# Cross multiplication method for linear equations

In this chapter, we will discuss cross multiplication method of solving simultaneous linear equation.

## Solving linear equation using Cross Multiplication method

Cross multiplication method works when two linear equations are given with two unknown variables.

It basically provide you with formula that will help you find value of unknown variables.

Let the two equations are;

\mathtt{a_{1} x\ +b_{1} y\ +\ c_{1} =0}\\\ \\ \mathtt{a_{2} x\ +b_{2} y\ +\ c_{2} =0}

Then the relation of the two variables are given as;

\mathtt{\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}}}

On cross multiplication, we get the formula for x & y variable as;

\mathtt{x\ =\ \frac{b_{1} c_{2} -b_{2} c_{1}}{a_{1} b_{2} -a_{2} b_{1}}}\\\ \\ \mathtt{y\ =\frac{c_{1} a_{2} -c_{2} a_{1}}{a_{1} b_{2} -a_{2} b_{1}}}

You have to remember the above formula to solve the question using this technique.

For x variable
The numerator is given by above cross multiplication.

i.e. \mathtt{b_{1} c_{2} -b_{2} c_{1}}

The denominator is given by below cross multiplication.

Hence, the formula becomes;

\mathtt{x\ =\ \frac{\mathtt{b_{1} c_{2} -b_{2} c_{1}}}{a_{1} b_{2} -a_{2} b_{1}}}

Y variable calculation

The numerator is given by below cross multiplication.

i.e. \mathtt{c_{1} a_{2} -c_{2} a_{1}}

Denominator is given as;

i.e. \mathtt{a_{1} b_{2} -a_{2} b_{1}}

Hence, the formula becomes;
\mathtt{y\ =\frac{\mathtt{c_{1} a_{2} -c_{2} a_{1}}}{\mathtt{a_{1} b_{2} -a_{2} b_{1}}}}

Note: These formulas will not work when the coefficients of variable x and y is 0.

### Cross Multiplication method of linear equations – Solved problems

Example 01
Solve the below equation using cross multiplication method.

x + y = 7
2x – 3y = 11

Solution
First arrange these equation into cross multiplication format.

x + y – 7 = 0
2x – 3y -11 = 0

Now applying the formula for cross multiplication.

\mathtt{x\ =\ \frac{1( -11) -( -3)( -7)}{1( -3) -( 2)( 1)}}\\\ \\ \mathtt{x\ =\ \frac{-11-21}{-3-2}}\\\ \\ \mathtt{x\ =\ \frac{-32}{-5}}\\\ \\ \mathtt{x\ =\frac{32}{5}}

Now finding value of y.

\mathtt{y\ =\frac{\mathtt{c_{1} a_{2} -c_{2} a_{1}}}{\mathtt{a_{1} b_{2} -a_{2} b_{1}}}}\\\ \\ \mathtt{y\ =\ \frac{( -7)( 2) -( -11)( 1)}{1( -3) -( 2)( 1)}}\\\ \\ \mathtt{y=\ \frac{-14+11}{-3-2}}\\\ \\ \mathtt{y=\frac{-3}{-5}}\\\ \\ \mathtt{y\ =\ \frac{3}{5}}

Hence, ( 32/5, 3/5) is the solution of given expression.

Example 02
Solve the below linear equation using cross multiplication

5x – 3y – 2 = 0
4x + 7y – 6 = 0

Solution
Finding x variable;

\mathtt{x\ =\ \frac{\mathtt{b_{1} c_{2} -b_{2} c_{1}}}{a_{1} b_{2} -a_{2} b_{1}}}\\\ \\ \mathtt{x\ =\ \frac{-3( -6) -( 7)( -2)}{5( 7) -( 4)( -3)}}\\\ \\ \mathtt{x\ =\ \frac{18+14}{35+12}}\\\ \\ \mathtt{x\ =\ \frac{32}{47}}

Now find y variable.

\mathtt{y\ =\frac{\mathtt{c_{1} a_{2} -c_{2} a_{1}}}{\mathtt{a_{1} b_{2} -a_{2} b_{1}}}}\\\ \\ \mathtt{y\ =\ \frac{( -2)( 4) -( -6)( 5)}{5( 7) -( 4)( -3)}}\\\ \\ \mathtt{y=\ \frac{-8+30}{35+12}}\\\ \\ \mathtt{y=\frac{22}{47}}

Hence, (32/47, 22/47) is the solution of given equations.

Example 03
Solve the simultaneous equations using cross multiplication method.

11 = 8x + 5y
3x – 4y = 10

Solution
First arrange the equation in cross multiplication format.

8x + 5y -11 = 0
3x – 4y -10 = 0

Finding the x variable.

\mathtt{x\ =\ \frac{\mathtt{b_{1} c_{2} -b_{2} c_{1}}}{a_{1} b_{2} -a_{2} b_{1}}}\\\ \\ \mathtt{x\ =\ \frac{5( -10) -( -4)( -11)}{8( -4) -( 3)( 5)}}\\\ \\ \mathtt{x\ =\ \frac{-50-44}{-32-15}}\\\ \\ \mathtt{x\ =\ \frac{-94}{-47}}\\\ \\ \mathtt{x\ =\ \frac{94}{47}}

Finding y variable

\mathtt{y\ =\frac{\mathtt{c_{1} a_{2} -c_{2} a_{1}}}{\mathtt{a_{1} b_{2} -a_{2} b_{1}}}}\\\ \\ \mathtt{y\ =\ \frac{( -11)( 3) -( -10)( 8)}{8( -4) -( 3)( 5)}}\\\ \\ \mathtt{y=\ \frac{-33+80}{-32-15}}\\\ \\ \mathtt{y=\frac{47}{-47}}\\\ \\ \mathtt{y\ =\ -1}

Hence, (94/47, -1) is the solution of given equations.

Example 04
Solve the below linear equations.

2x+ 5y + 3 = 0
6y – 3x = 18

Solution
First arrange the equations in cross multiplication format.

2x+ 5y + 3 = 0
-3x + 6y -18 = 0

Finding the variable x.

\mathtt{x\ =\ \frac{\mathtt{b_{1} c_{2} -b_{2} c_{1}}}{a_{1} b_{2} -a_{2} b_{1}}}\\\ \\ \mathtt{x\ =\ \frac{5( -18) -( 6)( 3)}{2( 6) -( -3)( 5)}}\\\ \\ \mathtt{x\ =\ \frac{-90-18}{12+15}}\\\ \\ \mathtt{x\ =\ \frac{-108}{27}}

Dividing numerator and denominator by 9.

\mathtt{x\ =\ \frac{-108\div 9}{27\div 9}}\\\ \\ \mathtt{x\ =\ \frac{-12}{3}}\\\ \\ \mathtt{x=\ -4}

Finding variable y

\mathtt{y\ =\frac{\mathtt{c_{1} a_{2} -c_{2} a_{1}}}{\mathtt{a_{1} b_{2} -a_{2} b_{1}}}}\\\ \\ \mathtt{y\ =\ \frac{( 3)( -3) -( -18)( 2)}{2( 6) -( -3)( 5)}}\\\ \\ \mathtt{y=\ \frac{-9+36}{12+15}}\\\ \\ \mathtt{y=\frac{27}{27}}\\\ \\ \mathtt{y\ =\ 1}

Hence, (-4, 1) is the solution of given equations