What is the value of x+y-2z?

To determine the value of the expression $x + y – 2z$, we need to know the values of $x$, $y$, and $z$. Without specific values or additional context, we can’t calculate an exact numerical answer. However, let’s explore how we can approach solving such an expression in different scenarios.

Scenario 1: Given Values

If we have specific values for $x$, $y$, and $z$, we can simply substitute them into the expression and perform the arithmetic operations.

Example

Suppose $x = 3$, $y = 5$, and $z = 2$

Substitute these values into the expression:

$x + y – 2z = 3 + 5 – 2(2)$

First, multiply $2$ by $2$:

$3 + 5 – 4$

Then, add $3$ and $5$:

$8 – 4$

Finally, subtract $4$ from $8$:

$4$

So, the value of $x + y – 2z$ is $4$ when $x = 3$, $y = 5$, and $z = 2$

Scenario 2: Solving Equations

Sometimes, we might be given equations that relate $x$, $y$, and $z$. In such cases, we can solve these equations to find the values of $x$, $y$, and $z$, and then substitute them into the expression.

Example

Suppose we have the following system of equations:

$x + y = 10$

$x – z = 3$

$2y + z = 12$

We can solve these equations step by step.

First, solve the first equation for $y$:

$y = 10 – x$

Next, substitute $y = 10 – x$ into the third equation:

$2(10 – x) + z = 12$

Simplify and solve for $z$:

$20 – 2x + z = 12$

$z = 12 – 20 + 2x$

$z = 2x – 8$

Now, substitute $z = 2x – 8$ into the second equation:

$x – (2x – 8) = 3$

Simplify and solve for $x$:

$x – 2x + 8 = 3$

$-x + 8 = 3$

$-x = 3 – 8$

$-x = -5$

$x = 5$

Now, substitute $x = 5$ back into the first and third equations to find $y$ and $z$:

$y = 10 – 5 = 5$

$z = 2(5) – 8 = 10 – 8 = 2$

So, $x = 5$, $y = 5$, and $z = 2$. Substitute these values into the expression $x + y – 2z$:

$5 + 5 – 2(2) = 5 + 5 – 4 = 10 – 4 = 6$

Thus, the value of $x + y – 2z$ is $6$ when $x = 5$, $y = 5$, and $z = 2$

Conclusion

The value of the expression $x + y – 2z$ depends on the values of $x$, $y$, and $z$. By substituting known values or solving related equations, we can determine the specific value of the expression in different scenarios. Always remember to follow the order of operations: parentheses, exponents, multiplication and division (from left to right), and addition and subtraction (from left to right).

Citations

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

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