How to Determine the Value of x?

Determining the value of x is a fundamental skill in algebra and is crucial for solving equations. Let’s explore some common methods to find x.

Solving Linear Equations

Example 1: Simple Linear Equation

Consider the equation $2x + 3 = 11$

  1. Isolate the variable: Subtract 3 from both sides:
    $2x + 3 – 3 = 11 – 3$
    $2x = 8$
  2. Solve for x: Divide both sides by 2:
    $x = frac{8}{2}$
    $x = 4$

Example 2: Linear Equation with Fractions

Let’s solve $frac{3x}{4} – 2 = 1$

  1. Eliminate the fraction: Multiply both sides by 4 to get rid of the denominator:
    $4 times frac{3x}{4} – 4 times 2 = 4 times 1$
    $3x – 8 = 4$
  2. Isolate the variable: Add 8 to both sides:
    $3x – 8 + 8 = 4 + 8$
    $3x = 12$
  3. Solve for x: Divide both sides by 3:
    $x = frac{12}{3}$
    $x = 4$

Solving Quadratic Equations

Example 3: Factoring Method

Consider the equation $x^2 – 5x + 6 = 0$

  1. Factor the quadratic: Find two numbers that multiply to 6 and add to -5. These numbers are -2 and -3:
    $(x – 2)(x – 3) = 0$
  2. Set each factor to zero:
    $x – 2 = 0$ or $x – 3 = 0$
  3. Solve for x:
    $x = 2$ or $x = 3$

Example 4: Quadratic Formula

Consider the equation $x^2 + 4x + 4 = 0$

  1. Identify coefficients: Here, $a = 1$, $b = 4$, and $c = 4$
  2. Apply the quadratic formula:
    $x = frac{-b , pm , sqrt{b^2 – 4ac}}{2a}$
    Plugging in the values:
    $x = frac{-4 , pm , sqrt{4^2 – 4 cdot 1 cdot 4}}{2 cdot 1}$
    $x = frac{-4 , pm , sqrt{16 – 16}}{2}$
    $x = frac{-4 , pm , 0}{2}$
    $x = frac{-4}{2}$
    $x = -2$

Solving Systems of Equations

Example 5: Substitution Method

Consider the system:
$begin{cases}
2x + y = 10
3x – y = 5
end{cases}$

  1. Solve one equation for one variable: From the first equation, solve for y:
    $y = 10 – 2x$
  2. Substitute into the second equation:
    $3x – (10 – 2x) = 5$
    $3x – 10 + 2x = 5$
    $5x – 10 = 5$
  3. Solve for x:
    $5x = 15$
    $x = 3$
  4. Find y: Substitute $x = 3$ back into $y = 10 – 2x$:
    $y = 10 – 2(3)$
    $y = 4$

Conclusion

By mastering these methods, you can solve for x in various types of equations. Practice makes perfect, so keep working on different problems to strengthen your skills!

Citations

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