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Reducing Linear Equations To Simpler Form: Certain mathematical equations are not in the form of linear equations, but they can be made into linear equations by using particular mathematical operations. Reducing an equation to a simpler form is known as reduction. It is the process of simplifying them and converting nonlinear ones to linear ones.
These equations can be solved, and the unknown value can be estimated after being reduced to linear form. Reducing such equations will ease the calculations and reduces the time required to solve them. Performing the simplification includes various mathematical approaches, and they vary as per the need of a particular equation.
An equation is any mathematical statement with an equality sign \(\left( = \right)\) between two algebraic expressions. The equations of degree one are known as linear equations. It’s the formula for a straight line. The answers to linear equations produce values that make the equation true when substituted for unknown values. There is just one solution in the case of a single variable equation.
A linear equation is an algebraic expression in which each term is either the product of variables and constants or just a constant. It contains the variables’ first-order power. The common representation for such equations is \(ax + b = 0,\) where \(a\) and \(b\) are constants and \(a \ne 0\).
The linear equations can be of the following types:
Some linear equations may have nonlinear representations in addition to this. To make the computation process easier, they must be reduced to linear form. Solving such equations is difficult as it necessitates simplification.
Reducing equations is a technique for solving a complicated equation and rewriting it in a simpler form. They can, however, be solved by converting them to linear equations and executing mathematical operations on them such as addition, subtraction, and cross-multiplication. These nonlinear equations can be solved, and the value of the variable obtained after they are reduced to linear form. Since not all equations are presented in a linear format, it is critical to simplify them further into more understandable forms.
The main goal of this simplification procedure is to make the computation process easier. Reducing complex equations to their linear form makes them easier to calculate and makes fewer mistakes.
Linear equations can be simplified by
Let us understand the process with an example.
Consider the equation \(\frac{{6x + 1}}{3} + 1 = \frac{{x – 3}}{6}\).
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The steps to reduce equations to a simpler form are given below:
Step 1: Since the equation \(\frac{{6x + 1}}{3} + 1 = \frac{{x – 3}}{6}\) is nonlinear, it cannot be solved directly. Let us first simplify the given equation.
\(\frac{{6x + 1}}{3} + 1 = \frac{{x – 3}}{6}\)
\( \Rightarrow \frac{{6x + 1 + 3}}{3} = \frac{{x – 3}}{6}\)
\( \Rightarrow \frac{{6x + 4}}{3} = \frac{{x – 3}}{6}\)
Step 2: Cross multiply the terms on both the sides of the equation, that is, multiply the denominator on one side to the numerator on the other.
Cross multiplication is one of the most common methods for simplifying nonlinear equations. This procedure can be used to multiply the numerator of one fraction with the denominator of another.
\( \Rightarrow 6\left( {6x + 4} \right) = \left( {x – 3} \right)3\)
Step 3: Simplify the brackets using the distributive law.
Following this cross multiplication, another mathematical operation must be performed to get a step closer to solving this equation. If there are any terms with brackets, they must be opened. This is done using the distributive law. The law states that any value between the brackets should be multiplied by the value outside of them.
\( \Rightarrow 6 \times 6x + 6 \times 4 = 3 \times x – 3 \times 3\)
\( \Rightarrow 36x + 24 = 3x – 9\)
Step 4: Arrange all the variables on one side of the equation (LHS) and the constants on the other side (RHS).
Following the distributive law application, the variables have to be arranged on one side and the constants on the other. While performing this step, keep in mind that moving any value from the RHS to the LHS will change a negative value to a positive one and vice versa.
\(36x – 3x = – 9 – 24\)
\( \Rightarrow 33x = – 33\)
Step 5: Solve the rest of the equation as a one-variable linear equation.
On completing the addition, subtraction, or multiplication with the same value on both sides, the value of the unknown variable will be obtained.
\(33x = – 33\)
\( \Rightarrow x = \frac{{ – 33}}{{33}}\)
\(\therefore x = – 1\)
Below are a few solved examples that can help in getting a better idea.
Q.1. Solve the equation \(\frac{{x + 1}}{{x + 3}} = \frac{2}{3}\).
Ans: Given: \(\frac{{x + 1}}{{x + 3}} = \frac{2}{3}\).
The denominator having the variable cannot be equal to zero.
\(x + 3 \ne 0\)
By doing the cross-multiplication, we get
\(3\left( {x + 1} \right) = 2\left( {x + 3} \right)\)
Opening the brackets, we get
\(3x + 3 = 2x + 6\)
Transposing numbers and variables, we get
\(3x – 2x = 6 – 3\)
\( \Rightarrow x = 3\)
Hence, the solution of the given equation is \(x = 3\).
Q.2. Solve \(\frac{{6x + 1}}{3} = \frac{{x + 3}}{6}\)
Ans: To solve the given equation, first cross multiply the denominators.
\(\Rightarrow 6\left( {6x + 1} \right) = 3\left( {x + 3} \right)\)
Opening the brackets, we get
\(6 \times 6x + 6 \times 1 = 3 \times x + 3 \times 3\)
\( \Rightarrow 36x + 6 = 3x + 9\)
Transposing numbers and variables, we get
\( \Rightarrow 36x – 3x = 9 – 6\)
\( \Rightarrow 33x = 3\)
\( \Rightarrow x = \frac{3}{{33}}\)
\(\therefore x = \frac{1}{{11}}\)
Hence, the solution of the given equation is \(x = \frac{1}{{11}}\).
Q.3. Solve the equation \(\frac{x}{2} – \frac{1}{5} = \frac{x}{3} + \frac{1}{4}\).
Ans: To solve the given equation, first, we should find the LCM of the denominators on both sides.
\(\frac{{5x – 2}}{{2 \times 5}} = \frac{{4x + 3}}{{3 \times 4}}\)
\( \Rightarrow \frac{{5x – 2}}{{10}} = \frac{{4x + 3}}{{12}}\)
Now, cross multiplying the denominators on both the sides
\(12\left( {5x – 2} \right) = 10\left( {4x + 3} \right)\)
Opening the brackets, we get
\(12 \times 5x – 12 \times 2 = 10 \times 4x + 10 \times 3\)
\( \Rightarrow 60x – 24 = 40x + 30\)
\( \Rightarrow 60x – 40x = 30 + 24\)
\( \Rightarrow 20x = 54\)
\( \Rightarrow x = \frac{{54}}{{20}}\)
\(\therefore x = 2.7\)
The solution of the given equation is \(x = 2.7\).
Q.4. Solve the equation \(x – 1 = \frac{x}{3} + \frac{3}{4}\).
Ans: To solve the given equation, first find the LCM of the denominators on the righthand side.
\(x – 1 = \frac{x}{3} + \frac{3}{4}\)
\( \Rightarrow x – 1 = \frac{{4x + 9}}{{12}}\)
Now cross multiplying the denominator on the RHS, we get
\( \Rightarrow 12\left( {x – 1} \right) = 4x + 9\)
Opening the brackets to simplify further, we get
\(12x – 12 = 4x + 9\)
\( \Rightarrow 12x – 4x = 12 + 9\)
\( \Rightarrow 8x = 21\)
\(\therefore x = \frac{{21}}{8}\)
Hence, the solution of the given equation is \(x = \frac{{21}}{8}\)
Q.5. Solve the given linear equation: \(\frac{{3x – 4}}{2} – \frac{{2x – 1}}{3} = \frac{1}{2} + x\)
Ans: Given: \(\frac{{3x – 4}}{2} – \frac{{2x – 1}}{3} = \frac{1}{2} + x\)
First, take the LCM of the terms on either side of the equation to solve the given equation.
\( \Rightarrow \frac{{3\left( {3x – 4} \right) – 2\left( {2x – 1} \right)}}{{2 \times 3}} = \frac{{1 + 2x}}{2}\)
\( \Rightarrow \frac{{3\left( {3x – 4} \right) – 2\left( {2x – 1} \right)}}{6} = \frac{{1 + 2x}}{2}\)
\( \Rightarrow \frac{{9x – 12 – 4x + 2}}{6} = \frac{{1 + 2x}}{2}\)
\( \Rightarrow \frac{{5x – 10}}{6} = \frac{{1 + 2x}}{2}\)
Now cross multiply the denominators on either side of the equation.
\( \Rightarrow 2\left( {5x – 10} \right) = 6\left( {1 + 2x} \right)\)
\( \Rightarrow 2 \times 5x – 2 \times 10 = 6 \times 1 + 6 \times 2x\)
\( \Rightarrow 10x – 20 = 6 + 12x\)
Now, bring all the variables to one side and constants to the other side.
\(10x – 12x = 6 + 20\)
\( \Rightarrow – 2x = 26\)
\( \Rightarrow x = \frac{{26}}{{ – 2}}\)
\(\therefore x = – 13\)
Hence, the solution of the given equation is \(x = – 13\).
Reducing equations to simpler forms is used to obtain linear equations. Not all the equations are in linear form, i.e., the equation of the first order. However, they can be solved by putting them into a linear equation through mathematical operations like cross-multiplication.
After reducing nonlinear equations into linear form, the value of the variable can be calculated easily. Reducing equations is an approach to solving a complicated equation and rewriting it in a simpler form. Although not all problems are in the form of linear equations, these nonlinear equations may be calculated, and the variable value is simply derived after they are reduced to linear form.
Students might be having many questions regarding the Reducing Linear Equations to Simpler Form. Here are a few commonly asked questions and answers.
Q.1. Can you reduce linear equations?
Ans: Yes. We can reduce the linear equations to simpler forms for easy calculations.
Q.2. What is a linear equation?
Ans: A mathematical statement with an equality sign \(\left( = \right)\) between the algebraic expressions is known as an equation. A linear equation is an algebraic expression in which each term is either the product of single variables and constants or remains constant. It contains the variables’ first-order power. \(ax + b = 0\) is a common representation for such equations, where \(a\) and \(b\) are constants and \(a \ne 0\)
Q.3. What does it mean to reduce an equation?
Ans: Reducing equations is an approach to solving a complicated equation and rewriting it in a simpler form. Reducing an equation means simplifying the equation into a simpler form which can be easily manipulated to find the unknown.
Q.4. How do you reduce a system of linear equations?
Ans: Linear equations can be simplified by
• Finding the LCM of the denominators
• Cross multiplication or removing the denominators of both the sides
• Opening the brackets
• Transposing the terms
Q.5. Is reducing subtraction?
Ans: No. Reducing is not subtraction. It is a process of simplifying the equation to a simpler linear form. Subtraction may also be involved in the reduction process.
We hope this information about the Reducing Linear Equations to Simpler Form has been helpful. If you have any doubts, comment in the section below, and we will get back to you.
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