1 Million Means: 1 million in numerical is represented as 10,00,000. The Indian equivalent of a million is ten lakh rupees. It is not a...

1 Million Means: 1 Million in Rupees, Lakhs and Crores

June 5, 2024It is the system of concurrent lines, which are in the form of straight lines. The highest degree of the linear equation in two variables is one. The general form of linear equation in two variable is \(ax + by + c = 0.\). A** Pair of Linear Equations in Two Variables**, which has a solution, is called a consistent pair of linear equations. There are an endless number of different common solutions to a pair of equivalent linear equations. A dependent pair of linear equations in two variables is such a pair.

The pair of linear equations in two variables is the system of concurrent lines. The geometrical representation of pair of linear equations in two variables is straight lines. A linear equation in two variables can be different forms, such as standard form, slope-intercept form and point-intercept form etc.

The standard form of linear equation in two variables is \(ax + by + c = 0,\) where both \(a\) and \(b\) are not equalled to zero. Similarly, the standard form of pair of linear equations in two variables is given by

\({a_1}x + {b_1}y + {c_1} = 0\)

\({a_2}x + {b_2}y + {c_2} = 0\)

Here**,** \({a_1},\;{a_2},\;{b_1},\;{b_2},\;{c_1},\;{c_2}\) are real numbers and \(a_1^2 + b_1^2 \ne 0\) and \(a_2^2 + b_2^2 \ne 0.\) Here, \({a_1},\;{a_2},\;{b_1},\;{b_2}\) and they are known as the co-efficient of variables \(x,\;y\) respectively.

Pair of linear equations in two variables are denoted graphically or geometrically in the form of straight lines. There are mainly three types of straight lines that are possible. They are

- Intersecting lines
- Parallel Lines
- Co-incident lines

Two lines, which are intersecting at only one point, is called the intersecting lines. There is a unique solution for the intersecting lines.

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 intersecting lines, if

\(\frac{{{a_1}}}{{{a_2}}} \ne \frac{{{b_1}}}{{{b_2}}}\)

The intersecting lines are consistent in nature. The below diagram represents the intersecting lines.

Two lines, which are collaborating or incident on each other, are called co-incident lines. The co-incident lines have infinitely many solutions.

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 co-incident lines, if

\(\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}\)

The co-incident lines are consistent in nature. The below diagram shows the co-incident lines.

Two lines, which are not touching each other and having the same distance between them, are known as parallel lines. There is no intersecting point for the parallel lines, so they have zero solutions or no solutions.

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 parallel lines, if

\(\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \ne \frac{{{c_1}}}{{{c_2}}}\)

The parallel lines are inconsistent in nature. The below diagram shows the pair of parallel lines**.**

The below table gives the conditions and graphs of different types of lines:

The pair of linear equations in two variables are solved algebraically or geometrically. The pair of linear equations have one or infinite or no solutions. The solution of pair of linear equations obtained by solving is expressed in the form of ordered pairs.

The linear equations are solved by using different methods. They are:

- Graphical method
- Elimination method
- Substitution method
- Cross multiplication method

In this method, we will solve equations graphically by using a graph. The steps used to solve the linear equations in two variables graphically as given below:

- Consider the graph paper by using the suitable scale.
- Write the given equations in the form of \(y = mx + c,\) by solving the equations.
- Consider the points, which satisfies the given equations and plot on the graph paper.
- For intersecting lines, observe the intersection point of two lines, which gives the solution of the given pair of linear equations.
- There are many infinite solutions for co-incident lines, and for the parallel lines, there is no solution.

**Example:**

The graph for pair of linear equations \(3x + 4y = 20\) and \(x – 2y = 0\) are given below:

From the above graph, the intersection point \({\rm{Q\;}}\left( {4,{\rm{\;}}2} \right)\) gives the solution of given lines.

**Example:**

The graph of the pair of equations \(x – 2y = \, – 4\) and \( – 3x + 6y = 0\) is given below:

From the graph, we can say that two lines are non-intersecting lines or known as parallel lines.

So, the given two lines have no solution.

The steps to be followed in the elimination method given as below:

- The steps to be followed in the elimination method given as below:
- If they have different variables, multiply the one equation or both the equations with the non-zero constant value to make the co-efficient of one variable \(\left( {x,\;y} \right)\) numerically equal. Add or subtract the equation for cancelling the variable.
- Solve the obtained equation of a single variable \(\left( {x,\;y} \right)\) to get its value.
- Substitute this value of the variable \(\left( {x,\;y} \right)\) in any one of the equations to get the value of another variable.

**Example:**

Solve the equations \(x + y = 5\) and \(2x – 3y = 4\) by using the elimination method as follows:

Let, \(x + y = 5\,\,\,\,…\left( 1 \right)\) and \(2x – 3y = 4\,\,\,\,…\left( 2 \right)\)

To make the co-efficient of variable \(x,\) multiply the equation \(\left( 1 \right)\) with \(2.\)

Thus, \(\left( 1 \right) \times 2 \Rightarrow 2x + 2y = 10\,\,\,\,…\left( 3 \right)\)

Subtract equation \(\left( 3 \right)\) from equation \(\left( 2 \right),\)

\(\Rightarrow y = \frac{6}{5}\)

Substitute the above value in the equation \(\left( 1 \right),\)

\(\Rightarrow x + \frac{6}{5} = 5\)

\(\Rightarrow x = 5 – \frac{6}{5} = \frac{{25 – 6}}{5}\)

\(\Rightarrow x = \frac{{19}}{5}\)

So, the values of \(x,\;y\) are \(\frac{{19}}{5},\frac{6}{5}\) respectively.

The steps to be followed in solving the equation by using the substitution method as follows:

- Solve any one equation, whichever is convenient and write the equation in one variable, say \(y\) in terms of \(x.\)
- Substitute the value of variable \(y\) obtained in the first step in the other equation.
- Solve the reducible equation obtained in the previous step to get the other variable \(x.\)

**Example:**

Solving the pair of linear equations \(7x – 15y = 2,\;\,x + 2y = 3\) by using the substitution method as follows:

Let, \(7x – 15y = 2\,\,\,\,…\left( 1 \right)\) and \(x + 2y = 3\,\,\,\,\,…\left( 2 \right)\)

Consider the equation \(x + 2y = 3\)

\(\Rightarrow x = 3 – 2y\)

Substitute the above value of \(x\) in the equation \(\left( 1 \right)\)

\(\Rightarrow 7\left( {3 – 2y} \right) – 15y = 2\)

\(\Rightarrow 21 – 14y – 15y = 2\)

\(\Rightarrow \, – 29y = 2 – 21 = – 19\)

\(\Rightarrow y = \frac{{19}}{{29}}\)

Substitute the above value in the equation \(\left( 3 \right)\)

\(\Rightarrow x = 3 – 2 \times \left( {\frac{{19}}{{29}}} \right)\)

\(\Rightarrow x = \frac{{87 – 38}}{{29}}\)

\(\Rightarrow x = \frac{{49}}{{29}}\)

Therefore, the solution of given lines is \(x = \frac{{49}}{{29}},\;y = \frac{{19}}{{29}}.\)

Consider two linear equations \({a_1}x + {b_1}y + {c_1} = 0\) and \({a_2}x + {b_2}y + {c_2} = 0.\) To solve the pair of equations by using the cross-multiplication method as follows:

- Write the coefficients of the variables \(\left( {x,\;y} \right)\) along with the constant values as shown below:

2. Write the equations in the following manner by cross-multiplying and subtracting the products as shown 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. By solving the above equations, we two 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. From the above equations, we get the values of variables \(\left( {x,\;y} \right)\)

\(x = \frac{{{b_1}{c_2} – {b_2}{c_1}}}{{{a_1}{b_2} – {b_1}{a_2}}};y = \frac{{{c_1}{a_2} – {c_2}{a_1}}}{{{a_1}{b_2} – {b_1}{a_2}}}\)

**Example:**

Solving the pair of equations \(2x + 3y = 46\) and \(3x + 5y = 74\) by using cross-multiplication method as follows:

The above equations can be written as follows:

\(2x + 3y – 46 = 0\) and \(3x + 5y – 74 = 0\)

Writing the co-efficient of the above two equations as mentioned below:

\(\Rightarrow \frac{x}{{3\left( { – 74} \right) – \left( 5 \right)\left( { – 46} \right)}} = \frac{y}{{\left( { – 46} \right)\left( 3 \right) – 2\left( { – 74} \right)}} = \frac{1}{{2\left( 5 \right) – 3\left( 3 \right)}}\)

\(\Rightarrow \frac{x}{{ – 222 + 230}} = \frac{y}{{ – 138 + 148}} = \frac{1}{{10 – 9}}\)

\(\Rightarrow \frac{x}{8} = \frac{y}{{10}} = 1\)

\(\Rightarrow \frac{x}{8} = 1;\,\frac{y}{{10}} = 1\)

\(\Rightarrow x = 8\) and \(y = 10\)

Hence, the solution of the given pair of equations is \(x = 8,\;y = 10.\)

The pair of linear equations is consistent if they have at least one solution and are inconsistent if they don’t have a solution.

The intersecting lines and co-incident lines are consistent as they have unique (one) solutions and infinitely many solutions, respectively.

The conditions for checking the consistent of pair of linear equations equations \({a_1}x + {b_1}y + {c_1} = 0\) and \({a_2}x + {b_2}y + {c_2} = 0\) is given below:

\(\frac{{{a_1}}}{{{a_2}}} \ne \frac{{{b_1}}}{{{b_2}}}\)

\(\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}\)

Theparallel lines are inconsistent as they have no solution, and the condition for inconsistent is given by:

\(\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \ne \frac{{{c_1}}}{{{c_2}}}\)

** Q.1. The difference between two numbers is \(12,\) and one number is four times the other. Find them.**: Let the number be \(x\) and \(y.\)

Ans

Given that, the difference between the two numbers is \(12.\)

\(x – y = 12\,\,\,\,…\left( 1 \right)\)

And, one number is four times the other.

\(x = 4y\,\,\,\,…\left( 2 \right)\)

Substitute equation \(\left( 2 \right)\) in equation \(\left( 1 \right)\) thus:

\(\Rightarrow 4y – y = 12\)

\(\Rightarrow 3y = 12\)

\(\Rightarrow y = \frac{{12}}{3}\)

\(\Rightarrow y = 4\)

Substitute the above value of \(y\) in the equation \(\left( 2 \right),\)

\(\Rightarrow x = 4\left( 4 \right)\)

\(\Rightarrow x = 16\)

Therefore, the numbers are \(16\) and \(4.\)

** Q.2. The difference between two digits of a number is \(2.\) The sum of a two-digit number and the number obtained by reversing the digits is \(66.\) Find the numbers.**: Let the two-digit number be \(10x + y.\)

Ans

The number obtained by reversing the digits is \($10y + x.\)

Given, the difference between the digits of the two-digit number is \(2.\)

\(x – y = 2\,\,\,\,…\left( 1 \right)\)

And given the sum of a two-digit number and the number obtained by reversing the digits is \(66.\)

\(\left( {10x + y} \right) + \left( {10y + x} \right) = 66\)

\(\Rightarrow 11x + 11y = 66\)

\(\Rightarrow 11\left( {x + y} \right) = 66\)

\(\Rightarrow x + y = \frac{{66}}{{11}}\)

\(\Rightarrow x + y = 6\,\,\,\,\,…\left( 2 \right)\)

Solving the above two equations by elimination method as follows:

\(\Rightarrow x = \frac{8}{2} = 4\)

Substituting the value of \(x\) Substituting the value of \(\left( 2 \right)\)

\(\Rightarrow y = 6 – 4 = 2\)

Hence, the required two-digit number is \(24\) or \(42.\)

** Q.3. In a river, a boat can travel \(51\) kilometres downstream in \(3\) hours. The same boat can travel \(30\) kilometres upstream in \(2\) hours. Find the speed of the boat still in the water.**: Let us consider the speed of a boat still in water \(= x\) kilometres per hour and the speed of water \(=y\) kilometres per hour.

Ans

Speed of the boat upstream is \(\left( {x – y} \right){\rm{\;kmph}}.\)

Speed of the boat downstream is \(\left( {x + y} \right){\rm{\;kmph}}.\)

Distance | Time | Speed | |

Upstream | \(30\) | \(2\) | \(x – y = \frac{{30}}{2} = 15\) |

Downstream | \(51\) | \(3\) | \(x + y = \frac{{51}}{3} = 17\) |

From the above table, \(x – y = 15\,\,\,\,\,…\left( 1 \right)\) and \(x + y = 17\,\,\,\,…\left( 2 \right)\)

Solving the above equations by elimination method as follows:

\(\Rightarrow x = \frac{{32}}{2} = 16\)

Hence, the speed of the boat is \(16\;{\rm{kmph}}.\)

** Q.4. Check if the given pair of linear equations are consistent or not?**: 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 \({a_2}x + {b_2}y + {c_2} = 0.\)

\(3x + 2y = 15;\,6x + 4y = 10\)

Ans

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 \(x + y + 1 = 0\) are having a unique solution.**: 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, if \(\frac{{{a_1}}}{{{a_2}}} \ne \frac{{{b_1}}}{{{b_2}}}.\)

Ans

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, \({a_1} = 4,\;{b_1} = p,\;{c_1} = 8,\;{a_2} = 1,\;{b_2} = 1,\;{c_2} = 1\)

\(\Rightarrow \frac{4}{p} \ne \frac{1}{1}\)

\(\Rightarrow p \ne 4\)

Therefore, all values of \(p,\) expect \(4,\) the given pair of equations have a unique solution.

This article discussed the pair of linear equations in two variable definitions, the standard form of pair of linear equations in two variables. In addition, we have discussed the different types of lines such as intersecting lines, parallel lines, and co-incident lines and studied different solutions such as unique, zero and infinite solutions.

Here, we studied the different methods of solving linear equations in two variables: elimination method, substitution method, graphical method, and cross-multiplication method.

** Q.1. What is a linear equation? Explain with example?** The linear equation is an equation whose highest degree is one. The graph of a linear equation is a straight line.

Ans:

Example: \(3x + 2y + 1 = 0,\;ax + b = 0\)

** Q.2. How many solutions are there for a pair of co-incident lines?**: Co-incident lines consist of infinitely many solutions.

Ans

*Q.3. How do you solve linear equations?*** Ans**: Pair of linear equations in two variables are solved by using the methods, such as:

a. Elimination method

b. Substitution method

c. Cross multiplication method

d. Graphical method

*Q.5. What is the standard form of pair of linear equations in two variables?*** Ans**: \({a_1}x + {b_1}y + {c_1} = 0\) and \({a_2}x + {b_2}y + {c_2} = 0,\) where \({a_1},\;{a_2},\;{b_1},\;{b_2},\;{c_1},\;{c_2}\) are real numbers and \(a_1^2 + b_1^2 \ne 0\) and \(a_2^2 + b_2^2 \ne 0.\)

** Q.**6.

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