An equation is a mathematical statement such that two expressions on both the left-hand side and right-hand side are equal. An algebraic equation is an equation that involves one or more variables. The graph of a linear equation is a straight line. The method of finding the value a variable satisfies in an equation is called solving a linear equation. The value so obtained is called the solution of the linear equation.

This article will discuss the definition of linear equation in one, two variables, the linear equation solution in one and two variables, and the solved examples.

Solution of a Linear Equation: An Overview

Algebraic expression: Algebraic expression is the combination of variables and constants connected by the four fundamental arithmetic operations such as \(+\), \(-\), \(×\), and \(÷\).
For example, \(3x + 2y – 8\).
Equation: An equation is a mathematical statement with an equality symbol \((=)\). The left-hand side (L.H.S.) is the expression to the left of the equality sign, and the right-hand side (R.H.S.) is the expression to the right of the equality sign (R.H.S.).
For example, \(3x + 2y = 8\).
Solution of an Equation: The expression values on the L.H.S. and R.H.S. are equal in an equation. It is valid only for specific values of the variables. These values are the solutions to the equation.
For example: For equation \(3x + 2 = 8\), \(x = 2\) is the solution of the equation.
Linear Equations: The equations in which the highest power of the variable appearing in the expression is \(1\) are called linear equations.
Note: A linear equation may have linear expressions on both sides of the equality sign.
Transposing: We may take any term of an expression to the other side with a change in its sign, and this process is called transposition or transposing.
For example \(5x + 9 = 3x + 1\)
\(⇒ 5x – 3x = 1 – 9\)
\(⇒ 2x = – 8 ⇒ x = – 4\)

Linear Equation in One Variable

The expressions, which form the equation containing only one variable, and the highest power of the variable appearing in the equation is \(1\), are called linear equations in one variable.
For example \(3x – 8 = 4\).

Solving Equations Which have Linear Expressions on One Side and Numbers on the Other Side

This type of problem involves the following rules:
Rule 1: Same quantity or number can be added to (or subtracted from) both sides of an equation without changing the equality.
Rule 2: Both sides of an equation can be multiplied or divided by the same non-zero number without changing the equality.
For example: Find the solution of the linear equation \(2x – 3 = 7\).
Consider \(2x – 3 = 7\)
Add \(3\) to both sides.
\(2x – 3 + 3 = 7 + 3\)
or \(2x = 10\)
Now, divide both sides by \(2\).
\(\frac{2x}{2} = \frac{10}{2}\) or \(x = 5\)

Solving the Equations having the Variable on Both Sides

This method is based on the following steps:
Step 1: Obtain the linear equation.
Step 2: Identify the variables (unknown quantities) and constants (numerals).
Step 3: Simplify L.H.S. and R.H.S. to their simplest form by removing the brackets.
Step 4: Transpose all terms containing variables to the L.H.S. and all constants to the R.H.S.
Note that the sign of the terms will change when moving these terms from L.H.S. to R.H.S. and vice versa.
Step 5: Reduce L.H.S. and R.H.S. to their simplest form, with only one term on each side.
Step 6: Now, divide both sides of the equation by the coefficient of the variable on the L.H.S. to solve the equation obtained in step \(5\).

For example, solve \(2x – 3 = x + 2\).
We have \(2x – 3 = x + 2\)
Transpose \(3\) from L.H.S. to R.H.S. and x from R.H.S. to L.H.S.
\(2x – x = 2 + 3\)
Hence, the required solution of the equation is \(x = 5\).

Reducing Equations to Linear Form

The process of multiplying the numerator on L.H.S. with the denominator on R.H.S. and equating it to the product of the denominator on L.H.S. and the numerator on R.H.S. is called cross-multiplication.
An equation of the form
\(\frac{ {ax + b}}{ {cx + d}} = \frac{m}{n}\)
can be converted by using cross-multiplication to a linear equation of the form
\(n(ax + b) = m (cx + d)\)
For example, solve \(\frac{ {2x + 1}}{ {3x – 2}} = \frac{9}{10}\).
We have \(\frac{ {2x + 1}}{ {3x – 2}} = \frac{9}{10}\).
By cross multiplication, we get
\(⇒ 10(2x + 1) = 9(3x – 2)\)
\(⇒ 20x + 10 = 27 x -18\)
Transposing \(27x\) to L.H.S. and \(10\) to R.H.S.
\(⇒ 20x – 27x = – 18 – 10\)
\(⇒ – 7x = – 28\)
Dividing both sides by \(- 7\)
\(⇒ – \frac{7x}{7} = -\frac{28}{7}\)
\(x = 4\)

Applications of Linear Equations to Real-World Problems

To solve a word problem, we must follow the below steps:
Step 1: Read out the problem carefully and note what is provided and what is required.
Step 2: Refer to the unknown quantity by certain letters say \(x, y, z\), etc.
Step 3: Translate the conditions in the problem to mathematical statements.
Step 4: When you use the condition(s) provided in the problem, form the equation.
Step 5: Solve the equation for the unknown.
Step 6: Check whether the solution satisfies the given equation.
For example, Sunil is now \(8\) years older than Satish. Find the age of each of them if their combined age is \(34\) years.
Let the age of Satish \(= x\) years
Then, the age of Sunil \(= (x + 8)\) years
According to the problem,
Sunil’s age \(+\) Satish’s age \(= 34\)
i.e., \(x + (x + 8) = 34\)
\( \Rightarrow \,\,2x + 8 = 34\)
\( \Rightarrow \,\,2x = 26\)
\( \Rightarrow \,\,x = 13\)
Hence, the age of Satish and Sunil are \(13\) and \(21\) years, respectively.

Linear Equations in Two Variables

If an equation is written in form \(ax + by + c = 0\), where \(a, b\), and \(c\) are integers and the coefficients of \(x\) and \(y\), i.e., \(a\) and \(b\), are non-zero, it is a linear equation in two variables.
\(5x + 4y = 7\) and \(- 4x + 7y = 9\) are examples of linear equations in two variables.

Solvability Conditions for a Pair of Two-Variable Simultaneous Linear Equations

Consider linear equation in two variables \(x\) and \(y\)
\(a_1 x + b_1 y + c_1 = 0\)
and \(a_2 x + b_2 y + c_2 = 0\)
Consistent System: If a system of simultaneous linear equations has at least one solution, it is said to be consistent.
1. When, \(\frac {a_1}{a_2} ≠ \frac {b_1}{b_2}\), we get a unique solution. In this case, the two lines representing the equations intersect each other.
2. When \(\frac{a_1}{a_2} = \frac {b_1}{b_2} = \frac{c_1}{c_2}\), there are infinitely many solutions. In this case, the two lines representing the equations overlap each other.
In-consistent system: If a system of simultaneous linear equations has no solution, it is said to be inconsistent.
When \(\frac{a_1}{a_2} = \frac {b_1}{b_2} ≠ \frac{c_1}{c_2}\), there is no solution. In this case, the two lines representing the equations are parallel to each other.

Solution of Linear Equations in Two Variables

Different approaches are used to solve linear equations. They are as follows:

1. Graphical Method

The steps involved to solve the linear equations in two variables graphically is given as follows:

1. Take a graph paper and mark the scale.
2. Transform the equations in the form of \(y = mx + c\).
3. Determine the points which satisfy the above equations and plot them on graph paper.
4. In intersecting lines, check for the point at which two lines intersect, which is the solution to the given pair of linear equations.
5. There are infinitely many solutions for co-incident lines but no solution for parallel lines.

Example: The graph for the system of linear equations \(3x + 4y = 20\) and \(x – 2y = 0\) are drawn below:
Therefore, we see that the intersection point \(Q(4, 2)\) gives the given system of linear equations.

2. Elimination Method

The following are the steps to be followed in the elimination method:

1. If the equations have the same co-efficient for any variable, add or subtract them to cancel the variable.
2. If the coefficients are different, multiply one or both equations by a non-zero constant value to numerically balance the co-efficient of one variable \((x\) or \(y)\). To cancel the variable, add or subtract the equation.
3. To find the value of a single variable \((x\) or \(y)\), solve the obtained equation.
4. To get the value of another variable, substitute this value of the variable \((x\) or \(y)\) in any of the equations.

Example: Solve the pair of linear equations \(x + y = 5\) and \(2x – 3y = 4\) by using the elimination method.
Given, \(x + y = 5\) —– (i)
and \(2x – 3y = 4\) —– (ii)
To make the co-efficient of variable \(x\) same, multiply the equation (i) with \(2\).
Thus, \((i) × 2 ⟹ 2x + 2y = 10\) —– (iii)
Subtract equation (iii) from equation (ii),
\(\left({2x – 3y = 4} \right)\, – \,\left({2x + 2y = 10} \right) = – 5y = – 6\)
\( \Rightarrow y = \frac{6}{5}\)
Now, substitute the above value in the equation (i),
\(\Rightarrow x + \frac{6}{5} = 5\)
\(\Rightarrow x = 5 – \frac{6}{5} = \frac{{25 – 6}}{5}\)
\(\Rightarrow x = \frac{{19}}{5}\)
So, the values of \(x, y\) are \(\frac {19}{5}, \frac {6}{5}\) respectively.

3. Substitution Method

The following are the steps to solve the equation using the substitution method:

1. Solve any one equation for either \(x\) or \(y\). Let us say we write \(y\) in terms of \(x\).
2. Now, substitute the value of variable \(y\) obtained in step \(1\) in another equation.
3. Solve the equation obtained in step \(2\) to get the value of the variable \(x\).

Example: Solve the system of linear equations \(7x – 15y = 2\), \(x + 2y = 3\) by using the substitution method.
Given, \(7x – 15y = 2\) …….(i)
and \(x + 2y = 3\) ……..(ii)
Consider the equation (ii), transpose the term \(2y\) to R.H.S., we get
\(\Rightarrow x = 3 – 2y\) ………(iv)
Now, substitute the above value of \(x\) in the equation (i),
\(\Rightarrow 7(3 – 2y) – 15y = 2\)
\(\Rightarrow 21 – 14y – 15y = 2\)
\(\Rightarrow -29y = 2 – 21 = -19\)
\( \Rightarrow y = \frac{{19}}{{29}}\)
Substitute the value of \(y\) in the equation (iv),
\( \Rightarrow x = 3 – 2 \times \left({\frac{{19}}{{29}}} \right)\)
\( \Rightarrow x = \frac{{87 – 38}}{{29}}\)
\( \Rightarrow x = \frac {49}{29}\)
Hence, the required solution is \(x = \frac{{49}}{{29}}\), \(y = \frac{{19}}{{29}}\)

4. Cross-Multiplication Method

Consider \(a_1 x + b_1 y + c_1 = 0\) and \(a_2 x + b_2 y + c_2 = 0\). Using the cross-multiplication method, we can solve the pair of equations as follows:
1. As indicated below, write the coefficients of the variables \((x, y)\) as well as the constant values:

2. Write the equations by cross-multiplying and subtracting the products in the order given by the arrow.
\(\frac{x}{{{b_1}{c_2} – {b_2}{c_1}}} = \frac{y}{{{c_1}{a_2} – {c_2}{a_1}}} = \frac{1}{{{a_1}{b_2} – {b_1}{a_2}}}\)

3. We get two equations by solving the previous equations.
\(\frac{x}{{{b_1}{c_2} – {b_2}{c_1}}} = \frac{1}{{{a_1}{b_2} – {b_1}{a_2}}};\frac{y}{{{c_1}{a_2} – {c_2}{a_1}}} = \frac{1}{{{a_1}{b_2} – {b_1}{a_2}}}\)

4. We can deduce the values of variables \((x, y)\) from the above equations
\(x = \frac{{{b_1}{c_2} – {b_2}{c_1}}}{{{a_1}{b_2} – {b_1}{a_2}}};y = \frac{{{c_1}{a_2} – {c_2}{a_1}}}{{{a_1}{b_2} – {b_1}{a_2}}}\)

Example: Solve the pair of linear equations in two variables \(2x + 3y = 46\) and \(3x + 5y = 74\) by using the cross-multiplication method.
Write the above equations as follows:
\(2x + 3y – 46 = 0\)
\(3x + 5y – 74 = 0\)
As indicated below, write the coefficients of the variables \((x, y)\) and the constant values.

\( \Rightarrow \frac{x}{{3( – 74) – (5)( – 46)}} = \frac{y}{{( – 46)(3) – 2( – 74)}} = \frac{1}{{2(5) – 3(3)}}\)
\(\Rightarrow \frac{x}{{ – 222 + 230}} = \frac{y}{{ – 138 + 148}} = \frac{1}{{10 – 9}}\)
\( \Rightarrow \frac{x}{8} = \frac{y}{{10}} = 1\)
\( \Rightarrow \frac{x}{8} = 1;\,\frac{y}{{10}} = 1\)
\(⟹ x = 8\) and \(y = 10\)
Hence, the required solution is \(x = 8\), \(y = 10\).

Solved Examples – Solution of a Linear Equation

Q.1. Solve \(2(5x – 2) = 4x + 8\).
Ans: The given equation is \(2(5x – 2) = 4x + 8\)
\(⟹ 10x – 4 = 4x + 8\)
\(⟹ 10x – 4x = 8 + 4\)
\(⟹ 6x = 12\)
\(⟹ x = 2\)
Hence, the value of \(x\) is \(2\).

Q.2. Solve \(\frac{{7y + 2}}{5} = \frac{{6y – 5}}{{11}}\)
Ans: The given equation is \(\frac{{7y + 2}}{5} = \frac{{6y – 5}}{{11}}\)
\(⟹ 11(7y + 2) = 5(6y – 5)\)
\(⟹ 77y + 22 = 30y – 25\)
\(⟹ 77y – 30y = -25 – 22\)
\(⟹ 45y = -45\)
\(⟹ y = -1\)
Hence, the value of y is \(-1\).

Q.3. Solve the pair of linear equations in two variables graphically \(x – 2y = -4\) and \(-3x + 6y = 0\).
Ans: Given, \(x – 2y = -4\) and \(-3x + 6y = 0\).
For the first equation \(x – 2y = -4\)

\(x\)	\(0\)	\(4\)	\(2\)
\(y\)	\(2\)	\(4\)	\(3\)

For the second equation \(-3x + 6y = 0\)

\(x\)	\(0\)	\(2\)	\(4\)
\(y\)	\(0\)	\(1\)	\(2\)

Plotting the points on the graph, we get

From the graph, we can say that two lines are non-intersecting lines or parallel lines.
So, the given linear equation has no solution.

Q.4. Check if the given pair of linear equations is consistent or not?
\(3x + 2y = 15; 6x + 4y = 10\)
Ans: Compare the given lines \(3x + 2y = 15\); \(6x + 4y = 10\) with standard pair of linear equations \(a_1 x + b_1 y + c_1 = 0\) and \(a_2 x + b_2 y + c_2 = 0\).
Thus, \(a_1 = 3\), \(b_1 = 2\), \(c_1 = -15\), \(a_2 = 6\), \(b_2 = 4\), \(c_2 = -10\)
Comparing the ratio of the coefficient of the linear equations, then
\( \Rightarrow \frac{{{a_1}}}{{{a_2}}} = \frac{3}{6} = \frac{1}{2}\) and \(\frac{{{b_1}}}{{{b_2}}} = \frac{2}{4} = \frac{1}{2}\) and \(\frac{{{c_1}}}{{{c_2}}} = – \frac{{15}}{{10}} = \frac{3}{2}\)
Two lines \(a_1 x + b_1 y + c_1 = 0\) and \(a_2 x + b_2 y + c_2 = 0\) are said to be inconsistent lines. If \(\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \ne \frac{{{c_1}}}{{{c_2}}}\)
And, here \(\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \ne \frac{{{c_1}}}{{{c_2}}}\). So the given lines are inconsistent.

Q.5. Find the value of \(p\), for which the pair of linear equations \(4x + py + 8 = 0\) and \(x + y + 1 = 0\) have a unique solution.
Ans: We know that two lines \(a_1 x + b_1 y + c_1 = 0\) and \(a_2 x + b_2 y + c_2 = 0\) are said to have a unique solution, \(\frac{{{a_1}}}{{{a_2}}} \ne \frac{{{b_1}}}{{{b_2}}}\)
Compare the given lines \(4x + py + 8 = 0\) and \(x + y + 1 = 0\) with standard pair of linear equations \(a_1 x + b_1 y + c_1 = 0\) and \(a_2 x + b_2 y + c_2 = 0\).
Thus, \(a_1 = 4\), \(b_1 = p\), \(c_1 = 8\), \(a_2 = 1\), \(b_2 = 1\), \(c_2 = 1\)
\(⟹ \frac {4}{p} ≠ \frac{1}{1}\)
\(⟹ p ≠ 4\)
Therefore, all values of \(p\), except \(4\), the given pair of equations have a unique solution.

Summary

In this article, we looked at the definition of a linear equation in one and two variables and how to solve a linear equation using two methods: transposing and substitution. Along with the solved examples, we also discussed the solution of linear equations in two variables using the graphical approach, elimination method, substitution method, and cross-multiplication method.

Frequently Asked Questions (FAQs) – Solution of a Linear Equation

Q.1. Can a system of linear equations have two solutions?
Ans: A system of linear equations can have no, unique, or infinitely many solutions. But a system of linear equations cannot have exactly two solutions.

Q.2. How to find the number of solutions in a linear equation?
Ans: We use different approaches to solve linear equations. They are as follows:
1. Graphical method
2. Elimination method
3. Substitution method
4. Cross multiplication method

Q.3. When does a system of the linear equation have no solution?
Ans: When the graphs are parallel, then the system of linear equations has no solution.

Q.4. Where do we find the solution to a linear equation?
Ans: The point where the lines representing the linear equations intersect is the solution to a system of linear equations. In the \(XY\)-plane, two lines can cross once, never cross, or entirely overlap.

Q.5. When will a system of the linear equation have infinite solutions?
Ans: When the graphs are the same line, then the system of linear equations has infinite solutions.

We hope this detailed article on the solution of a linear equation helped you in your studies. If you have any doubts or queries regarding this topic, feel to ask us in the comment section. Happy learning!