Math balancing equations: Balancing Math Equations

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Solving Two Step Equations using the Balance Method

Solving Two Step Equations using the Balance Method —

April 5, 2023

An equation is when one expression, or term, is equal to another.  To solve an equation means to find the value of the variable (represented by a letter) that makes the two expressions equal.  There are two types of equations for secondary school mathematics, linear and none-linear.  In this blog I write about how I introduce solving two step equations using the balance method.

Solving Equations Involving One Step

The start of the lesson recaps solving one step equations.  At this stage it is important students understand y/3 means y divided by 3 and 7y means 7 multiplied by y.  Next, we discuss inverse operations, i.e., the opposite of dividing by 3 is multiplying by 3 and the opposite of multiplying by 7 is dividing by 7.   I will often use function machines to demonstrate this.

Solving Two Step Equations using the Balance Method

When solving two step equations using the balance method we arrange the terms, so the unknown is on one side of the scale and the numbers on the other.    Using scales as part of the working helps students to see how the equation remains balanced throughout this process.  Function machines help students identify the individual operations but, in my opinion, lack the perception of balance.

The terms move according to the inverse of the order of operations.  For instance, with 6x + 2 = 26, the addition of 2 is moved to the other side before the multiplication of 6.  The equation remains balanced by doing the same operation to both sides of the scale (or equals sign).

Developing Algebraic Reasoning

As we progress through the lesson I consolidate and extend the learning by increasing the level of challenge.  I do this by:

  • writing the equation on the right side of the equals sign rather than the left;
  • including divisions as fractions;
  • requiring students to simplify an equation before they attempt to solve it;
  • making the unknown term negative, eg., 0 =6 – 2x.

These mini-extensions all help to develop student’s algebraic fluency.

Solving Two Step Equations Plenary

In the plenary students match the equation to its solution.  This involves attempting a variety of questions across the ability range.   This activity takes about 10 minutes with students working independently on mini-whiteboards.   I assess student’s progress against these differentiated learning objectives:

  • All students should be able solve equations involving a multiplication and addition.
  • Most students should be able to solve equations involving divisions and multiplications.
  • Some students should be able to solve equations where the unknown term is presented as a negative.

As we progress through the scheme of work students learn how to solve equations involving brackets and begin to form their own equations from known geometrical properties.

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Chemical reactions are represented by balanced chemical equations, which show the molecules that are reacting on the left side and the molecules that are produced on the right side. We have to make sure that the reactant and product sides have identical amounts of atoms – for example, if we react molecules containing 2 atoms of hydrogen then we must produce molecules that have 2 atoms of hydrogen as well. This is known as the law of conservation of mass: matter cannot be created or destroyed.

As an example, let’s look at the unbalanced reaction of water (H2O) with sodium (Na):

There is a problem, because the left side has 2 hydrogen atoms but the right side has 3. To fix this, we can change the coefficients that come before the molecules. We can’t change the subscripts (for example, we can’t change the 2 in H2O). The correct balanced equation looks like this:

We had to add three new coefficients to balance the equation. Now we have 4 hydrogens on the left and 4 hydrogens on the right. Don’t forget that when you are counting atoms that you must multiply the molecule’s coefficient with the atom’s subscript – in this case, 2 H2O has 4 H total because each individual H2O molecule has 2 H.

Basic steps for balancing equations

  1. Count all the elements on both sides of the reaction. Note which elements are currently unbalanced.
  2. Start by balancing elements that only appear in one molecule on each side. Do this by changing the coefficients, not the subscripts!
  3. Double check the balance and note the new count of each element on both sides of the equation.
  4. Balance the remaining elements.
  5. Check that everything is balanced! Make sure you are taking into account the coefficients and the subscripts.

Tip: You can use fractions to help get the right ratio of atoms, but you must convert these to whole numbers before you finish balancing. You can eliminate fractions by multiplying the entire reaction by the denominator.

e.g. Let’s try balancing this equation using the above steps:

First: we count that the left side has 2 C, 6 H, and 2 O, while the right side has 1 C, 2 H, and 3 O. All three elements are unbalanced.

Second: we note that oxygen appears in two molecules on the right side, which makes it a bit trickier to balance, so we will save it for later. To balance the C and H atoms, we can change the coefficients of CO2 and H2O on the right side.

Third: we count our atoms again. The left side has 2 C, 6 H, and 2 O. The right side has 2 C, 6 H, and 7 O. The only element that is not balanced is O.

Fourth: we balance the oxygens. This is where the above tip about using fractions becomes useful, because we have 2 O on the left and 7 O on the right. We cannot multiply O2 by a whole number to get 7, but we can multiply it by a fraction to get 7.

Now we have 7 oxygens on each side. However, we need all the coefficients to be whole numbers. We can fix this by multiplying the entire equation by the denominator (2) to get rid of the fraction:

Fifth: double check that all the elements have the same count on both sides. The left side and right side both have 4 C, 12 H, and 14 O, so we’re finished!





The main groups of equations included in the mathematical description of the process

part of the mathematical description,
developed on the basis of physical
the nature of the object under study,
select the following groups of equations:

1. Equations
mass and energy balances recorded with
taking into account the hydrodynamic structure
flow movements. This group of equations
characterizes the distribution in flows
temperatures, compositions and related
properties: density, heat capacity, etc.

2. Equations
process kinetics. These include
descriptions of elementary processes
the course of a chemical reaction,
heat transfer, mass transfer. For the record
equations are used hypotheses
additional links between the main
process variables.

3. Theoretical,
empirical or semi-empirical
relationships between different parameters
process — equations for physical
or hydrodynamic parameters:
heat capacity, viscosity, coefficients
heat transfer, etc.

4. Restrictions
on process parameters, for example,
temperature in the catalyst zone
may be higher than some given
otherwise the catalyst will lose its properties

Group data
equations are part of the mathematical
models of any chemical-technological
process. They represent analytical
theory of the study of physical and chemical
processes. In addition, the model is built
so that it best reflects
the nature of the behavior of the flows of matter and
energy at the same time enough
simple mathematical description.
Simplicity of mathematical description
determined by the possibility of solving
computer models.

relation of mathematical description
processes in the apparatus is a common
material balance drawn up for
individual flow components.

If concentrations
substances are not the same at different points
apparatus, then the equation of the material
balance make up for elemental
volume. For the device volume element
material balance is recorded in
according to the law of conservation

substances entering the elementary
volume for a given period of time


substances emerging from the elemental
volume over a period of time


substances accumulated in
considered elementary volume
for a given period of time

the law of conservation of mass can be represented
in the form of the following ratio


where M in ,
M out — mass
flows (kg/s) at inlet and outlet
considered elementary volume,
t —
elementary interval of time (time
observations), M
is the accumulation of mass (kg) in the considered
volume in time t.

If you can accept
that the concentrations of substances at each point
volume of the apparatus are the same (good
mixing), equation (1) is related to
throughout the apparatus. In this case
the equation for the law of conservation of mass can be
present in differential form.
To do this, we divide the equation by and
let’s move on to the aisle at

Given that


those. speed
accumulation of the mass of matter in some
volume is equal to the difference of mass flows
(kg/s) at the inlet and outlet of this volume.

When compiling
material balance equations
researched chemical-technological
process use either equation (1) —
receive a mathematical description in
form of differential equations in
partial derivatives, or equation (2)
— receive ordinary differential
equations. It should also be noted
that equation (2) is a special case
equations (1).

heat balance is compiled using
law of energy conservation. For elementary
volume, the conservation law is written in
increments, i.e.


If you can accept
that temperature and related
flow properties at each point in total
the volume of the apparatus are the same, then the law
saves are recorded for everything
considered volume in the differential


Here Q
(J) — accumulation of energy in the elementary
volume; Q in , Q out
(J/s) — heat fluxes at the inlet and outlet
considered volume,

— the rate of energy accumulation in the studied

Equations (3.4)
used in the preparation of thermal
balances of chemical and technological
processes. Just like equation (1),
relation (3) leads to the mathematical
description in the form of differential
partial differential equations, and
equation (4) — to ordinary
differential equations. Type
used mathematical apparatus
determines the structure of the future mathematical
models. To establish structure
models need to identify features
hydrodynamic structure of flows,
which manifests itself in the character
particle residence time distribution
flow in the apparatus, since the movement
particles determines the transfer conditions
energy and mass.

In real
continuous system due to the stochastic
the nature of the movement of its particles at the microlevel
there is always an imbalance
time distribution of particles
stay in the device. To the most
significant sources of unevenness
distribution of flow elements by
residence time in industrial
devices include: uneven
velocity profile, flow turbulence,
molecular diffusion, the presence of stagnant
areas in flows, bypass and
cross currents, temperature
moving media gradients, etc.
The listed reasons exist in
technological apparatus and operating
in various combinations, cause
specific nature of unevenness
in each specific case.

For rate
of non-uniformity of flows, a series is introduced
statistical distribution functions,
since the nature of the distribution of particles
subject to statistical laws.
The distribution functions are found by
type of signal passing through the system.
This signal is used
supply of substance — indicator to the input
apparatus in the form of perturbation. Based
statistical processing of experimental
indicator distribution data by
the machine is set to function

statistical distribution functions
for devices of various designs
receive various models of hydrodynamic
particle distribution patterns that
and are used to compile
material and heat balances
process, i.e. constructing a mathematical

«Material balance equation», Mathematics, chemistry, physics

Abstract Help in writing Find out the cost of my work tana С7Н16 + 11О2> 7СО2 + 8Н2О Here, according to the law conservation of mass, the number of atoms of each element in the left and right sides of the chemical equation must be the same. These written equations of chemical reactions of combustion show only the initial and final state of the system, i.e., which initial substances enter into the reaction and which products … Read more >


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When substances burn in air, a rapid chemical reaction occurs between the fuel substance and oxygen in the air with intense heat release. As a result of this reaction, mainly oxides of the elements that make up the combustible substance are formed. During the combustion of hydrocarbons consisting of C and H atoms, CO 9 is formed0066 2 (carbon dioxide) and H 2 O (water). With incomplete combustion, along with them, CO (carbon monoxide) and C (solid carbon) are formed in the form of soot. When writing the combustion equation and compiling the material balance, as a rule, only the products of complete oxidation of CO 2 and H 2 O are recorded. If there is sulfur (S) in the composition of the combustible substance, then SO 2 is formed during combustion. Nitrogen (N), which is part of a combustible substance, does not oxidize when burning in air, but is released in the form of free nitrogen (N 2 ). This is explained by the fact that during combustion in air, the combustion temperature is relatively low (1500–2000 K) and at this temperature nitrogen oxides are not formed. Examples of combustion reactions:

combustion of methane CH 4 + 2O 2 > CO 2 + 2H 2 O combustion of p-heptane C 7 H 16 + 11О 2 >7СО 2 + 8H 2 O Here, according to the law of conservation of mass, the number of atoms of each element in the left and right sides of the chemical equation must be the same.

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