 1.94 x 10¹⁰ 13 e⁻aq + O₂ → O₂⁻ 1.9 x 10¹⁰ 14 e⁻aq + O₂ → HO₂⁻ + OH⁻ 1.3 x 10¹⁰ 15 e⁻aq + HO₂ 2.0 x 10¹⁰ 16 e⁻aq + H₂O₂ 1.1 x 10¹⁰ 17 e⁻aq + HO₂ → OH + OH⁻ 1.3 x 10¹⁰ 18 e⁻aq + H⁺ → H 2.3 x 10¹⁰ 19 e⁻aq + e⁻aq → H₂ + OH⁻ + OH⁻ 2.5 x 10⁹ 20 HO₂ + O₂ → O₂ + HO₂ 1.3 x 10⁹ 21 HO₂ + HO₂ → O₂ + H₂O₂ 8.3 x 10⁵ 22 HO₂ + HO₂ → O₂ + OH + H₂O 3.7 23 HO₂ + HO₂ → O₂ + O₂ + OH + H₂O 7 x 10⁵ s⁻¹ 24 H⁺ + O₂⁻ → HO₂ 4.5 x 10¹⁰ 25 H⁺ + O₂⁻ → O₂ 2.0 x 10¹⁰ 26 H⁺ + OH⁻ 1.4 x 10¹¹ 27 H⁺ + HO₂⁻ 2 x 10¹⁰ 28 H₂O₂ → HO₂ + H⁺ + OH⁻ 2.5 x 10⁻⁵ s⁻¹ 29 H₂O₂ → H⁺ + OH⁻ 1.4 x 10⁻⁷ s⁻¹ 30 O₂ + O₂ → O₂ + HO₂ + OH⁻ 0.3 31 O₂ + H₂O₂ → O₂ + OH + OH 16 32

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H2O + O → 2 OH k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) OH + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-8 s^-1) φ