Method To Solve Simultaneous Linear Equations: Out of all the methods for solving linear equations, the elimination method in an algebraic method of solving is the most widely used. Using fundamental arithmetic operations, we can eliminate any of the variables and then simplify the equation to get the value of the other variable. Then we can put that number into most equations to find out what it is.

Pair of Linear Equations

Two linear equations in the same two variables are called pairs of linear equations in two variables. The most general form of linear equations is:

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

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

where \({a_1},\,{a_2},\,{b_1},\,{b_2},\,{c_1},\,{c_2}\) are real numbers, such that \(a_1^2 + b_1^2 \ne 0,\,a_2^2 + b_2^2 \ne 0\).

A pair of linear equations in two variables can be represented and solved by the:

1. Graphical Method
2. Algebraic Method

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Simultaneous Linear Equations

Simultaneous linear equations are two linear equations in two variables together.

The ordered pair \(\left( {x,\,y} \right)\) which satisfies both linear equations, is the solution of a system of simultaneous linear equations.

To find the values of letters within two or more equations, simultaneous equations require algebraic knowledge. Simultaneous equations are so-called because the equations are solved at the same time.

Three Methods of Solving Simultaneous Equations

There are three algebraic methods to solve simultaneous equations. Those are:

1. Substitution method

One of the algebraic methods for solving simultaneous linear equations is the substitution method. It involves moving the value of any of the variables from one equation to the other.

2. Elimination method

As per the elimination method definition, it is eliminating one of the terms containing any of the variables that makes the calculations easier.

3. Cross-multiplication method

Cross-multiplication is a method for finding the solution to two-variable linear equations. It is the fastest way to solve a set of linear equations.

Solving Simultaneous Equations by Substitution Method

The substitution method is a convenient method for solving linear equations algebraically and determining the variable’s solutions. It helps to determine the value of the \(x\)-variable in terms of the \(y\)-variable and then substituting or replacing the value of the \(x\)-variable in the second equation, as the names suggest. We can solve and find the value of the \(y\)-variable this way. Finally, we can put the value of \(y\) in an equation.

Example: Solve the following pair of equations by substitution method:

\(7x – 15y = 2\)

\(x + 2y = 3\)

Step-1: Let us consider the given equations as,

\(7x – 15y = 2 \ldots \ldots (1)\)

\(x + 2y = 3 \ldots \ldots (2)\)

Step-2: We can pick either of the equation and write one variable in terms of the other.

Let us consider equation \((2)\),

\(x + 2y = 3\)

\(x = 3 – 2y \ldots \ldots (3)\)

Step-3: Substitute the value of \(x\) in equation \((1)\). We get:

\(7(3 – 2y) – 15y = 2\)

\( \Rightarrow 21 – 14y – 15y = 2\)

\( \Rightarrow – 29y = – 19\)

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

Step-4:  Substitute the value of \(y\) in equation \((3)\) to get the value of \(x\).

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

\(x = \frac{{87 – 38}}{{29}} = \frac{{49}}{{29}}\)

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

Hence, the solution \(x = \frac{{49}}{{29}}\) and \(y = \frac{{19}}{{29}}\).

Elimination Method

The elimination method of algebraically solving a system of linear equations is the most commonly used of all the methods for solving linear equations. We can eliminate any of the variables using basic arithmetic methods, then simplify the equation to have the value of the other variable. Then we may use any of the equations to find out what that number is.

Steps to Solve Simultaneous Linear Equations by Elimination Method

When solving linear equations with two or three variables, the elimination method comes in handy. This method can also be used to solve three equations. Therefore, it can only be used on two equations at once. Let’s look at how to use the elimination method to solve a system of equations.

Step-1: Multiplying or dividing the linear equations with a non-zero number to get a  common coefficient of any of the variables in both equations as numerically equal.

Step-2: Then add or subtract both the equations such that the same terms will get eliminated.

Step-3: Simplify the result to get a final answer of the left out variable (let’s say \(y\)) such that we will only get an answer in the form of \(y = c\), where \(c\) is any constant.

Step-4: At last, substitute this value in any of the given equations to find the value of the other given variable.

Solving Simultaneous Equations by Elimination method

Example: Solve the pair of linear equations by the elimination method.

\(3x + 7y = 27\)
\(5x + 2y = 16\)

Let the given equations be

\(3x + 7y = 27 \ldots .(1)\)

\(5x + 2y = 16\,\,….\left( 2 \right)\)

Multiply the equation \((1)\) with \(5\) and equation \((2)\) with \(3\) to equal the co-efficients of \(x\). Then,

\(15x + 35y = 135 \ldots .(3)\)

\(15x + 6y = 48\)

Subtracting equation \((4)\) from equation \((3)\),

\(15x – 15x + 35y – 6y = 135 – 48\)

\( \Rightarrow 29y = 87\)

\( \Rightarrow y = \frac{{87}}{{29}}\)

\( \Rightarrow y = 3\)

Substitute the \(y\) value in equation \((1)\)

\(3x + 7(3) = 27\)

\( \Rightarrow 3x = 27 – 21\)

\( \Rightarrow 3x = 6\)

\( \Rightarrow x = 3\)

Hence, the solution are \(x = 3\) and \(y = 3\).

Solved Examples – Method of Elimination to Solve Simultaneous Linear Equations

Q.1. Solve the pair of linear equations by the elimination method.
\(2x + 3y = 5\)
\(3x + 4y = 7\)
Ans: Let the given equations be \(2x + 3y = 5 \ldots (1)\)
\(3x + 4y = 7 \ldots \left( 2 \right)\)
Multiply the equation \((1)\) with \(3\) and equation \((2)\) with \(2\) to equal the co-efficients of \(x\). Then,
\(6x + 9y = 15 \ldots (3)\)
\(6x + 8y = 14 \ldots .(4)\)
Subtracting equation \((4)\) from equation \((3)\),
\( \Rightarrow 6x – 6x + 9y – 8y = 15 – 14\)
\( \Rightarrow y = 1\)
Substitute the \(y\) value in equation \((1)\),
\( \Rightarrow 2x + 3(1) = 5\)
\( \Rightarrow 2x = 5 – 3\)
\( \Rightarrow 2x = 2\)
\( \Rightarrow x = 1\)
Hence, the solution are \(x = 1\) and \(y = 1\).

Q.2. Diya was given two equations \(5a – 2b = 17\) and \(3a + b = 8\), and asked to find the value of  \(a\) and \(b\). Can you help him in finding the value of  \(a\) and \(b\) using the elimination method?
Ans: The given equations are:
\(5a – 2b = 17 \ldots (1)\)
\(3a + b = 8 \ldots (2)\)
Multiply the equation \((1)\) with \(1\) and equation \((2)\) with \(2\) to equal the coefficients of \(x\). Then,
\(5a – 2b = 17 \ldots (3)\)
\(6a + 2b = 16 \ldots (4)\)
Add the equation \((3)\) and \((4)\),
\( \Rightarrow 6a + 5a – 2b + 2b = 17 + 16\)
\( \Rightarrow 11a = 33\)
\( \Rightarrow a = 3\)
Substitute the \(a\) value in equation \((1)\),
\( \Rightarrow 5(3) – 2b = 17\)
\( \Rightarrow 15 – 2b = 17\)
\( \Rightarrow – 2b = 17 – 15\)
\( \Rightarrow – 2b = 2\)
\( \Rightarrow b = \frac{2}{{ – 2}} = – 1\)
Hence, the solution are \(a = 3\) and \(b = – 1\).

Q.3. Anu has two numbers such that their sum is \(8\) and the difference between them is \(6\). Then, find the numbers by the elimination method?
Ans: Let the greater number be \(a\) and the smaller number be \(b\).
It is given that \(a + b = 8 \ldots (1)\) and \(a – b = 6 \ldots (2)\)
Add both the equations \((1)\) and \((2)\) to eliminate \(b\) variable terms. We get,
\( \Rightarrow a + a + b – b = 8 + 6\)
\( \Rightarrow 2a = 14\)
\( \Rightarrow a = \frac{{14}}{2} = 7\)
\( \Rightarrow a = 7\)
Now, substitute the value of \(a\) in equation \((1)\).
So,  \(b = 8 – 7 = 1\)
Hence, the numbers are \(7\) and \(1\).

Q.4. Solve the following pair of equations by substitution method:
\(x + y = 14\)
\(x – y = 4\)
Ans: Step-1: Let us consider the given equations as:
\(x + y = 14\,\,…\left( 1 \right)\)
\(x – y = 4\,\,…\left( 2 \right)\)
Step-2: We can pick either of the equation and write one variable in terms of the other.
Let us consider equation \((2)\),
\(x – y = 4\)
\( \Rightarrow x = 4 + y\,\,…\left( 3 \right)\)
Step-3: Put the value of \(x\) in equation \((1)\). We get,
\( \Rightarrow 4 + y + y = 14\)
\( \Rightarrow 2y = 14 – 4 = 10\)
\( \Rightarrow y = \frac{{10}}{2} = 5\)
Step-4:  Substitute the value of \(y\) in equation \((3)\) to get the value of \(x\).
\(x = 4 + y = 4 + 5 = 9\)
Hence, the solution \(x = 9\) and \(y = 5\).

Q.5. Using the elimination method to find all the possible solutions of the following pair of linear equations.
\(2x + 3y = 8\)
\(4x + 6y = 7\)
Ans: The given equations are:
\(2x + 3y = 8 \ldots (1)\)
\(4x + 6y = 7 \ldots (2)\)
Multiply the equation \((1)\) with \(2\) and equation \((2)\) with \(1\) to equal the coefficients of \(x\). Then,
\(4x + 6y = 16 \ldots (3)\)
\(4x + 6y = 7 \ldots (4)\)
Subtract the equation \((4)\) from \((3)\),
\( \Rightarrow 4x – 4x + 6y – 6y = 16 – 7\)
\( \Rightarrow 0 = 9\)
So, it’s a false statement. In this case, the two given equations representing the two straight lines are parallel lines. They never meet each other. Hence, the given pair of equations has no solution.

Summary

Among all the methods for solving linear equations, the elimination method of algebraically solving a system of linear equations is the most popular. Using basic arithmetic operations, we can eliminate any of the variables, then simplify the equation to get the value of the other variable. This article includes the definitions of linear, pair of linear, simultaneous linear equations. We focused our discussion on mainly the elimination method, its steps to solve the simultaneous linear equations and examples.

FAQs On Method To Solve Simultaneous Linear Equations

Let’s look at some of the commonly asked questions about method to solve simultaneous linear equations:

Q.1. How do you eliminate simultaneous equations?
Ans: Make the coefficients of one of the variables the same value in both equations to solve the simultaneous equations. Then, to build a new equation with only one variable, either add the equations together and subtract one equation from the other (whatever is appropriate). This is referred to as “elimination of the variable.”

Solve the resulting equation. Then, for the variable, substitute the value found.

Q.2. What is the elimination method?
Ans: The elimination method of algebraically solving a system of linear equations is the most commonly used of all the methods for solving linear equations. We can eliminate any of the variables using basic arithmetic methods, then simplify the equation to have the value of the other variable. Then we may use any of the equations to find out what that number is.

Q.3: What are the four steps for solving systems of equations by elimination?
Ans:  Four steps for solving systems of equations by elimination method are:
Step-1: Multiplying or dividing the linear equations with a non-zero number to get a common coefficient of any variables in both equations as numerically equal.
Step-2: Then add or subtract both the equations such that the same terms will get eliminated.
Step-3: Simplify the result to get a final answer of the left out variable (let’s say \(y\)) such that we will only get an answer in the form of \(y = c\), where \(c\) is any constant.
Step-4: Then, substitute this value in any of the given equations to find the value of the other given variable.

Q.4: Which are the three algebraic methods of solving simultaneous equations.
Ans: The three algebraic methods of solving simultaneous equations are
1. Substitution method
2. Elimination method
3. Cross multiplication method

Q.5: How do you solve an algebraic equation by elimination?
Ans: To get an equation in one variable, we could add or subtract the equations using the elimination method. When the coefficients of one variable are opposites, the equations are added to delete the variable. The equations are subtracted to eliminate the variable when the coefficients of one variable are equal.

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