Solve each of the following systems of equations by the method of cross-multiplication :

Question:

Solve each of the following systems of equations by the method of cross-multiplication :

$5 a x+6 b y=28$

$3 a x+4 b y=18$

Solution:

GIVEN:

$5 a x+6 b y=28$

$3 a x+4 b y=18$

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

$5 a x+6 b y-28=0$

$3 a x+4 b y-18=0$

By cross multiplication method we get

$\frac{x}{(-18 \times 6 b)-(4 b \times(-28))}=\frac{-y}{(5 a) \times(-18)-((3 a) \times-(28))}=\frac{1}{20 a b-18 a b}$

$\frac{x}{(-108 b)-(-112 b)}=\frac{-y}{(-90 a)-(-84 a)}=\frac{1}{2 a b}$

$\frac{x}{4 b}=\frac{-y}{-6 a}=\frac{1}{2 a b}$

$\frac{x}{4 b}=\frac{y}{6 a}=\frac{1}{2 a b}$

Consider the following to calculateĀ x

$\frac{x}{4 b}=\frac{1}{2 a b}$

$x=\frac{4 b}{2 a b}$

$\Rightarrow x=\frac{2}{a}$

AndĀ 

$\frac{y}{6 a}=\frac{1}{2 a b}$

$\frac{y}{6 a}=\frac{1}{2 a b}$

$\Rightarrow y=\frac{6 a}{2 a b}=\frac{3}{b}$

Hence we get the value of $x=\frac{2}{a}$ and $\mathrm{y}=\frac{3}{b}$

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now