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.

Below diagram can also help you remember the formula.

**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**