Solution of an Equation: A mathematical equation is a statement that two expressions are equal. 4 + 4=8 is a mathematical expression. An algebraic equation is a mathematical formula that has one or more variables. As a result, (2x + 5 = 35) is a one-variable linear equation. It has the first-degree variable (x), and its graph is a straight line.

Solving an equation is the process of determining the number that a variable represents. The obtained value is referred to as the equation’s solution. A mathematical language that asserts that two algebraic expressions must be equal in nature is known as an equation.

Define Solution of an Equation

The solution of an equation is the array of all values that, when replaced for unknowns, make an equation true. For equations requiring one unknown, raised to a power one, two basic algebra rules involving the additive property and the multiplicative property are used to decide its solutions.

Solution of an Equation Notes

The solution of an equation notes are explained below:

Methods to Solve Linear Equation in One Variable

1. Transposition method for solving linear equations in one variable
2. Cross-multiplication method for a solution to the linear equation in one variable

Transposition Method for Solving Linear Equations in One Variable

Sometimes the two sides of an equation contain variable (unknown quantity) and constants (numerals). In such cases, we first simplify two sides in their easiest forms and then transpose (shift) terms containing variable on \(RHS\) to \(LHS\) and constant terms on \(LHS\) to \(RHS\) By transposing a term from one side to another, we mean changing its sign and carrying it to the other side.

The transposition method involves the following steps:

Step I: Get the linear equation.
Step II: Recognize the variables and constants.
Step III: Shorten the \(LHS\) and \(RHS\) to their most simple forms by eliminating brackets.
Step IV: Transpose all terms, which includes the variable on \(LHS\) and constant terms on \(RHS.\)
Step V: Simplify \(LHS\) and \(RHS\) in the most basic form so that every side contains just one term.
Step VI: Solve the equation received in step \(V\) by dividing the two sides by the coefficient of the variable on \(LHS.\)
For example, \(3x + 3 = 2x + 4\)
\( \Rightarrow 3x – 2x = 4 – 3 \Rightarrow x = 1\)

Cross-multiplication Method for a Solution to the Linear Equation in One Variable

The procedure of multiplying a numerator on \(LHS\) with a denominator on \(RHS\) and equating it to the product of a common denominator on \(LHS\) with a numerator on \(RHS\) is called cross-multiplication.
By utilizing cross-multiplication, we can convert an equation of the form
\(\frac{{ax + b}}{{cx + d}} = \frac{m}{n}\)
to a linear equation \(n\left({ax + b} \right) = m\left({cx + d} \right.)\)
For example,\(\frac{{x + 1}}{{2x + 3}} = \frac{2}{3}\)
Answer: \(3\left({x + 1} \right) = 2\left({2x + 3} \right)\)
\( \Rightarrow 3x + 3 = 4x + 6\)
\( \Rightarrow x = – 3\)

Rules for a solution to the Linear Equations in One Variable

Solving an equation means determining its roots, i.e., determining the variable’s value, which satisfies it.

Rules for a solution to the linear equations in one variable are

Rule 1: Same quantity can be added to both sides of an equation without having to change the equality.
Rule 2: Same quantity can be subtracted from either side of an equation without changing the equality.
Rule 3: Both sides of an equation may be multiplied by an identical non-zero number without changing the equality.
Rule 4: Both sides of an equation can be divided by an identical non-zero number without changing the equality.
It should be noted that some complex equations can be solved by using two or more of these rules collectively.
For example, \(\frac{{2x}}{3} + 2 = \frac{3}{2}\)
Answer: \(\frac{{2x}}{3} + 2 – 2 = \frac{3}{2} – 2\) (Subtracting both the sides by \(2\))
\( \Rightarrow \frac{{2x}}{3} = \frac{{3 – 4}}{2}\)
\( \Rightarrow \frac{{2x}}{3} \times 3 = \frac{{ – 1}}{2} \times 3\) (Multiplying both the sides by \(3\))
\( \Rightarrow 2x = \frac{{ – 3}}{2}\)
\( \Rightarrow \frac{{2x}}{2} = \frac{{ – 3}}{4}\) (Dividing both the sides by \(2\))
\( \Rightarrow x = \frac{{ – 3}}{4}\)

Applications of Linear Equations to Practical Problems

In this section, we will study the formulation and solution of some practical problems. These problems involve relations between the unknown quantities, and numerals are often stated in words.
The following steps must be followed to find a solution to a word problem:
Step I: Read out the problem thoroughly and note what is provided and what is required.
Step II: Refer to the unknown quantity by certain letters say \(x,y,z,\) etc.
Step III: Translate the declarations made of the problem into mathematical words.
Step IV: When you use the condition (s) provided in the problem, form the equation.
Step V: Solve the equation for the unknown.
Step VI: Check whether the solution fulfils the equation.
For example,
Ravi is now \(7\) years older than Mahesh. If the sum of their age is \(33\) years, find the age of each.
Answer: Let Mahesh’s age \( = x\) years
Then, Ravi’s age \( = \) Mahesh’s age \( + 7 = \left({x + 7} \right)\) years
According to the problem, Ravi’s age \( + \) Mahesh’s age \( = 33.\) i.e. \(x + \left({x + 7} \right) = 33\)
\(\Rightarrow 2x + 7 = 33 \Rightarrow 2x = 26 \Rightarrow x = 13\)
Therefore, Mahesh’s age \( = 13\) years, Ravi’s age \( = 20\) years.

Conditions for the solvability of a pair of simultaneous linear equations in two variables.

Consistent system: A system of simultaneous linear equations is consistent if it has at least one solution.

1. When, \(\frac{{{a_1}}}{{{a_2}}} \ne \frac{{{b_1}}}{{{b_2}}},\) 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: A system of simultaneous linear equations is thought to be inconsistent if it has no solution.

1. When \(\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \ne \frac{{{c_1}}}{{{c_2}}},\) there is no solution. In this case the two lines representing the equations are parallel to each other.

Solution of Simultaneous Linear Equations in Two Variables

A couple of linear equations in two variables is said to form a simultaneous linear equation.

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

The most used algebraic methods of solving simultaneous linear equations in two variables are:

1. Method of elimination by substitution.
2. Method of elimination by equating the coefficients.
3. Method of cross-multiplication.

Method of Elimination by Substitution: In this technique, we express one of the variables in terms of the other variable from each of the two equations. This expression is put in the other equation to obtain an equation in one variable, as explained in the following algorithm.
Step I: Get the two equations.
Let the equations be \({a_1}x + {b_1}y + {c_1} = 0 – – – (1)\) and \({a_2}x + {b_2}y + {c_2} = 0 – – – (2)\)
Step II: Choose either of the two equations, say \(\left( 1 \right)\) and find the value of one variable, say \(y,\) in terms of the other, i.e. \(x.\)
Step III: Substitute the value of \(y,\) received in step \(II,\) in the other equation, i.e.\(\left( 2 \right),\) to get an equation in \(x.\)
Step IV: Solve the equation received in step \(III\) to get the value of \(x.\)
Step V: Substitute the value of \(x\) obtained in step \(IV\) in the expression for \(y\) in terms of \(x\) obtained in step \(II\) to get the value of \(y.\)
Step VI: The values of \(x\) and \(y\) obtained in steps \(IV\) and \(V,\) respectively, constitute the solution of the given system of two linear equations.

(Refer to question-4 of Solved Examples below)

Method of Elimination by Equating the Coefficients

In this technique, we remove one of the two variables to get an equation in one variable, which can easily be solved. Then, putting the value of this variable in any one of the specified equations, the value of the other variable can be obtained. The following algorithm explains the procedure.
Step I: Obtain the two equations.
Step II: Multiply the equations to make the coefficients of the variable to be eliminated equally.
Step III: Add or subtract the equations obtained in step \(II\) according to the terms having the same coefficients are of opposite or of the same sign.
Step IV: Solve equation in one variable, which got in step \(III.\)
Step V: Substitute the value found in step \(IV\) in any one of the given equations and find the value of the other variable.
The values of the variables in steps \(IV\) and \(V\) form the solution of the system of equations.
For example,
Solve for \(x\) and \(y,\)in \(2x + 3y = 5\) and \(3x – y = 0\)
Answer: Let \(2x + 3y = 5 – – – (1)\) and \(3x – y = 0 – – – (2)\)
Now, by multiplying the equation \(\left( 2 \right)\) by \(3,\) we get \(9x – 3y = 0 – – – – – \left( 3 \right)\)
By adding equation \(\left( 1 \right)\) and \(\left( 3 \right)\) we have \(2x + 3y + \left({9x – 3y} \right) = 5\)
\( \Rightarrow 11x = 5 \Rightarrow x = \frac{5}{{11}}\) Then, \(3 \times \frac{5}{{11}} – y = 0 \Rightarrow y = \frac{{15}}{{11}}\)
Hence, \(x = \frac{5}{{11}}\) and \(y = \frac{{15}}{{11}}.\)

Method of Cross-multiplication

Let \({a_1}x + {b_1}y + {c_1} = 0\) and \({a_2}x + {b_2}y + {c_2} = 0\) be a system of simultaneous linear equations in two variables \(x\) and \(y.\)
To find solutions using the cross-multiplication method we have,
\(x = \frac{{\left({{b_1}{c_2} – {b_2}{c_1}} \right)}}{{\left({{a_1}{b_2} – {a_2}{b_1}} \right)}}\) and \(y = \frac{{\left({{c_1}{a_2} – {c_2}{a_1}} \right)}}{{\left({{a_1}{b_2} – {a_2}{b_1}} \right)}}\)

Solution of a Quadratic Equation

Quadratic Equation: If \(p\left( x \right)\) is a quadratic function, then \(p\left( x \right) = 0\) is called a quadratic equation.
The general form of a quadratic equation is \(a{x^2} + bx + c = 0,\) where \(a,b,c \in R\) and \(a \ne 0.\)

Roots of a Quadratic Equation: Let \(p\left( x \right) = 0\) be a quadratic equation, then the roots of the function \(p\left( x \right)\) are called the roots of the equation \(p\left( x \right) = 0.\)
Thus, \(x = k\) is a root of \(p\left( x \right) = 0\) if and only if \(p\left( k \right) = 0.\)
Finding the roots of a quadratic equation is known as the solution to the quadratic equation.

Solution of Quadratic Equation by Splitting the Middle Term

Let the quadratic equation be \({x^2} + px + q = 0\)
If any factor is common in all terms of the given expression, take this factor out and write the remaining expression in brackets.
Find \(p,\) which is the coefficient of \(x,\) and \(q,\) which is the constant term.
Find out factors \(a\) and \(b\) of \(q\) whose sum is \(p.\)
Then, write \({x^2} + px + q = {x^2} + (a + b)x + q = \left({{x^2} + ax} \right) + (bx + q)\) and take out common factors in each bracket.
How to find \(a\) and \(b:\)
\(p\) will be split into \(a\) and \(b\) in such a way that we have \(p = a + b\) and \(q = a \times b\)
If the sign of \(q\) is positive, then both the factors \(a\) and \(b\) of \(q\) will have the sign same as that of \(p,\) i.e., if \(p\) is positive, then both \(a\) and \(b\) will be positive, but if \(p\) is negative, then both \(a\) and \(b\) will be negative.
If \(q\) is negative, then find \(\left| q \right|\) (Numerical value of \(q \)), then the numerically greater factor of \(q\) will have sign same as that of \(p,\) and smaller factor will have sign opposite to that of \(p.\)

Solution of Quadratic Equation by Using the Quadratic Formula

We can find the zeros of a quadratic equation \(\left({a{x^2} + bx + c = 0} \right)\) we have the quadratic formula.
That is, \(x = \frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}\)
For example,
Find the roots of \({x^2} + x – 1 = 0.\)
Answer: Here, \(a = 1,b = 1\) and \(c = – 1\)
We know that, \(x = \frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}\)
By substituting \(a,b,c\) in quadratic formula, we get
\(x = \frac{{ – 1 \pm \sqrt {{1^2} – 4 \times 1 \times \left({ – 1} \right)} }}{{2 \times 1}}\)
\(\Rightarrow x = \frac{{ – 1 \pm \sqrt 5 }}{2}\)
\( \Rightarrow x = \frac{{ – 1 + \sqrt 5 }}{2}\) or \(x = \frac{{ – 1 – \sqrt 5 }}{2}\)
Hence, the roots of the given quadratic equation is \(x = \frac{{ – 1 + \sqrt 5 }}{2}\) and \(x = \frac{{ – 1 – \sqrt 5 }}{2.}\)

Solution of Quadratic Equation by Completing Square Method

We may use the following algorithm to obtain the roots of a quadratic equation by using the method of completing squares.
Step I: Obtain the quadratic equation. Let the quadratic equation be \(a{x^2} + bx + c = 0,a = 0.\)
Step II: Make the coefficients of \({x^2}\) unity by dividing throughout by it if it is not unity, i.e., obtain \({x^2} + \frac{b}{a}x + \frac{c}{a} = 0.\)
Step III: Shift the constant term \(\frac{c}{a}\) on \(RHS\) to get \({x^2} + \frac{b}{a}x + \frac{c}{a} = 0.\)
Step IV: Add the square of half of the coefficient of \(x\) i.e. \({\left({\frac{b}{{2a}}} \right)^2}\) on both sides to obtain
\({x^2} + 2\left({\frac{b}{{2a}}} \right)x + {\left({\frac{b}{{2a}}} \right)^2} = {\left({\frac{b}{{2a}}} \right)^2} – \frac{c}{a}\)
Step V: Write \(LHS\) as the perfect square of a binomial expression and simplify \(RHS\) to get
\({\left({x + \frac{b}{{2a}}} \right)^2} = \frac{{{b^2} – 4ac}}{{4{a^2}}}\)
Step VI: Take the square root of both sides to get \(x + \frac{b}{{2a}} = \pm \sqrt {\frac{{{b^2} – 4ac}}{{4{a^2}}}} \)
Step VII: Obtain the values of \(x\) by shifting the constant term \(\frac{b}{{2a}}\) on \(RHS.\)

(Refer to question-3 of Solved Examples below)

Solved Examples – Solution of an Equation

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

Q.2. Solve \(\frac{{7y + 2}}{5} = \frac{{6y – 5}}{{11}}\)
Ans: The given equation is \(\frac{{7y + 2}}{5} = \frac{{6y – 5}}{{11}}\)
\( \Rightarrow 11\left({7y + 2} \right) = 5\left({6y – 5} \right)\)
\( \Rightarrow 77y + 22 = 30y – 25\)
\( \Rightarrow 77y – 30y = – 25 – 22\)
\( \Rightarrow 47y = – 47\)
\( \Rightarrow y = – 1\)
Hence, the value of \(y\) is \( – 1.\)

Q.3. Find the roots of the equation \(5{x^2} – 6x – 2 = 0\) by the method of completing the square.
Ans: We have, \(5{x^2} – 6x – 2 = 0\)
\( \Rightarrow {x^2} – \frac{6}{5}x – \frac{2}{5} = 0\) [Dividing throughout by \(5\)]
\( \Rightarrow {x^2} – 2\left({\frac{3}{5}} \right)x + {\left({\frac{3}{5}} \right)^2} = \frac{2}{5} + {\left({\frac{3}{5}} \right)^2}\)
\( \Rightarrow {\left({x – \frac{3}{5}} \right)^2} = \frac{{19}}{{25}}\)
\(\Rightarrow x – \frac{3}{5} = \pm \frac{{\sqrt {19} }}{5}\)
\( \Rightarrow x = \frac{3}{5} \pm \frac{{\sqrt {19} }}{5} = \frac{{3 \pm \sqrt {19} }}{5}\)
\( \Rightarrow x = \frac{{3 + \sqrt {19} }}{5}\) or \(x = \frac{{3 – \sqrt {19} }}{5}\)
Hence, the roots of the given equation are \(x = \frac{{3 + \sqrt {19} }}{5}\) and \(x = \frac{{3 – \sqrt {19} }}{5.}\)

Q.4. Solve the following system of linear equations \(x – y = 1\) and \(2x + y = 8.\)
Ans: Let \(x – y = 1 – – – (1)\) and \(2x + y = 8 – – – (2)\)
From equation \(\left( 1 \right),\) by solving \(x\) value in terms of \(y,\) we have
\(x = 1 + y – – – (3)\)
By substituting equation \(\left( 3 \right)\) in equation \(\left( 2, \right)\) we get
\(2\left({1 + y} \right) + y = 8\)
\( \Rightarrow 2 + 2y + y = 8\)
\( \Rightarrow 3y = 6\)
\( \Rightarrow y = 2\)
By substituting \(y = 2\) in the equation \(\left( 3, \right)\) we get
\(x = 3\)
Hence, \(x = 3\) and \(y = 2.\)

Q.5. Solve the quadratic equation \({x^2} – 9 = 0\) by factorisation method.
Ans: We have \({x^2} – 9 = 0\)
The given equation can be written as \({x^2} – {3^2} = 0\)
\(\Rightarrow \left({x + 3} \right)\left({x – 3} \right) = 0\)
\( \Rightarrow \left({x + 3} \right) = 0\) or \(\left({x – 3} \right) = 0\)
\( \Rightarrow x = – 3\) or \(x = 3\)
Thus, \(x = 3\) and \(x = -3\) are the roots of the given equation.

Summary

You can use math equations to solve an equation or a system of equations. The equation is a statement that states that two expressions are equal. In the vast majority of cases, accurate solutions to the presented math equations may be found. However, most of the time it is not feasible to obtain an exact answer, but approximate solutions that equal the precision of the precise solution can be found. It deals with quadratic math problems and specific instructions. A mathematical language that asserts that two algebraic expressions must be equal in nature is known as an equation.

In this, we learnt about the definition of the solution of an equation, notes on the solution of an equation, rules for solving linear equations in one variable, applications of linear equations to practical problems, solution of simultaneous linear equations in two variables, solution of a quadratic equation, solved examples on the solution of an equation.

The learning outcome of this article helps to translate a word problem in the form of an equation is known as the formulation of the problem. Thus, the procedure of solving a word problem is made up of two parts, such as formulation and solution.

Frequently Asked Questions – Solution of an Equation

Q.1. What is a solution to an algebraic equation?
Ans: The solution of an equation is the array of all values that, when replaced for unknowns, make an equation true. For equations requiring one unknown, raised to a power one, two basic algebra rules involving the additive property and the multiplicative property are used to decide its solutions.

Q.2. How do you solve equations with a variable?
Ans: One of the methods to solve equations with a variable is the transposition method. The transposition method involves \(6\) steps.
Step I: Get the linear equation.
Step II: Recognize the variables and constants.
Step III: Shorten the \(LHS\) and \(RHS\) to their most simple forms by eliminating brackets.
Step IV: Transpose all terms which includes the variable on \(LHS\) and constant terms on \(RHS.\)
Step V: Simplify \(LHS\) and \(RHS\) in the most basic form so that every side contains just one term.
Step VI: Solve the equation received in step \(V\) by dividing the two sides by the coefficient of the variable on \(LHS.\)

Q.3. What are the \(4\) rules to solving an equation?
Ans: Rule 1: Same quantity can be added to both sides of an equation without having to change the equality.
Rule 2: Same quantity can be subtracted from either side of an equation without changing the equality.
Rule 3: Both sides of an equation may be multiplied by an identical non-zero number without changing the equality.
Rule 4: Both sides of an equation can be divided by an identical non-zero number without changing the equality.
It should be noted that some complex equations can be solved by using two or more of these rules collectively.

Q.4. What is the formula of no solution?
Ans: A system of simultaneous linear equations is thought to be inconsistent if it has no solution.
When \(\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \ne \frac{{{c_1}}}{{{c_2}}},\) the two lines are parallel and hence, there is no solution.

Q.5. How do you solve pair of simultaneous equations with two variables?
Ans: One of the methods to solve the equation with two variables is the elimination method.
Step I: Obtain the two equations.
Step II: Multiply the equations to make the coefficients of the variable to be eliminated equally.
Step III: Add or subtract the equations obtained in step \(II\) according to the terms having the same coefficients are of opposite or of the same sign.
Step IV: Solve equation in one variable, which got in step \(III.\)
Step V: Substitute the value found in step \(IV\) in any one of the given equations and find the value of the other variable.
The values of the variables in steps \(IV\) and \(V\) form the solution of the system of equations.

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