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In Maths, we use the cross multiplication method for solving linear equations in two variables. This is the simplest method to solve them and gives us the accurate value of the two variables. However, the method of cross multiplication is applicable only when we have a pair of linear equations in two variables.
Cross Multiplication of Equations
Cross multiplication is used in arithmetic operations to solve the equations or for finding the value of the variable. In this process, we multiply the numerator of one side with the denominator of the other side.
We have to multiply the numerator and denominator of one fraction by the denominator of the second fraction. Secondly, we can also multiply the numerator and denominator of the second fraction by the denominator of the first fraction. This will help to get an answer for the given question.
Consider that a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are two equations that need to be solved. By the method of cross-multiplication, we would find the values of the x and y variables.
x = \[\frac{b_{1}c_{2}-b_{2}c_{1}}{b_{2}a_{1}-b_{1}a_{2}}\] and y = \[\frac{c_{1}a_{2}-c_{2}a_{1}}{b_{2}a_{1}-b_{1}a_{2}}\]
Here, b2a1 – b1a2 ≠ 0
The final solution is,
\[\frac{x}{b_{1}c_{2}-b_{2}c_{1}}\] = \[\frac{y}{c_{1}a_{2}-c_{2}a_{1}}\] = \[\frac{1}{b_{2}a_{1}-b_{1}a_{2}}\]
This method given above is known as the ‘Cross-Multiplication Method’ as this technique is very useful when it comes to simplifying the solution. For memorizing the method of cross-multiplication and solving the linear equation in two variables, the diagram given below is helpful.
The arrows in the diagram indicate the multiplication of the values that are connected through the arrow. We can see that the second product is then subtracted from the first one. The result that we get is substituted as the denominator of the given variables and 1, as we can see above the arrow. The entire values that are obtained are then equated to form an equation as follows.
\[\frac{x}{b_{1}c_{2}-b_{2}c_{1}}\] = \[\frac{y}{c_{1}a_{2}-c_{2}a_{1}}\] = \[\frac{1}{b_{2}a_{1}-b_{1}a_{2}}\]
From here, we can evaluate the values of x and y, provided that b2a1 – b1a2 ≠ 0. Hence, we call this method cross-multiplication.In this method, the condition for the consistency of the given pair of linear equations in two variables has to be checked. These conditions are as follows:
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If \[\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}\], the result is a unique solution and the given pair of the linear equations in two variables are known to be consistent.
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If \[\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}\] , there would be infinite solutions and the pair of the lines would be coincident and hence, consistent and dependent.
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If \[\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\], then there would be no existing solution and the pair of the given linear equations in two variables are inconsistent.
Solved Examples
Let us consider the following examples.
Example 1: Solve the following pair of linear equations by using the cross multiplication method.
3x − 4y = 2
y − 2x = 7
Solution:
We can rewrite the above-given equation as:
3x − 4y = 2
−2x + y = 7
By the method of the cross multiplication,
\[\frac{x}{b_{1}c_{2}-b_{2}c_{1}}\] = \[\frac{y}{c_{1}a_{2}-c_{2}a_{1}}\] = \[\frac{1}{b_{2}a_{1}-b_{1}a_{2}}\]
When we substitute the values from the above given equation,
\[\frac{x}{28+2}\] = \[\frac{y}{4+21}\] = \[\frac{1}{13-8}\]
\[\frac{x}{30}\] = \[\frac{x}{25}\] = -15
Hence, x = -6 and y = -5
Example 2: Determine the value of the given variables that satisfy the following equation: 2x + 5y = 20 and 3x+6y =12.
Solution:
We have,
2x+5y = 20
3x + 6y = 12
By the cross multiplication method, we know that
\[\frac{x}{b_{1}c_{2}-b_{2}c_{1}}\] = \[\frac{y}{c_{1}a_{2}-c_{2}a_{1}}\] = \[\frac{1}{b_{2}a_{1}-b_{1}a_{2}}\]
Hence, by substituting the values that we have in the above equation, we get,
\[\frac{x}{ [(5).(12) – (6).(20)] }\] = \[\frac{y}{ [(20).(3) – (12).(2)] }\] = \[\frac{1}{ [(5).(3) – (6).(2)] }\]
\[\frac{x}{(60-120)}\] = \[\frac{y}{(60-24)}\] = \[\frac{1}{(15-12)}\]
\[\frac{x}{(-60)}\] = \[\frac{y}{36}\] = \[\frac{1}{3}\]
On solving, we get,
\[\frac{x}{-60}\] = \[\frac{1}{3}\]
Hence, x = -20
And solving for y,
\[\frac{y}{36}\] = \[\frac{1}{3}\]
Hence, y = 12
Therefore, we have, x= -20 and y = 12, which is the point at which the given two equations would intersect.
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