Cross Multiplication Method
Cross multiplication method is used in solving linear equations in two variables. The simplest and easiest method of solving linear equations in two variables is done by the method of cross-multiplication. This method is mostly used when we have a pair of variables in a linear equation. Let us learn more about the cross multiplication method, the definition, the derivation, and solve a few examples for a better understanding.
Definition of Cross Multiplication Method
Cross-multiplication is a technique to determine the solution of linear equations in two variables. It proves to be the fastest method to solve a pair of linear equations.For a given pair of linear equations in two variables:
(a)1x + (b)1y + (c)1 = 0
(a)2x + (b)2y + (c)2 = 0
By using cross multiplication, the values x and y will be:
Cross Multiplication Formula
To solve linear equations with two variables, we use the cross multiplication formula:
\[\frac{x}{{{b_1}{c_2} - {b_2}{c_1}}} = \frac{{ - y}}{{{a_1}{c_2} - {a_2}{c_1}}} = \frac{1}{{{a_1}{b_2} - {a_2}{b_1}}}\]
Derivation of Cross Multiplication Method
To understand this technique, consider an arbitrary pair of linear equations (that is, with any coefficients)
(a)1x + (b)1y + (c)1 = 0
(a)2x + (b)2y + (c)2 = 0
Let us write down the coefficients in the original pair of equations as follows, in a grid fashion
(a)1 (b)1 (c)1
(a)2 (b)2 (c)2
We now will simply ignore the coefficients of \(x\) in our grid, and cross-multiply the coefficients in the remaining two columns, and subtract them:
Thus, the first part of our solution equality becomes
\[\frac{x}{{{b_1}{c_2} - {b_2}{c_1}}}\]
Next, we consider the expression below negative y, which is \({a_1}{c_2} - {a_2}{c_1}\). To write this, we ignore the column of y coefficients, and cross-multiply the coefficients in the remaining two columns, and subtract them:
Thus, the second part of our solution equality becomes,
\[\frac{{ - y}}{{{a_1}{c_2} - {a_2}{c_1}}}\]
Finally, we consider the expression below 1, which is \({a_1}{b_2} - {a_2}{b_1}\). To write this, we ignore the column of constants, and cross-multiply the coefficients in the remaining two columns, and subtract them:
Thus, the last part of our solution equality becomes
\[\frac{1}{{{a_1}{b_2} - {a_2}{b_1}}}\]
Combining all the three parts, our complete solution to the pair of linear equations becomes:
\[\frac{x}{{{b_1}{c_2} - {b_2}{c_1}}} = \frac{{ - y}}{{{a_1}{c_2} - {a_2}{c_1}}} = \frac{1}{{{a_1}{b_2} - {a_2}{b_1}}}\]
Linear Equation by Cross Multiplication Method
Linear equations with two pairs are easily solved by the cross multiplication method. As we have already have a formula, let us see how to solve linear equations with two variables by taking an example.
Suppose that we have to solve the following pair of equations:
\[\begin{array}{l}2x + 3y - 11 = 0\\3x + 2y - 9 = 0\end{array}\]
Our solution equality will be of the following form, where we have to figure out the question marks:
\[\frac{x}{?} = \dfrac{{ - y}}{?} = \frac{1}{?}\]
To figure out the term below x, we do
Thus, the first part of the solution equality is: \(\dfrac{x}{{ - 5}}\). Next, to figure out the term below negative \(y\), we do
Thus, the second part of the solution equality is: \(\dfrac{{ - y}}{{15}}\). Finally, to figure out the term below 1, we do
Thus, the last part of the solution equality is:\(\dfrac{1}{{ - 5}}\). Combining all the three, our complete solution is
\[\begin{align}&\frac{x}{{ - 5}} = \frac{{ - y}}{{15}} = \frac{1}{{ - 5}}\\ &\Rightarrow \;\;\;\left\{ \begin{gathered}\!\!\!\!\!\!\!\!\!\!\!x = \frac{{ - 5}}{{ - 5}} = 1\\y = - \left( {\frac{{15}}{{ - 5}}} \right) = 3\end{gathered} \right.\\ &\Rightarrow \;\;\;x = 1,\;y = 3\end{align}\]
Thus, the solution is
\[\begin{align}&\frac{x}{{ - 138}} = \frac{{ - y}}{{184}} = \frac{1}{{ - 23}}\\ &\Rightarrow \;\;\;\left\{ \begin{gathered}\!\!\!\!\!\!\!\!\!\!\!\!x = \frac{{ - 138}}{{ - 23}} = 6\\y = - \left( {\frac{{184}}{{ - 23}}} \right) = 8\end{gathered} \right.\\ &\Rightarrow \;\;\;x = 6,\;y = 8\end{align}\].
Unique Solution by Cross Multiplication Method
On solving the linear equations in two variables by cross multiplication method we get a unique solution, inconsistently, and infinitely many solutions. Let us understand the main points step by step.
\[\begin{array}{l}{a_1}x + {b_1}y + {c_1} = 0\\{a_2}x + {b_2}y + {c_2} = 0\end{array}\]
To obtain a unique solution:
\[\frac{{{a_1}}}{{{a_2}}} \ne \frac{{{b_1}}}{{{b_2}}}\]
Thus,
\[\begin{align}&{a_1}{b_1} \ne {a_2}{b_1}\\&\Rightarrow \;\;\;{a_1}{b_2} - {a_2}{b_1} \ne 0\end{align}\]
This means that when we write the solution equality by the cross-multiplication technique, as long as the term below 1 is non-zero, there will be a unique solution:
Inconsistent:
The slopes of the two lines should be equal, but the y intercepts should be unequal. In terms of coefficients, the required conditions are:
\[\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \ne \frac{{{c_1}}}{{{c_2}}}\]
If you do not remember this, refer again to the discussion where we derived these conditions. Now, these conditions mean that
\[\begin{array}{l}{a_1}{b_2} - {a_2}{b_1} = 0\\{b_1}{c_2} - {b_2}{c_1} \ne 0\\{a_1}{c_2} - {a_2}{c_1} \ne 0\end{array}\]
This further means that when we write the solution equality by the cross-multiplication technique, if the term below 1 is zero, and either or both of the terms below x or negative y are non-zero, the pair will be inconsistent:
Infinitely many solutions:
\[\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}\]
Thus,
\[\begin{array}{l}{a_1}{b_2} - {a_2}{b_1} = 0\\{b_1}{c_2} - {b_2}{c_1} = 0\\{a_1}{c_2} - {a_2}{c_1} = 0\end{array}\]
This means that when we write the solution equality by the cross-multiplication technique, if all the three terms (below x, below negative y and below 1) are zero, the pair will have infinitely many solutions:
Let us apply these facts to an example.
Example: Consider the following pair of linear equations:
\[\begin{array}{l}2x + 7y - 1 = 0\\ - 3x + 4y + 3 = 0\end{array}\]
Using the cross-multiplication technique, decide whether this pair is consistent or not.
We have
Thus, in the solution equality, in the last term, we will have a non-zero denominator, which means that this pair is consistent and has a unique solution.
Related Topics
Listed below are a few topics that are related to the cross multiplication method, take a look.
Cross Multiplication Method Examples
-
Example 1: Help Fredie to solve the following pair of linear equations by cross-multiplication \[\begin{array}{l}2x + 5y - 52 = 0\\3x - 4y + 14 = 0\end{array}\]
Solution: The terms below x, negative y, and 1 are calculated below
Thus, the solution is
\[\begin{align}&\frac{x}{{ - 138}} = \frac{{ - y}}{{184}} = \frac{1}{{ - 23}}\\ &\Rightarrow \;\;\;\left\{ \begin{gathered}\!\!\!\!\!\!\!\!\!\!\!\!x = \frac{{ - 138}}{{ - 23}} = 6\\y = - \left( {\frac{{184}}{{ - 23}}} \right) = 8\end{gathered} \right.\\ &\Rightarrow \;\;\;x = 6,\;y = 8\end{align}\]
Therefore, x = 6 and y = 8.
-
Example 2: Joel is unable to solve the following pair of linear equations by cross-multiplication. \[\begin{array}{l}3x + 4y = 32\\6x - 7y = - 11\end{array}\]. Can you help him?
Solution: First, we rewrite the equations in standard form, that is, in the form \(ax + by + c = 0\), since it is using this form that we developed our expressions to be used in the cross-multiplication technique:
\[\begin{array}{l}3x + 4y - 32 = 0\\6x - 7y + 11 = 0\end{array}\]
Now, we calculate the terms below x, negative y and 1 which will come in the solution equality.
Thus, the solution is
\[\begin{align}&\frac{x}{{ - 180}} = \frac{{ - y}}{{225}} = \frac{1}{{ - 45}}\\& \Rightarrow \;\;\;\left\{ \begin{gathered}\!\!\!\!\!\!\!\!\!\!x = \frac{{ - 180}}{{ - 45}} = 4\\y = - \left( {\frac{{225}}{{ - 45}}} \right) = 5\end{gathered} \right.\\ &\Rightarrow \;\;\;x = 4,\;y = 5\end{align}\]
Therefore, x = 4 and y = 5.
FAQs on Cross Multiplication Method
What is the Cross Multiplication Method?
In a cross multiplication method, the numerator of one fraction is multiplied to the denominator of another and the denominator of the first term to the numerator of another term.
How Do We Solve Using the Cross Multiplication Method?
Linear equations can be solved by using the cross multiplication formula. Using cross multiplication the solution to the pair of linear equations in two variables becomes:
\[\frac{x}{{{b_1}{c_2} - {b_2}{c_1}}} = \frac{{ - y}}{{{a_1}{c_2} - {a_2}{c_1}}} = \frac{1}{{{a_1}{b_2} - {a_2}{b_1}}}\]
What is the Condition to Get a Unique Solution?
\[\begin{align}&{a_1}{b_1} \ne {a_2}{b_1}\\&\Rightarrow \;\;\;{a_1}{b_2} - {a_2}{b_1} \ne 0\end{align}\]
As long as the term below 1 is non-zero, there will be a unique solution.
How Do You Solve Cross Multiplication?
Using cross multiplication the solution to the pair of linear equations in two variables becomes:
\[\frac{x}{{{b_1}{c_2} - {b_2}{c_1}}} = \frac{{ - y}}{{{a_1}{c_2} - {a_2}{c_1}}} = \frac{1}{{{a_1}{b_2} - {a_2}{b_1}}}\]
The above equation is also called the cross multiplication method formula.
Why Does Cross Multiply Work?
Cross multiplying fractions helps to compare them. We can use this equality to find out the value of the unknown for solving linear equations in two variables.
What Do You Mean by the Mathematical Rule of Three?
The mathematical rule of three is a method that uses proportions. By using the cross multiplication method, we can find the values of unknown variables using proportion.
How Many Solutions are There for Linear Equations in Two Variables?
There are infinitely many solutions for linear equations in two variables.
What is a Two-Variable Equation?
A linear equation in two variables is an equation in the form of \(ax + by + c=0\), where \(a, b\) & \(c\) are real numbers and the coefficients of \(x\) and \(y\), i.e \(a\) and \(b\) are not equal to zero.