Equations Reducible to the Linear Form: Explanation & Examples

Equations Reducible to the Linear Form: In Mathematics, an equation is a mathematical statement stating that the value of two mathematical expressions is equal. An equation can be classified as linear, quadratic, cubic, etc., based on the degree of the variables. A linear equation is an equation with degree \(1.\) The equations reducible to the linear form is defined as the process of converting non-linear equations to linear equations.

Certain mathematical equations are not linear equations, but they may be made into linear equations using specific mathematical operations. These equations can be solved, and the unknown value is easily computed when reduced to linear form. In this article, we will discuss the equations reducible to linear form in detail.

Linear Equation in One Variable

Linear equations are those in which the maximum power of the variable appearing in the statement is \(1.\) The degree of a linear equation is \(1.\) Linear expressions can appear on both sides of the equality sign in a linear equation. If there is only one variable in a linear equation, then it is said as a linear equation in one variable.

The general form of a linear equation with only one variable \(x,\) is \(ax + b = 0,\) where \(a\) and \(b\) are constants and \(a \ne 0.\)

What is Reducing Equation?

The process of converting non-linear equations to linear equations is known as reducing equations. Because not all equations are provided plainly and straightforwardly, splitting them down to solve them is necessary.

On both sides, solving these equations requires various mathematical operations such as cross multiplication, division, etc. It helps in the conversion of complex equations to their linear forms. Following this change, determining the value of the variables becomes simple.

Reducing Equations to a Simpler Form

Suppose we are given equations of the form \(\frac{{ax}}{b} = \frac{{cx}}{d}\) where \(x\) is the variable and \(a,b,c,d\) are the constants. Let us understand this through an example.

Example: Solve the equation \(\frac{{6x + 1}}{3} = \frac{{x + 3}}{6}\)

At first sight, we may feel that this equation is not linear. But it can be brought to a simple linear form through the following steps.

1. First, take the LCM of the denominators and multiply both sides by it
LCM of \(3\& \,6 = 6\)
\( \Rightarrow \frac{{6x + 1}}{0} \times 6 = \frac{{x + 3}}{6} \times 6\)
\( \Rightarrow 12x + 2 = x + 3\)

2. Then, transpose the variables to one side and constants or numbers to the other side.
\( \Rightarrow 12x + 2 = x + 3\)

3. Now, simplify the new equation
\( \Rightarrow 11x = 1\)

4. Finally, find out the required solution
\(x = \frac{1}{{11}}\)

5. After finding the variable’s value, we have to verify the solution by substituting the variable’s value back in the equation.
\({\text{LHS}} = \frac{{6x + 1}}{3} = \frac{{6 \times \frac{1}{{11}} + 1}}{3} = \frac{{6 + 11}}{{33}} = \frac{{17}}{{33}}\)
\({\text{RHS}} = \frac{{x + 3}}{6} = \frac{{\frac{1}{{11}} + 3}}{6} = \frac{{1 + 33}}{{66}} = \frac{{34}}{{66}} = \frac{{17}}{{33}}\)
Hence, \({\text{LHS=RHS}}\)

Steps for Reducing Equations to Linear Form

Suppose we are given equations of the form \(\frac{{ax + p}}{{bx + q}} = \frac{c}{d}\) Where \(x\) is the variable and \(a,b,c,d,p,q\) are the constants. We will follow these steps to solve such equations.

1. As the given equation is non-linear, it cannot be solved directly. So, we must first use the cross multiplication method to simplify the following problem.
2. Both sides of the equation are cross multiplied, meaning the denominator on \({\text{LHS}}\) is multiplied by the numerator on the \({\text{RHS}}.\)
3. To open the brackets, use the distributive law.
4. Bring all the variables to one side of the equation \(\left({{\text{LHS}}} \right)\) and the constants to the other \(\left({{\text{RHS}}} \right)\) and solve the variable’s value.

Equations Reducible to the Linear Form

Let us understand equations reducilble to the linear form through an example.

Example: Solve the equation \(\frac{{x + 1}}{{2x + 3}} = \frac{3}{8}\)
As the given equation is non-linear, it cannot be solved directly. So, we must first use the cross multiplication method to simplify the following problem.
\(8x\left({x + 1} \right) = 3x\left({2x + 3}\right)\)
Open the brackets and multiply, we get
\(8{x^2} + 8x = 6{x^2} + 9x\)
\(8x + 8 = 6x + 9\)
Transpose the numbers and variables, we get
\(8x – 6x = 9 – 8\)
Simplifying the new equation, we get
\(2x = 1\)
Find out the required solution
\(x = \frac{1}{2}\)
Verify the solution
\({\text{LHS}} = \frac{{x + 1}}{{2x + 3}} = \frac{{\frac{1}{2} + 1}}{{2 \times \frac{1}{2} + 3}} = \frac{{1 + 2}}{{2 + 6}} = \frac{3}{8} = {\text{RHS}}\)

Example: The sum of a two-digit number’s digits is \(13.\) If the new number obtained after reversing the digits is \(45\) greater than the original number, determine the original number.
Answer: Let us assume that the original number in the units place is \(x.\)
According to the question,
Units digit \( + \) tens digit \( = 13\)
\( \Rightarrow x + {\text{tens’s}}\,{\text{digit}} = 13\)
\( \Rightarrow {\text{tens’s}}\,{\text{digit}} = 13 – x\)
Therefore, the original number \( = 10 \times \left({{\text{ten’s}}\,{\text{digit}}} \right) + {\text{unit’s}}\,{\text{digit}}\)
\( = 10 \times \left({13 – x} \right) + x\)
\( = 130 – 10x + x = 130 – 9x\)
On reversing the digits, the digit in the units place will be in the ten’s place and that in the ten’s place will be in the units place.
Therefore, units place \( = 13 – x\)
Ten’s place \( = x\)
Hence, the new number \( = 10 \times x + \left({13 – x} \right)\)
\( = 10x + 13 – x\)
\( = 9x + 13\)
According to the condition, the new number becomes \(45\) more than the original.
Therefore, the new number \( = \) original number \( + 45\)
\( \Rightarrow 9x + 13 = 130 – 9x + 45\)
\( \Rightarrow 9x + 13 = 175 – 9x\)
\( \Rightarrow 9x + 9x = 175 – 13\)
\( \Rightarrow 18x = 162\)
\( \Rightarrow x = \frac{{162}}{{18}} = 9\)
Therefore, the original number \( = 130 – 9x\)
\( = 130 – 9 \times 9\)
\( = 130 – 81 = 49\)

Solved Examples on Equations Reducible to the Linear Form

Q.1. When you deduct 4 from each of two integers in a 3:5 ratio, the ratio becomes 5:9. Find the numbers.
Ans: Suppose the first number \( = 3x\)
And the second number \( = 5x\)
On subtracting \(4\) from the first number, we get \(3x – 4.\)
Subtracting \(4\) from the second number, we get \(5x – 4.\)
According to the condition, the ratio obtained after subtracting \(4\) from each number \( = \frac{5}{9}\)
Therefore, \(\frac{{3x – 4}}{{5x – 4}} = \frac{5}{9}\)
On cross multiplication, we get
\( \Rightarrow 9 \times \left({3x – 4}\right) = 5 \times \left({5x – 4} \right)\)
\( \Rightarrow 27x – 36 = 25x – 20\) (on solving bracket)
\( \Rightarrow 27x – 25x = – 20 + 36\) (on transposing terms)
\( \Rightarrow 2x = 16\)
\( \Rightarrow x = \frac{{16}}{2}\)
\( \Rightarrow x = 8\)
Therefore, the first number \( = 3x = 3 \times 8 = 24\)
Second number \( = 5x = 5 \times 8 = 40\)

Q.2. The father is 20 years older than the boy. After three years, the son and father’s age ratio is 19:39. Then determine the son’s current age.
Ans: Suppose the present age of son \( = x\) years
Then father’s age \( = x + 20\) years
After \(3\) years, the son’s age would be \(\left({x + 3} \right)\) years
After \(3\) years, the father’s age would be \(\left({x + 20 + 3} \right) = \left({x + 23} \right)\) years
According to the question, After \(3\) years, the ratio of the ages of son and father \( = 19:39\)
Thus \(\left({x + 3} \right):\left({x + 23} \right) = 19:39\)
\( \Rightarrow \frac{{x + 3}}{{x + 23}} = \frac{{19}}{{39}}\)
\( \Rightarrow \left({x + 3} \right)39 = 19\left({x + 23} \right)\)
\( \Rightarrow 39x + 117 = 19x + 437\)
\( \Rightarrow 39x – 19x = 437 – 117\)
\( \Rightarrow 20x = 320\)
\( \Rightarrow x = \frac{{320}}{{20}} = 16\)
Therefore, the present age of the son \( = 16\) years
and the present age of the father \( = 16 + 20\)
\(= 36\) years

Q.3. The sum of a two-digit number’s digits is 12. If the new number obtained after reversing the digits is 18 greater than the original number, determine the original number.
Ans: Suppose the original number in the units place is \(x.\)
According to the question,
Units digit \( + {\rm{ten}}'{\rm{s}}\,{\rm{digit}} = 12\)
\( \Rightarrow x + {\text{ten’s}}\,{\text{digit}} = 12\)
\( \Rightarrow {\text{ten’s}}\,{\text{digit}} = 12 – x\)
Therefore, the original number \( = 10 \times \left({{\text{ten’sdigit}}} \right) + {\text{unit’s}}\,{\text{digit}}\)
\( = 10 \times \left({12 – x} \right) + x\)
\( = 120 – 10x + x = 120 – 9x\)
On reversing the digits, the digit in the units place will be in the ten’s place and that in the ten’s place will be in the units place.
Therefore, units place \( = 12 – x\)
Ten’s place \( = x\)
Hence, the new number \( = 10 \times x + \left({12 – x} \right)\)
\( = 10x + 12 – x\)
\( = 9x + 12\)
According to the condition, the new number becomes \(18\) more than the original.
Therefore, the new number \( = \) original number \( + 18\)
\( \Rightarrow 9x + 12 = 120 – 9x + 18\)
\( \Rightarrow 9x + 12 = 138 – 9x\)
\( \Rightarrow 9x + 9x = 138 – 12\)
\( \Rightarrow 18x = 126\)
\( \Rightarrow x = \frac{{126}}{{18}} = 7\)
Therefore, the original number \( = 120 – 9x\)
\( = 120 – 9 \times 7\)
\(120 – 63 = 57\)

Q.4. The denominator of a rational number is greater than its numerator by 4. On adding 6 to the numerator and subtracting 3 from the denominator, the fraction obtained is \(\frac{3}{2}.\) Find out the rational number.
Ans: Consider the numerator as \(x.\)
Then denominator \( = x + 4\)
The rational number \( = \frac{x}{{x + 4}}\)
Now on adding \(6\) to the numerator, we get, new numerator \( = x + 6\)
On subtracting \(3\) from the denominator, we get, new denominator \( = \left({x + 4} \right) – 3 = x + 1\)
The new rational number \( = \frac{{x + 6}}{{x + 1}}\)
According to the condition, the new number becomes \(\frac{3}{2},\) then \(\frac{{x + 6}}{{x + 1}} = \frac{3}{2}\)
On cross-multiplication, we get
\( \Rightarrow 3\left({x + 1} \right) = 2\left({x + 6} \right)\)
\( \Rightarrow 3x + 3 = 2x + 12\)
On transposing the terms, we get
\( \Rightarrow 3x – 2x = 12 – 3\)
\( \Rightarrow x = 9\)
Therefore, the required number is \(\frac{x}{{x + 4}} = \frac{9}{{9 + 4}} = \frac{9}{{13}}\)

Q.5. The sum of the two numbers is 35, and their ratio is 1:4. Find out the numbers.
Ans: Suppose the first number is \(x.\)
According to the first condition, the sum of the two numbers \( = 35.\)
\( \Rightarrow x + {2^{{\text{nd }}}}{\text{number}} = 35\)
\( \Rightarrow {2^{{\text{nd }}}}{\text{number}} = 35 – x\)
According to the second condition,
The ratio of the numbers \( = 1:4\)
Therefore, \(\frac{x}{{35 – x}} = \frac{1}{4}\)
On cross-multiplication, we get
\(4x = 35 – x\)
On transposing the terms, we get
\(4x + x = 35\)
\( \Rightarrow 5x = 35\)
\( \Rightarrow x = \frac{{35}}{5} = 7\)
The first number \( = 7\)
And the second number \( = 35 – 7 = 28\)

Summary

The process of converting non-linear equations to linear equations is referred to as reducing equations. Solving these equations needs several mathematical operations like cross multiplication, division, etc. After reducing these equations in a linear form, they can be solved, and the unknown value can be calculated simply. Furthermore, linear equations are defined as equations whose maximum power of the variable appearing in the statement is \(1.\) The general form of a linear equation with only one variable \(x,\) is \(ax + b = 0,\) where \(a\) and \(b\) are constants and \(a \ne 0.\)

FAQs on Equations Reducible to the Linear Form

Q.1. How do you reduce a linear equation?
Ans: The steps to reduce a linear equation are as follows:
1. If the given equation is non-linear, it cannot be solved directly. So, we must first use the cross multiplication method to simplify the problem.
2. Both sides of the equation are cross multiplied, meaning the denominator on \({\text{LHS}}\) is multiplied by the numerator on the \({\text{RHS}}.\)
3. To open the brackets, use the distributive law.
4. Bring all the variables to one side of the equation \(\left({{\text{LHS}}} \right)\) and the constants to the other \(\left({{\text{RHS}}} \right)\) and solve the variable’s value.

Q.2. What is a reducible equation?
Ans: Reducing an equation is a method of rewriting the equation in a more straightforward form. Certain mathematical equations are not linear equations but can be put in linear equations by performing mathematical operations. After reducing these equations in a linear form, they can be solved, and the unknown value can be calculated easily.

Q.3. What is LCM?
Ans: The least common multiple, or \({\text{LCM,}}\) is the lowest multiple between two numbers. It is generated from a value that can equally divide any two integers. The least common divisor is another name for it. The \({\text{LCM}}\) of \(4\) and \(6\) is \(24,\) and thus \(24\) is divisible by both \(4\) and \(6.\)

Q.4. What is a linear equation?
Ans: Linear equations are those in which the maximum power of the variable appearing in the statement is \(1.\)

Q.5. What is a linear equation in one variable?
Ans: Linear equations are those in which the maximum power of the variable appearing in the statement is \(1.\) The degree of the linear equation is \(1.\) If there is only one variable in a linear equation, then it is said as a linear equation in one variable.
The general form of a linear equation with only one variable \(x,\) is \(ax + b = 0,\) where \(a\) and \(b\) are constants and \(a \ne 0.\)

Learn About Solution of a Linear Equation

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