How to Solve for a in Equations?

Solving for a variable in an equation is a fundamental skill in algebra. Today, we’ll focus on how to isolate and solve for the variable ‘a’ in different types of equations. We’ll go through linear equations, quadratic equations, and more complex scenarios. Let’s dive in!

Linear Equations

A linear equation is an equation of the form $ax + b = c$. To solve for ‘a’, we need to isolate it on one side of the equation.

Example 1: $ax + b = c$

  1. Subtract b from both sides:

    $ax + b – b = c – b$

    Simplifies to:

    $ax = c – b$

  2. Divide both sides by x:

    $a = frac{c – b}{x}$

Example 2: $3a + 5 = 20$

  1. Subtract 5 from both sides:

    $3a + 5 – 5 = 20 – 5$

    Simplifies to:

    $3a = 15$

  2. Divide both sides by 3:

    $a = frac{15}{3}$

    Simplifies to:

    $a = 5$

Quadratic Equations

A quadratic equation is an equation of the form $ax^2 + bx + c = 0$. Solving for ‘a’ involves a different approach.

Example 3: $2a^2 + 3a – 5 = 0$

  1. Use the quadratic formula: The quadratic formula is given by:

    $a = frac{-b , pm , sqrt{b^2 – 4ac}}{2a}$

    Here, we have $x^2$ instead of ‘a’, so let’s solve for ‘a’.

  2. Identify coefficients: In this equation, $a = 2$, $b = 3$, and $c = -5$

  3. Substitute into the quadratic formula:

    $a = frac{-3 , pm , sqrt{3^2 – 4(2)(-5)}}{2(2)}$

    Simplifies to:

    $a = frac{-3 , pm , sqrt{9 + 40}}{4}$

    $a = frac{-3 , pm , sqrt{49}}{4}$

    $a = frac{-3 , pm , 7}{4}$

    Therefore, we have two solutions:

    $a = frac{4}{4}$ and $a = frac{-10}{4}$

    Simplifies to:

    $a = 1$ and $a = -2.5$

Solving for ‘a’ in Systems of Equations

Sometimes, you’ll encounter systems of equations where you need to solve for ‘a’. Let’s look at an example.

Example 4: Solving a System of Equations

Given:

$2a + 3b = 6$

$4a – b = 5$

  1. Isolate one variable in one equation: Let’s isolate ‘b’ in the first equation.

    $3b = 6 – 2a$

    $b = frac{6 – 2a}{3}$

  2. Substitute into the second equation:

    $4a – frac{6 – 2a}{3} = 5$

  3. Clear the fraction by multiplying by 3:

    $12a – (6 – 2a) = 15$

    Simplifies to:

    $12a – 6 + 2a = 15$

    $14a – 6 = 15$

  4. Add 6 to both sides:

    $14a = 21$

  5. Divide by 14:

    $a = frac{21}{14}$

    Simplifies to:

    $a = 1.5$

Solving for ‘a’ in Exponential Equations

Exponential equations have the form $a^x = b$. Solving for ‘a’ requires logarithms.

Example 5: $a^3 = 27$

  1. Take the cube root of both sides:

    Simplifies to:

    $a = 3$

Example 6: $2^a = 8$

  1. Rewrite 8 as a power of 2:

    $2^a = 2^3$

  2. Since the bases are the same, the exponents must be equal:

    $a = 3$

Solving for ‘a’ in Logarithmic Equations

Logarithmic equations have the form $log_a(b) = c$. Solving for ‘a’ involves exponentiation.

Example 7: $log_a(8) = 3$

  1. Rewrite in exponential form:

    $a^3 = 8$

  2. Take the cube root of both sides:

    Simplifies to:

    $a = 2$

Conclusion

Solving for ‘a’ can vary depending on the type of equation you’re dealing with. Whether it’s a linear, quadratic, system of equations, exponential, or logarithmic equation, the key is to isolate ‘a’ using algebraic techniques and sometimes logarithms. Practice these methods, and you’ll become proficient in solving for ‘a’ in any equation!

Citations

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

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