Equations Reducible to a Pair of Linear Equations in Two Variables: The equation in the form of \(ax + b = 0\) or \(ax = b,\) where a and b are real numbers and \(a \ne 0\) is called a linear equation in one variable \(x.\) Similarly, an equation of form \(ax + by + c = 0\) or \(ax + by = c\) is called a linear equation in two variables.

When substituted in an equation, a value of the variable, making LHS \( = \) RHS, satisfying the equation is called a solution or a root of the equation. This article will study linear equations in two variables, simultaneous linear equations, methods of solving simultaneous linear equations in two variables, and the equations that can be reduced to a pair of linear equations in two variables.

Linear Equations in Two Variables

The general form of linear equations in two variables is \(ax + by + c = 0\) or \(ax + by = c,\) where \(a,b,c\) are real numbers, \(a,b \ne 0\) and \(x,y\) are variables.
Example: \(x + 2y = 3, – 5x + 4y = 3\) are linear equations in two variables.
Any pair of values of \(x\) and \(y\) which satisfies the equation \(ax + by + c = 0\) or \(ax + by = c\) is called its solution. For example, \(x = 2\) and \(y = 5\) is a solution of equation \(4x – y = 3.\)

Graphs: The graph of a linear equation in one variable is a straight line parallel to the \(x\)-axis or \(y\)-axis according to as the equation is of the form \(ay = b\) or \(ax = b,\) where \(a \ne 0.\)
Example: The graph of \(2x + 9 = 0\) is shown below.

The graph of a linear equation in two variables is also a straight line. The coordinates of every point on the line representing a linear equation determine a solution of the equation. Every solution of the linear equation is represented by a point on the line represented by it. Thus, there is a one-to-one correspondence between the solutions of a linear equation and points lying on the straight line represented by it.

Example: The graph of \(x + 2y = 7\) is given below:

Simultaneous Linear Equations in Two Variables

A pair of linear equations in two variables is said to form a system of simultaneous linear equations. Below is the example of the system of two simultaneous linear equations in two variables:

i) \(x + 3y = 2\)
\(3x – y = 4\)
ii) \(2a + b – 1 = 0\)
\(3a + 6 – 2 = 0\)
The general form of a pair of linear equations in two variables \(x\) and \(y\) is
\({a_1}x + {b_1}y + {c_1} = 0\) and \({a_2}x + {b_2}y + {c_2} = 0\)

Solution: A pair of values of the variables \(x\) and \(y\) satisfying each one of the equations in a given system of two simultaneous linear equations in \(x\) and \(y\) is called a solution of the system.

(i) Consistent System: A system of simultaneous linear equations is consistent if it has at least one solution.
(ii) Inconsistent System: A system of simultaneous linear equations is inconsistent if it has no solutions.

There are two methods of solving the given system of simultaneous linear equations. One is the graphical method, and another one is the algebraic method.

Algebraic Methods of Solving Simultaneous Linear Equations in Two Variables

Most of the time, the graphical method is not convenient, particularly when the point of intersection of the lines represented by two given equations has coordinates as rational numbers. In such situations, the graphical method does not give an accurate solution. For example, if the solution to a system of two linear equations is \(x = – \frac{4}{5},y = \frac{5}{{7′}}\) then by graphical method, the point of intersection would be \(\left({ – \frac{4}{5},\frac{5}{7}} \right).\)

This point is close to \(\left({ – 0.8,0.7} \right)\) that on graph paper, it is not convenient to distinguish these two points, and while reading the coordinates of the point of intersection, we are likely to make an error. Hence, it is necessary to use precise methods. The algebraic methods given below determine the accurate answer.

The most commonly used algebraic methods of solving simultaneous linear equations in two variables are as follows.

1. Substitution method of solving the linear equations in two variables
2. Elimination method of solving the linear equations in two variables
3. Cross Multiplication of solving the linear equations in two variables

In all the above methods, the linear equations will be in the form

\({a_1}x + {b_1}y + {c_1} = 0\) and \({a_2}x + {b_2}y + {c_2} = 0\)

But if we have the equations in the form \(\frac{1}{{x – 1}} + \frac{2}{{y – 2}} = 2\) and \(\frac{3}{{x – 1}} – \frac{3}{{y – 2}} = 1,\) we need to reduce these equations into the standard form like \({a_1}x + {b_1}y + {c_1} = 0\) and \({a_2}x + {b_2}y + {c_2} = 0\) and then need to find the solution. The below section will discuss how to reduce the given equations into simultaneous linear equations in two variables.

Equations Reducible to a Pair of Linear Equations in Two Variables

This section will discuss the solution of such pairs of equations that are not linear but can be reduced to linear form by making some suitable substitutions.

Below are the steps to solve a pair of equations that are reducible to linear equations:

1. Find the expressions that are common in both equations. Substitute them with the suitable variable, say \(x\) and \(y.\)
2. Find the solution of the new pair of linear equations for the new variables.
3. Substitute back the values of the new variables and solve for the initial variables.

The below section will solve some example problems based on the equations reducible to a pair of linear equations in two variables to understand the concept. Also, after reducing the equations to a pair of linear equations, we can use the most common algebraic methods like substitution or elimination to find the solution of the given equations.

Solved Examples

Q.1. Solve: \(\frac{1}{{x – 1}} + \frac{2}{{y – 2}} = 2\) and \(\frac{3}{{x – 1}} – \frac{3}{{y – 2}} = 1\)
Ans: In both the equations, \(\frac{1}{{x – 1}}\) and \(\frac{1}{{y – 2}}\) are common. So, let us substitute new variables say \(u\) and \(v\) in place of \(\frac{1}{{x – 1}}\) and \(\frac{1}{{y – 2}}\) respectively to make it as a general form \({a_1}x + {b_1}y + {c_1} = 0\) and \({a_2}x + {b_2}y + {c_2} = 0\)
So, let \(\frac{1}{{x – 1}} = u\) and \(\frac{1}{{y – 2}} = v\)
Now, the given equation becomes \(u + 2v = 2\) and \(3u – 3v = 1.\)
Let, \(u + 2v = 2 \ldots \ldots \left( i \right)\)
\(3u – 3v = 1 \ldots \ldots \left({ii} \right)\)
We can solve the above equations using the elimination method.
Multiplying equation \(\left( i \right)\) by \(3,\) we get
\(3u + 6v = 6 \ldots \ldots \left({iii} \right)\)
Subtracting \(\left({ii} \right)\) from \(\left({iii} \right),\) we get
\(9v = 5\)
\( \Rightarrow v = \frac{5}{9}\)
Substituting \(v\) value in equation \(\left({ii} \right),\) we get
\(3u – 3\left({\frac{5}{9}} \right) = 1\)
\( \Rightarrow 3u – \frac{5}{3} = 1\)
\( \Rightarrow 9u – 5 = 3\)
\( \Rightarrow 9u = 8\)
\( \Rightarrow u = \frac{8}{9}\)
Now, we will resubstitute the values of \(u\) and \(v\) in the original equations.
\( \Rightarrow \frac{1}{{x – 1}} = \frac{8}{9}\) and \(\frac{1}{{y – 2}} = \frac{5}{9}\)
\( \Rightarrow 8x – 8 = 9\) and \(5y – 10 = 9\)
\( \Rightarrow 8x = 17\) and \(5y = 19\)
\( \Rightarrow x = \frac{{17}}{8}\) and \(y = \frac{{19}}{5}\)

Q.2. Solve the following system of equations:
\(\frac{1}{{2x}} – \frac{1}{y} = – 1\)
\(\frac{1}{x} + \frac{1}{{2y}} = 8,\) where \(x \ne 0,y \ne 0\)
Ans: Substituting \(\frac{1}{x} = u\) and \(\frac{1}{y} = v\) in the above equations, we get
\(\frac{u}{2} – v = – 1 \Rightarrow u – 2v = – 2 \ldots \ldots \left( i \right)\) and,
\(u + \frac{v}{2} = 8 \Rightarrow 2u + v = 16 \ldots \ldots .\left({ii} \right)\)
Let us eliminate \(u\) from equations \(\left( i \right)\) and \(\left( i \right).\) Multiply equation \(\left( i \right)\) by \(2,\) we get
\(2u – 4v = – 4 \ldots \ldots \left({iii} \right)\)
Subtracting \(\left( ii \right)\) from \(\left( iii \right),\) we get
\( – 5v = – 20\)
\( \Rightarrow v = 4\)
Substituting \(v = 4\) in equation \(\left( i \right),\) we get
\(u – 8 = – 2\)
\( \Rightarrow u = 6\)
Resubstituting the \(u\) and \(v\) values in the original equation, we get
\(x = \frac{1}{u} = \frac{1}{6}\) and \(y = \frac{1}{v} = \frac{1}{4}\)
So, the solution of the given system of equation is \(x = \frac{1}{6}\) and \(y = \frac{1}{4}\)

Q.3. Solve \(\frac{2}{x} + \frac{2}{{3y}} = \frac{1}{6}\) and \(\frac{3}{x} + \frac{2}{y} = 0\)
Ans: Put, \(\frac{1}{x} = u\) and \(\frac{1}{y} = v,\) we get
\(2u + \frac{2}{3}v = \frac{1}{6} \Rightarrow 12u + 4v = 1 \ldots \ldots \left( i \right)\) and,
\(3u + 2v = 0 \ldots \ldots \left({ii} \right)\)
Multiplying equation \(\left({ii} \right)\) by \(2,\) we get
\(6u + 4v = 0 \ldots \ldots \left({iii} \right)\)
Subtracting \(\left({iii} \right)\) from \(\left({i} \right),\) we get
\(6u = 1 \Rightarrow u = \frac{1}{6}\)
Substituting \(u = \frac{1}{6}\) in \(\left({i} \right),\) we get
\(12\left({\frac{1}{6}} \right) + 4v = 1\)
\( \Rightarrow 4v = – 1\)
\( \Rightarrow v = – \frac{1}{4}\)
Resubstituting \(u\) and \(v\) values in the original equation, we get
\(x = \frac{1}{u} = 6\) and \(y = \frac{1}{v} = – 4\)

Q.4. Solve \(\frac{5}{{x – 1}} + \frac{1}{{y – 2}} = 2\) and \(\frac{6}{{x – 1}} – \frac{3}{{y – 2}} = 1\)
Ans: Put, \(\frac{1}{{x – 1}} = u\) and \(\frac{1}{{y – 2}} = v,\) we get
\(5u + v = 2 \ldots \ldots \left( i \right)\) and,
\(6u – 3v = 1 \ldots \ldots \left({ii} \right)\)
Multiplying equation \(\left( i \right)\) by \(3,\) we get
\(15u + 3v = 6 \ldots \ldots \left({iii}\right)\)
Adding \(\left({ii} \right)\) and \(\left({iii} \right),\) we get
\(21u = 7 \Rightarrow u = \frac{1}{3}\)
Substituting \(u = 3\) in equation \(\left({i} \right),\) we get
\(5\left({\frac{1}{3}} \right) + v = 2\)
\( \Rightarrow v = 2 – \frac{5}{3}\)
\( \Rightarrow v = \frac{1}{3}\)
Resubstituting \(u\) and \(v\) values in the original equation, we get
\(\frac{1}{{x – 1}} = \frac{1}{3} \Rightarrow 3 = x – 1 \Rightarrow x = 4\) and, \(\frac{1}{{y – 2}} = \frac{1}{3} \Rightarrow 3 = y – 2 \Rightarrow y = 5\)
Therefore, \(x = 4\) and \(y = 5\)

Q.5. Solve \(\frac{1}{{2\left({2x + 3y} \right)}} + \frac{{12}}{{7\left({3x – 2y} \right)}} = \frac{1}{2}\) and \(\frac{7}{{\left({2x + 3y}\right)}} + \frac{4}{{\left({3x – 2y} \right)}} = 2\)
Ans: Put, \(\frac{1}{{\left({2x + 3y} \right)}} = u\) and \(\frac{1}{{\left({3x – 2y} \right)}} = v,\) we get
\(\frac{1}{2}u + \frac{{12}}{7}v = \frac{1}{2} \Rightarrow 7u + 24v = 7 \ldots \ldots .\left( i \right)\)
\(7u + 4v = 2 \ldots \ldots \left({ii} \right)\)
Subtracting \(\left({ii} \right)\) from \(\left({i} \right),\) we get
\(20v = 5 \Rightarrow v = \frac{1}{4}\)
Substituting \(v = \frac{1}{4}\) in equation \(\left({i} \right),\) we get
\(7u + 24\left({\frac{1}{4}} \right) = 7 \Rightarrow 7u = 1 \Rightarrow u = \frac{1}{7}\)
Resubstituting \(u\) and \(v\) values in the original equation, we get
\(\frac{1}{7} = \frac{1}{{\left({2x + 3y} \right)}}\) and \(\frac{1}{{\left({3x – 2y} \right)}} = \frac{1}{4}\)
\( \Rightarrow 2x + 3y = 7 \ldots \ldots \left( i \right)\) and,
\(3x – 2y = 4 \ldots \ldots \left({ii}\right)\)
Multiplying \(\left( i \right)\) by \(3\) and \(\left( ii \right)\) by \(2,\) we get
\(6x + 9y = 21 \ldots …\left({iii} \right)\) and,
\(6x – 4y = 8 \ldots \ldots \left({iv} \right)\)
Subtracting \(\left({iv} \right)\) from \(\left({iii} \right),\) we get
\(13y = 13 \Rightarrow y = 1\)
Substituting \(y = 1\) in equation \(\left({iii} \right),\) we get
\(6x + 9 = 21 \Rightarrow 6x = 12 \Rightarrow x = 2\)
Therefore, \(x = 2\) and \(y = 1\)

Summary

In this article, we have learned the definition of linear equations in two variables, simultaneous linear equations in two variables, algebraic methods of solving simultaneous linear equations in two variables, and the meaning of equations reducible to a pair of linear equations in two variables and solved some example problems on equations reducible to a pair of linear equations in two variables.

Frequently Asked Questions (FAQs)

Q.1. How do you reduce a pair of linear equations?
Ans: Below are the steps to solve a pair of equations that are reducible to linear equations:
1. Find the expressions that are common in both equations. Substitute them with the suitable variable, say \(x\) and \(y.\)
2. Find the solution of the new pair of linear equations for the new variables.
3. Substitute back the values of the new variables and solve for the initial variables.

Q.2. What is the reducible equation?
Ans: The equation which can be reduced to a linear equation is known as a reducible equation.
Example: \(\frac{2}{x} + \frac{2}{{3y}} = \frac{1}{6}\)

Q.3. How do you solve linear equations in two variables?
Ans: The most commonly used algebraic methods of solving simultaneous linear equations in two variables are:
1. Substitution Method of solving the linear equations in two variables
2. Elimination method of solving the linear equations in two variables
3. Cross Multiplication of solving the linear equations in two variables
Also, we can solve the linear equations in two variables by the graphical method.

Q.4. When two linear equations are in two variables, what are they called?
Ans: When two linear equations are in two variables, they are called simultaneous linear equations in two variables.

Q.5. What are solutions in simultaneous linear equations in two variables?
Ans: A pair of values of the variables \(x\) and \(y\) satisfying each one of the equations in a given system of two simultaneous linear equations in \(x\) and \(y\) is called a solution of the system.

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