How to Find the Value of q?

Finding the value of a variable, such as $q$, is a fundamental skill in algebra and other areas of mathematics. Whether you’re solving linear equations, quadratic equations, or working with inequalities, the process generally involves isolating the variable on one side of the equation. Let’s explore different methods to find the value of $q$

Solving Linear Equations

Example Problem

Consider the linear equation:
$3q + 5 = 20$

Step-by-Step Solution

  1. Isolate the variable term: Subtract 5 from both sides of the equation.
    $3q + 5 – 5 = 20 – 5$
    $3q = 15$
  2. Solve for $q$: Divide both sides by 3.
    $q = frac{15}{3}$
    $q = 5$

Another Example

Let’s solve another linear equation:
$2q – 7 = 13$

  1. Isolate the variable term: Add 7 to both sides.
    $2q – 7 + 7 = 13 + 7$
    $2q = 20$
  2. Solve for $q$: Divide both sides by 2.
    $q = frac{20}{2}$
    $q = 10$

Solving Quadratic Equations

Quadratic equations are of the form $ax^2 + bx + c = 0$. To find the value of $q$ in a quadratic equation, you can use the quadratic formula:

$q = frac{-b , text{±} , sqrt{b^2 – 4ac}}{2a}$

Example Problem

Consider the quadratic equation:
$q^2 – 4q + 4 = 0$

Step-by-Step Solution

  1. Identify coefficients: Here, $a = 1$, $b = -4$, and $c = 4$
  2. Apply the quadratic formula:
    $q = frac{-(-4) , text{±} , sqrt{(-4)^2 – 4 cdot 1 cdot 4}}{2 cdot 1}$
    $q = frac{4 , text{±} , sqrt{16 – 16}}{2}$
    $q = frac{4 , text{±} , 0}{2}$
    $q = frac{4}{2}$
    $q = 2$

Another Example

Let’s solve another quadratic equation:
$2q^2 + 3q – 2 = 0$

  1. Identify coefficients: Here, $a = 2$, $b = 3$, and $c = -2$

  2. Apply the quadratic formula:
    $q = frac{-3 , text{±} , sqrt{3^2 – 4 cdot 2 cdot (-2)}}{2 cdot 2}$
    $q = frac{-3 , text{±} , sqrt{9 + 16}}{4}$
    $q = frac{-3 , text{±} , sqrt{25}}{4}$
    $q = frac{-3 , text{±} , 5}{4}$

    This gives us two solutions:
    $q = frac{2}{4} = 0.5$
    $q = frac{-8}{4} = -2$

Solving Inequalities

When solving inequalities, the goal is similar to solving equations, but you must be mindful of the direction of the inequality sign.

Example Problem

Consider the inequality:
$4q – 3 < 9$

Step-by-Step Solution

  1. Isolate the variable term: Add 3 to both sides.
    $4q – 3 + 3 < 9 + 3$
    $4q < 12$
  2. Solve for $q$: Divide both sides by 4.
    $q < frac{12}{4}$
    $q < 3$

Another Example

Let’s solve another inequality:
$-2q + 5 leq 1$

  1. Isolate the variable term: Subtract 5 from both sides.
    $-2q + 5 – 5 leq 1 – 5$
    $-2q leq -4$
  2. Solve for $q$: Divide both sides by -2 (remember to flip the inequality sign).
    $q geq 2$

Systems of Equations

Sometimes you’ll need to find the value of $q$ in a system of equations. This can be done using substitution or elimination methods.

Example Problem

Consider the system of equations:
$2q + r = 10$
$q – r = 2$

Step-by-Step Solution

  1. Solve one equation for one variable: From the second equation, solve for $r$
    $r = q – 2$
  2. Substitute into the other equation:
    $2q + (q – 2) = 10$
    $3q – 2 = 10$
  3. Solve for $q$:
    $3q = 12$
    $q = 4$

Another Example

Let’s solve another system of equations:
$3q + 2r = 14$
$q + 4r = 10$

  1. Solve one equation for one variable: From the second equation, solve for $q$
    $q = 10 – 4r$
  2. Substitute into the other equation:
    $3(10 – 4r) + 2r = 14$
    $30 – 12r + 2r = 14$
    $30 – 10r = 14$
  3. Solve for $r$:
    $-10r = -16$
    $r = 1.6$
  4. Substitute back to find $q$:
    $q = 10 – 4(1.6)$
    $q = 10 – 6.4$
    $q = 3.6$

Conclusion

Finding the value of $q$ involves various methods depending on the type of equation or inequality. Understanding these basic principles and practicing different types of problems will help you become proficient in solving for any variable. Remember, the key is to isolate the variable and perform operations systematically.

Citations

  1. 1. Khan Academy – Solving Equations
  2. 2. Purplemath – Solving Quadratic Equations
  3. 3. Math is Fun – Solving Inequalities
  4. 4. Paul’s Online Math Notes – Systems of 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