Why Do Simultaneous Equations Have the Same Y Value?

Simultaneous equations are a set of equations with multiple variables that share common solutions. These solutions are the points where the graphs of the equations intersect. One of the fundamental reasons why simultaneous equations have the same y value at their intersection points is because they represent the same point in the coordinate system. Let’s break this down with some examples and explanations.

Understanding Simultaneous Equations

Consider two linear equations in two variables, x and y:

  1. $2x + 3y = 6$
  2. $x – y = 2$

To find the solution to these simultaneous equations, we need to determine the values of x and y that satisfy both equations at the same time. This is done by finding the point where the two lines intersect on a graph.

Graphical Interpretation

When we graph these equations, each equation represents a line on the coordinate plane. The point where these lines intersect is the solution to the simultaneous equations. At this intersection point, both equations yield the same y value for the same x value. This is because the intersection point is common to both lines.

Example

Let’s solve the given equations graphically:

  1. $2x + 3y = 6$
  2. $x – y = 2$

  1. Graph the First Equation

    To graph $2x + 3y = 6$, we need to find two points that satisfy the equation. For example:

    • When $x = 0$, $2(0) + 3y = 6$ ⟹ $y = 2$
    • When $y = 0$, $2x + 3(0) = 6$ ⟹ $x = 3$

    Plotting these points (0, 2) and (3, 0) on the graph and drawing a line through them gives us the first line.

  1. Graph the Second Equation

    To graph $x – y = 2$, we also find two points that satisfy the equation. For example:

    • When $x = 0$, $0 – y = 2$ ⟹ $y = -2$
    • When $y = 0$, $x – 0 = 2$ ⟹ $x = 2$

    Plotting these points (0, -2) and (2, 0) on the graph and drawing a line through them gives us the second line.

  1. Find the Intersection Point

    The intersection point of the two lines is where they cross each other. In this case, the lines intersect at the point (3, 1). This means that the solution to the simultaneous equations is $x = 3$ and $y = 1$

Algebraic Solution

We can also solve simultaneous equations algebraically using methods such as substitution or elimination.

Substitution Method

  1. Solve one of the equations for one variable in terms of the other. Let’s solve $x – y = 2$ for $x$:

    $x = y + 2$

  2. Substitute this expression for $x$ in the other equation:

    $2(y + 2) + 3y = 6$

  3. Simplify and solve for $y$:

    $2y + 4 + 3y = 6$

    $5y + 4 = 6$

    $5y = 2$

    $y = frac{2}{5}$

  4. Substitute the value of $y$ back into the expression for $x$:

    $x = frac{2}{5} + 2$

    $x = frac{2}{5} + frac{10}{5}$

    $x = frac{12}{5}$

So, the solution to the simultaneous equations is $x = frac{12}{5}$ and $y = frac{2}{5}$

Elimination Method

  1. Multiply each equation by a suitable number so that the coefficients of one of the variables are the same:

    Multiply the second equation by 3:

    $3(x – y) = 3(2)$

    $3x – 3y = 6$

  2. Subtract the second equation from the first equation:

    $(2x + 3y) – (3x – 3y) = 6 – 6$

    $2x + 3y – 3x + 3y = 0$

    $-x + 6y = 0$

  3. Solve for $x$ or $y$:

    $-x + 6y = 0$

    $x = 6y$

  4. Substitute $x = 6y$ into one of the original equations:

    $2(6y) + 3y = 6$

    $12y + 3y = 6$

    $15y = 6$

    $y = frac{6}{15}$

    $y = frac{2}{5}$

  5. Substitute $y = frac{2}{5}$ back into $x = 6y$:

    $x = 6 times frac{2}{5}$

    $x = frac{12}{5}$

So, the solution to the simultaneous equations is $x = frac{12}{5}$ and $y = frac{2}{5}$

Conclusion

Simultaneous equations have the same y value at their intersection points because they represent the same point in the coordinate system. Whether solved graphically or algebraically, the solutions to simultaneous equations are the points where the equations’ graphs intersect, sharing both x and y values. Understanding this concept is crucial for solving problems involving multiple variables and equations.

Citations

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