The statement that two expressions are equal is known as the equation. There can be one or more variables in the expressions. Finding out the variables’ values that satisfy (make both sides equal) the given equation is known as solving an equation.

The cross multiplication method of finding the solutions is applicable only for a pair of equations involving two variables. It is not applicable for linear equations in one variable. In this article, we shall discuss how to derive the solution using this cross multiplication method.

Derivation of Cross Multiplication Method

Generally, the pair of the linear equations in two variables can be shown as:
\({a_1}x + {b_1}y + {c_1} = 0\)
\({a_2}x + {b_2}y + {c_2} = 0\)
Be a system of simultaneous linear equations in two variables \(x\) and \(y\) such that \(\frac{{{a_1}}}{{{a_2}}} \ne \frac{{{b_1}}}{{{b_2}}}\) i.e., \({a_1}{b_2} – {a_2}{b_1} \ne 0\) then the system has a unique solution given by
\(x = \frac{{\left( {{b_1}{c_2} – {b_2}{c_1}} \right)}}{{\left( {{a_1}{b_2} – {a_2}{b_1}} \right)}}\) and \(y = \frac{{\left( {{c_1}{a_2} – {c_2}{a_1}} \right)}}{{\left( {{a_1}{b_2} – {a_2}{b_1}} \right)}}\)
The given system of equations is
\({a_1}x + {b_1}y + {c_1} = 0\)…..(i)
\({a_2}x + {b_2}y + {c_2} = 0\)….(ii)

Multiplying equation (i) by \({b_2},\) (ii) by \({b_1}\) and subtracting, we get
\({b_2}\left( {{a_1}x + {b_1}y + {c_1}} \right) – {b_1}\left( {{a_2}x + {b_2}y + {c_2}} \right) = 0\)
\( \Rightarrow x\left( {{a_1}{b_2} – {a_2}{b_1}} \right) = \left( {{b_1}{c_2} – {b_2}{c_1}} \right)\)
\( \Rightarrow x = \frac{{\left( {{b_1}{c_2} – {b_2}{c_1}} \right)}}{{\left( {{a_1}{b_2} – {a_2}{b_1}} \right)}}\,\left[∵ {\left( {{a_1}{b_2} – {a_2}{b_1}} \right) \ne 0} \right]\)
Multiplying equation (i) by \({{a_2}},\) (ii) by \({{a_1}},\) and subtracting, we get
\({a_2}\left( {{a_1}x + {b_1}y + {c_1}} \right) – {a_1}\left( {{a_2}x + {b_2}y + {c_2}} \right) = 0\)
\( \Rightarrow y\left( {{a_2}{b_1} – {a_1}{b_2}} \right) + \left( {{c_1}{a_2} – {c_2}{a_1}} \right) = 0\)
\( \Rightarrow y\left( {{a_2}{b_1} – {a_1}{b_2}} \right) = \, – \left( {{c_1}{a_2} – {c_2}{a_1}} \right)\)
\( \Rightarrow y = \frac{{ – \left( {{c_1}{a_2} – {c_2}{a_1}} \right)}}{{ – \left( {{a_1}{b_2} – {a_2}{b_1}} \right)}}\,\left[ ∵{\left( {{a_1}{b_2} – {a_2}{b_1}} \right) \ne 0} \right]\)
Hence, \(x = \frac{{\left( {{b_1}{c_2} – {b_2}{c_1}} \right)}}{{\left( {{a_1}{b_2} – {a_2}{b_1}} \right)}}\) and \(y = \frac{{\left( {{c_1}{a_2} – {c_2}{a_1}} \right)}}{{\left( {{a_1}{b_2} – {a_2}{b_1}} \right)}}\)

Remark 1: The above solution is generally written as \(\frac{x}{{{b_1}{c_2} – {b_2}{c_1}}} = \frac{y}{{{c_1}{a_2} – {c_2}{a_1}}} = \frac{1}{{{a_1}{b_2} – {a_2}{b_1}}}\) Or, \(\frac{x}{{{b_1}{c_2} – {b_2}{c_1}}} = \frac{{ – y}}{{{a_1}{c_2} – {a_2}{c_1}}} = \frac{1}{{{a_1}{b_2} – {a_2}{b_1}}}.\) Remark 2: The following procedure is very helpful m determining the solution without remembering the above formula:
1. Obtain the two equations.
2. Shift all terms on LHS in the two equations to introduce zeros on RHS, i.e., write the two equations in the following form:
\({a_1}x + {b_1}y + {c_1} = 0\)
\({a_2}x + {b_2}y + {c_2} = 0\)
3. In the above system of equations, there are three columns viz. column containing x i.e. \(\left( {\frac{{{a_1}}}{{{a_2}}}} \right)\) column containing y i.e., \(\left( {\frac{{{b_1}}}{{{b_2}}}} \right)\) and column containing constant terms i.e. \(\left( {\frac{{{c_1}}}{{{c_2}}}} \right).\) To obtain the solution, write \(x, y\) and \(1\) separated by equality signs as shown below:
In the denominator of \(x\) leave column containing \(x\) and write remaining two columns in the same order, in the denominator of \(-y\) leave column containing \(y\) and write the remaining two columns. Similarly, in the denominator of one, write columns containing \(x\) and \(y.\)
Mark crossed-arrows pointing downward from top to bottom and pointing upwards from bottom to top as shown above.
The arrows between two numbers indicate that the numbers are to be multiplied.
4. To obtain the denominators of \(x, y\) and \(1,\) multiply the numbers with downward arrow and from their product subtract the product of the numbers with upward arrow. Applying this, we get
\(\frac{x}{{{b_1}{c_2} – {b_2}{c_1}}} = \frac{y}{{{c_1}{a_2} – {c_2}{a_1}}} = \frac{1}{{{a_1}{b_2} – {a_2}{b_1}}}\)
5. Obtain the value of \(x\) by equating the first and third expressions in step IV. The value of \(y\) is obtained by equating second and third expressions in step IV.

Cross-multiplication Method Quadratic

The cross multiplication method is also applicable for deriving the common solutions (common root) for a set of the quadratic equation also.
If we have two quadratic equations given by \({a_1}{x^2} + {b_1}x + {c_1} = 0\) and \({a_2}{x^2} + {b_2}x + {c_2} = 0\) and, if they have a common solution \(\alpha ,\) using the cross multiplication method, we can derive the value of this common solution \(\alpha .\)
In this case, we can write \(\frac{{{\alpha ^2}}}{{{b_1}{c_2} – {b_2}{c_1}}} = \frac{\alpha }{{{c_1}{a_2} – {c_2}{a_1}}} = \frac{1}{{{a_1}{b_2} – {a_2}{b_1}}}.\)
We can find the value of the common solution \(\alpha \) using either pair of the relations.

Solving Linear Equations by Cross Multiplication Method

We can provide an example of how to solve the linear equation by using the cross multiplication method.

Example-1: Assume
\(2x + 3y – 11 = 0\)
\(3x + 2y – 9 = 0\)
Here the solution equality will be the following form, where we have to figure out the question marks:
\(\frac{x}{?} = \frac{{ – y}}{?} = \frac{1}{?}\)
To identify the term below \(x,\) we have to do Now, the first part of the solution equality is \(\frac{x}{{ – 5}}.\) Now, to find out the term below the negative \(y,\) we have to do Hence, the second part of the solution equality is \(\frac{{ – y}}{{15}}.\) Now, to find out the term below \(1,\) we have to do

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Thus, the last part of the solution equality is \(\frac{1}{{ – 5}}.\) When we combine all the three, we get
\(\frac{x}{{ – 5}} = \frac{{ – y}}{{15}} = \frac{1}{{ – 5}}\)
\( \Rightarrow \left\{ {\begin{array}{*{20}{c}} {x = \frac{{ – 5}}{{ – 5}} = 1}\\ {y = \, – \left( {\frac{{15}}{{ – 5}}} \right) = 3} \end{array}} \right.\)
\( \Rightarrow x = 1,\,y = 3\)
Example-2: If the two equations are
\(2x + 5y – 52 = 0\)
\(3x + 4y – 14 = 0\)
Then, their solutions are derived as follows. Hence, the solution will be as shown below:
\(\frac{x}{{ – 138}} = \frac{{ – y}}{{184}} = \frac{1}{{ – 23}}\)
\( \Rightarrow \left\{ {\begin{array}{*{20}{c}} {x = \frac{{ – 138}}{{ – 23}} = 6}\\ {y = \, – \left( {\frac{{184}}{{ – 23}}} \right) = 8} \end{array}} \right.\)
\( \Rightarrow x = 6,\,y = 8\)

Cross-multiplication Method Calculator

To cross multiply the equations, you have to multiply the numerator in the first fraction and the denominator in the second fraction, and then you have to write that number down. Later, you have to multiply the numerator of the second fraction and the number in the denominator of the first fraction, and you have to write the number you have obtained down.

Solved Examples: Cross Multiplication for Solving Linear Equations

Q.1. Solve the following system of equations in \(x\) and \(y.\)
\(\left( {a – b} \right)x + \left( {a + b} \right)y = {a^2} – 2ab – {b^2}\)
\(\left( {a + b} \right)\left( {x + y} \right) = {a^2} + {b^2}\)
Ans: The given system of equations may be written as
\(\left( {a – b} \right)x + \left( {a + b} \right)y – \left( {{a^2} – 2ab – {b^2}} \right) = 0\)
\(\left( {a + b} \right)x + \left( {a + b} \right)y – \left( {{a^2} + {b^2}} \right) = 0\)
By cross-multiplication, we have
\(\frac{x}{{\left( {a + b} \right) \times – \left( {{a^2} + {b^2}} \right) – \left( {a + b} \right) \times – \left( {{a^2} – 2ab – {b^2}} \right)}} – \frac{y}{{\left( {a – b} \right) \times – \left( {{a^2} + {b^2}} \right) – \left( {a + b} \right) \times – \left( {{a^2} – 2ab – {b^2}} \right)}} = \frac{1}{{\left( {a – b} \right)\left( {a + b} \right) – {{\left( {a + b} \right)}^2}}}\)
\( \Rightarrow \frac{x}{{ – \left( {a + b} \right)\left( {{a^2} + {b^2}} \right) + \left( {a + b} \right)\left( {{a^2} – 2ab – {b^2}} \right)}} = \frac{{ – y}}{{ – \left( {a – b} \right)\left( {{a^2} + {b^2}} \right) + \left( {a + b} \right) + \left( {{a^2} – 2ab – {b^2}} \right)}} = \frac{1}{{ – \left( {a – b} \right) + \left( {a + b} \right){{\left( {a + b} \right)}^2}}}\)
\( \Rightarrow \frac{x}{{ – \left( {a + b} \right)\left\{ { – \left( {{a^2} + {b^2}} \right) + \left( {{a^2} – 2ab – {b^2}} \right)} \right\}}}\frac{{ – y}}{{ – \left( {a – b} \right)\left( {{a^2} – 2ab – {b^2}} \right) – \left( {a – b} \right)\left( {{a^2} + {b^2}} \right)}} = \frac{1}{{\left( {a + b} \right)\left( { – a – b – a – b} \right)}}\)
\( \Rightarrow x = \frac{{ – \left( {a + b} \right)\left\{ { – \left( {{a^2} + {b^2}} \right) + \left( {{a^2} – 2ab – {b^2}} \right)} \right\}}}{{ – 2{{\left( {a + b} \right)}^2}}}\)
and \(y = \frac{{ – \left( {a + b} \right)\left( {{a^2} – 2ab – {b^2}} \right) – \left( {a – b} \right)\left( {{a^2} + {b^2}} \right)}}{{ – 2{{\left( {a + b} \right)}^2}}}.\)

Q.2. Solve the given linear equation using the cross multiplication method: \(3x – 4y = 2,\,y – 2x = 7\)
Ans: The above equation you can write as:
\(3x – 4y = 2 \Rightarrow 3x – 4y – 2 = 0\)
\( – 2x + y = 7 \Rightarrow – 2x + y – 7 = 0\)
By the method of cross-multiplication,
\(\frac{x}{{{b_1}{c_2} – {b_2}{c_1}}} = \frac{y}{{{c_1}{a_2} – {c_2}{a_1}}} = \frac{1}{{{b_2}{a_1} – {b_1}{a_2}}}\)
Now, you have to substitute the value in the above equation;
\(\frac{x}{{28 + 2}} = \frac{y}{{4 + 21}} = \frac{1}{{3 – 8}}\)
\( \Rightarrow \frac{x}{{30}} = \frac{y}{{25}} = \, – \frac{1}{5}\)
\( \Rightarrow x = \, – 6,\,y = \, – 5.\)

Q.3. Find the value of variables that satisfies the given equation: \(2x + 5y = 20\) and \(3x + 6y = 12.\)
Ans: Given,
\(2x + 5y = 20 \Rightarrow 2x + 5y – 20 = 0\)
\(3x + 6y = 12 \Rightarrow 3x + 6y – 12 = 0\)
By using the cross multiplication method, we know that,
\(\frac{x}{{{b_1}{c_2} – {b_2}{c_1}}} = \frac{y}{{{c_1}{a_2} – {c_2}{a_1}}} = \frac{1}{{{b_2}{a_1} – {b_1}{a_2}}}\)
Therefore, by substituting the values in the above equation, we get,
\(\frac{x}{{\left[ {\left( 5 \right).\left( { – 12} \right) – \left( 6 \right).\left( { – 20} \right)} \right]}} = \frac{y}{{\left[ {\left( { – 20} \right).\left( 3 \right) – \left( { – 12} \right).\left( 2 \right)} \right]}} = \frac{1}{{\left[ {\left( 2 \right).\left( 6 \right) – \left( 3 \right).\left( 5 \right)} \right]}}\)
\(\frac{x}{{\left( { – 60 + 120} \right)}} = \frac{y}{{\left( { – 60 + 24} \right)}} = \frac{1}{{\left( {12 – 15} \right)}}\)
\(\frac{x}{{60}} = \frac{y}{{ – 36}} = \frac{1}{{ – 3}}\)
Now, \(\frac{x}{{60}} = \frac{1}{{ – 3}}\)
\(x = \, – 20\)
Ans \(\frac{y}{{ – 36}} = \frac{1}{{ – 3}}\)
\(y = 12\)
Hence, \(x = \, – 20\) and \(y = 12\) is the point where the given equations intersect.

Q.4. Find the value of \(x\) in the given expression, result obtained by using the cross multiplication: \(\frac{x}{2} = \frac{1}{4}\)
Ans: Given,
\(\frac{x}{2} = \frac{1}{4}\)
\(x = \frac{2}{4}\)
\(x = \frac{1}{2}\)
Hence, the value of \(x\) is \(1.\)

Q.5. Find the value of \(x\) in the given expression, result obtained by using the cross multiplication: \(\frac{x}{3} = \frac{1}{{18}}\)
Ans: Given,
\(\frac{x}{3} = \frac{1}{{18}}\)
\(x = \frac{3}{{18}}\)
\(x = \frac{1}{6}\)
Hence, the value of \(x\) is \(\frac{1}{6}.\)

Summary

In the given article, we discussed the term cross multiplication for solving the linear equation, the cross multiplication method, and then we solved some examples. We also talked about the derivative cross multiplication method followed by the cross multiplication method quadratic. We discussed solving the linear equation using the cross multiplication method and briefed about the calculator. We have provided some of the solved examples along with a few FAQs.

Frequently Asked Questions (FAQs)

Q.1. How do you do cross multiplication in a linear equation in one variable?
Ans: In the case of linear equations, the cross multiplication method of solving is applicable for linear equations in two variables only. It is not appliable for linear equations in one variable.

Q.2. How do you cross multiply equations?
Ans: In the arithmetic operations, we use the process known as cross multiplication to ease the equations or to identify the value of the variable. This term is the process of multiplying the numerator of one side to the denominator of the other side, this effectively crossing the terms over.

Q.3. What is the formula of the cross multiplication method of pair of linear equations in two variables?
Ans: To figure out the solution of the pair of linear equations, you have to use the cross multiplication method.
If \({a_1}x + {b_1}y + {c_1} = 0\) and \({a_2}x + {b_2}y + {c_2} = 0\) are the two linear equations, so we can figure out the value of \(x\) and \(y\) by using this method. The solution is as follows:
\(\frac{x}{{{b_1}{c_2} – {b_2}{c_1}}} = \frac{y}{{{c_1}{a_2} – {c_2}{a_1}}} = \frac{1}{{{b_2}{a_1} – {b_1}{a_2}}}.\)

Q.4. How do you do cross multiply multiplication?
Ans: You need to multiply the numerator and the denominator of the first fraction by the denominator of the second number.
Secondly, you have to multiply the numerator and the denominator of the second fraction by the denominator of the first fraction. Finally, you acquire the answer.

Q.5. Why do we use the cross multiplication method in linear algebra?
Ans: To figure out the pair of linear equation solutions, you have to use the cross multiplication method.
If \({a_1}x + {b_1}y + {c_1} = 0\) and \({a_2}x + {b_2}y + {c_2} = 0\) are the two linear equations, so we can figure out the value of \(x\) and \(y\) by using this method. The solutions are
\(\frac{x}{{{b_1}{c_2} – {b_2}{c_1}}} = \frac{y}{{{c_1}{a_2} – {c_2}{a_1}}} = \frac{1}{{{b_2}{a_1} – {b_1}{a_2}}}.\)

Q.6. How do you cross multiply on a calculator?
Ans: The process of how to use the cross-multiplication calculator is given below:
1. First, you have to enter the fractions with the unknown value \(x\) in the respective input field.
2. Secondly, you have to click on the button calculate \(x\) to receive the output.
3. Later, the unknown value \(x\) will be displayed in the output field \(x.\)
4. Finally, \(\frac{a}{b} = \frac{c}{d}.\)

Q.7. What does it mean to cross simplify?
Ans: When a numerator or the denominator gets simplified, you will cross it out with the slash and write the new numerator or the denominator next to it. The number you divide by the other number does not get indicated in any way.

Q.8. Why can we cross multiply?
Ans: By comparing fractions using cross-multiplication, you can lose the concept of finding the equivalent fractions; that is ​where the ​cross-multiplication works. This property states that if you multiply both the sides of an equation or the inequality by the same number, each side’s values remain equal.

We hope this detailed article on cross multiplication to solve linear equations helped you in your studies. If you have any doubts, queries or suggestions regarding this article, feel to ask us in the comment section and we will be more than happy to assist you. Happy learning!