Solving for

When solving for $x – y$ in a system of equations, we typically deal with two linear equations involving two variables, $x$ and $y$. Let’s go through the process step-by-step with examples to make it clear.

Step-by-Step Guide

  1. Write Down the Equations

    Consider the following system of linear equations:

    1. $2x + 3y = 12$
    2. $4x – y = 5$

  1. Align the Equations

    To make our calculations easier, let’s write the equations one below the other:

    1. $2x + 3y = 12$
    2. $4x – y = 5$

  1. Eliminate One Variable

    To find $x – y$, we need to eliminate one of the variables. This can be done using the method of elimination or substitution. Let’s use elimination here.

    First, we need to make the coefficients of $y$ in both equations equal. To do this, we can multiply the second equation by 3:

    1. $2x + 3y = 12$
    2. $3(4x – y) = 3(5)$

    This simplifies to:

    1. $2x + 3y = 12$
    2. $12x – 3y = 15$

  1. Add or Subtract the Equations

    Next, we add the two equations to eliminate $y$:

    $(2x + 3y) + (12x – 3y) = 12 + 15$

    This simplifies to:

    $14x = 27$

  1. Solve for $x$

    Now, we can solve for $x$:

    $x = frac{27}{14}$

  1. Substitute $x$ Back into One of the Original Equations

    We substitute $x = frac{27}{14}$ back into one of the original equations to find $y$. Let’s use the first equation:

    $2bigg(frac{27}{14}bigg) + 3y = 12$

    Simplify and solve for $y$:

    $frac{54}{14} + 3y = 12$

    $3.857 + 3y = 12$

    $3y = 12 – 3.857$

    $3y = 8.143$

    $y = frac{8.143}{3}$

    $y = 2.714$

  1. Calculate $x – y$

    Now that we have $x$ and $y$, we can find $x – y$:

    $x – y = frac{27}{14} – 2.714$

    $x – y = 1.929 – 2.714$

    $x – y = -0.785$

    So, the value of $x – y$ is approximately $-0.785$

Another Example

Let’s consider another system of equations:

  1. $3x + 4y = 20$
  2. $5x – 2y = 10$

  1. Write Down the Equations

    1. $3x + 4y = 20$
    2. $5x – 2y = 10$

  1. Align the Equations

    1. $3x + 4y = 20$
    2. $5x – 2y = 10$

  1. Eliminate One Variable

    To eliminate $y$, we can multiply the second equation by 2:

    1. $3x + 4y = 20$
    2. $2(5x – 2y) = 2(10)$

    This simplifies to:

    1. $3x + 4y = 20$
    2. $10x – 4y = 20$

  1. Add or Subtract the Equations

    Adding the two equations to eliminate $y$:

    $(3x + 4y) + (10x – 4y) = 20 + 20$

    This simplifies to:

    $13x = 40$

  1. Solve for $x$

    $x = frac{40}{13}$

  1. Substitute $x$ Back into One of the Original Equations

    Substitute $x = frac{40}{13}$ back into the first equation:

    $3bigg(frac{40}{13}bigg) + 4y = 20$

    Simplify and solve for $y$:

    $frac{120}{13} + 4y = 20$

    $9.231 + 4y = 20$

    $4y = 20 – 9.231$

    $4y = 10.769$

    $y = frac{10.769}{4}$

    $y = 2.692$

  1. Calculate $x – y$

    $x – y = frac{40}{13} – 2.692$

    $x – y = 3.077 – 2.692$

    $x – y = 0.385$

    So, the value of $x – y$ is approximately $0.385$

Conclusion

Solving for $x – y$ involves a systematic approach of eliminating one variable, solving for the other, and then substituting back. This method ensures that we find accurate values for both $x$ and $y$, and subsequently $x – y$. Practice with different sets of equations can help you become more comfortable with these steps.

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