How to Isolate the Variable t in an Equation

Isolating a variable in an equation is a fundamental skill in algebra that allows you to solve for that variable. In this guide, we’ll walk through the steps to isolate the variable $t$ in various types of equations. Let’s start with some basics and gradually move to more complex examples.

Basic Linear Equations

Step-by-Step Process

  1. Identify the equation: Let’s start with a simple linear equation, for example, $3t + 5 = 20$
  2. Move constants to the other side: Subtract 5 from both sides to get $3t = 15$
  3. Divide by the coefficient of $t$: Finally, divide both sides by 3 to isolate $t$, giving $t = 5$

Example

Consider the equation $2t – 4 = 10$

  1. Add 4 to both sides: $2t – 4 + 4 = 10 + 4$ which simplifies to $2t = 14$
  2. Divide both sides by 2: $t = frac{14}{2} = 7$

Equations with Fractions

Step-by-Step Process

  1. Identify the equation: Let’s consider $frac{t}{3} + 2 = 5$
  2. Move constants to the other side: Subtract 2 from both sides to get $frac{t}{3} = 3$
  3. Multiply by the denominator: Multiply both sides by 3 to isolate $t$, giving $t = 9$

Example

Consider the equation $frac{t}{4} – 1 = 3$

  1. Add 1 to both sides: $frac{t}{4} – 1 + 1 = 3 + 1$ which simplifies to $frac{t}{4} = 4$
  2. Multiply both sides by 4: $t = 4 times 4 = 16$

Equations with Multiple Variables

Step-by-Step Process

  1. Identify the equation: For example, $2t + 3x = 12$
  2. Move terms with other variables to the other side: Subtract $3x$ from both sides to get $2t = 12 – 3x$
  3. Divide by the coefficient of $t$: Divide both sides by 2 to isolate $t$, giving $t = frac{12 – 3x}{2}$

Example

Consider the equation $4t – 5y = 20$

  1. Add $5y$ to both sides: $4t – 5y + 5y = 20 + 5y$ which simplifies to $4t = 20 + 5y$
  2. Divide both sides by 4: $t = frac{20 + 5y}{4}$

Quadratic Equations

Step-by-Step Process

  1. Identify the equation: For example, $t^2 – 4t + 4 = 0$
  2. Factor the quadratic equation: This can be factored to $(t – 2)^2 = 0$
  3. Solve for $t$: Set the factor equal to zero, giving $t – 2 = 0$, so $t = 2$

Example

Consider the equation $t^2 – 6t + 9 = 0$

  1. Factor the quadratic equation: $(t – 3)^2 = 0$
  2. Solve for $t$: $t – 3 = 0$, so $t = 3$

Exponential Equations

Step-by-Step Process

  1. Identify the equation: For example, $5^t = 125$
  2. Rewrite the equation using logarithms: Take the logarithm of both sides to get $log(5^t) = log(125)$
  3. Use logarithm properties: Apply the power rule to get $t times log(5) = log(125)$
  4. Solve for $t$: Divide both sides by $log(5)$ to isolate $t$, giving $t = frac{log(125)}{log(5)}$. Since $125 = 5^3$, $t = 3$

Example

Consider the equation $2^t = 16$

  1. Rewrite the equation using logarithms: $log(2^t) = log(16)$
  2. Apply the power rule: $t times log(2) = log(16)$
  3. Solve for $t$: $t = frac{log(16)}{log(2)}$. Since $16 = 2^4$, $t = 4$

Logarithmic Equations

Step-by-Step Process

  1. Identify the equation: For example, $log(t) = 3$
  2. Rewrite the equation in exponential form: $t = 10^3$
  3. Solve for $t$: $t = 1000$

Example

Consider the equation $log(t) = 4$

  1. Rewrite the equation in exponential form: $t = 10^4$
  2. Solve for $t$: $t = 10000$

Conclusion

Isolating the variable $t$ involves understanding the type of equation you’re dealing with and applying the appropriate algebraic techniques. Whether it’s a linear, fractional, multiple variable, quadratic, exponential, or logarithmic equation, the key is to systematically apply inverse operations to isolate $t$

Citations

  1. 1. Khan Academy – Solving Equations
  2. 2. Purplemath – Solving Linear Equations
  3. 3. Math is Fun – Solving Equations

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