Solve the following systems of equations by the method of cross-multiplication:
$3 x + 2 y + 25 = 0$
$2 x + y + 10 = 0$

Open in App

Solution

$3 x + 2 y + 25 = 0 . . . (1)$
$2 x + y + 10 = 0 . . . (2)$
From equation $(1) : a_{1} = 3, b_{1} = 2$ and $c_{1} = 25$
From equation $(2) : a_{2} = 2, b_{2} = 1$ and $c_{2} = 10$

Using cross multiplication method:

$\frac{x}{20 - 25} = \frac{y}{50 - 30} = \frac{1}{3 - 4}$

$= \frac{x}{- 5} = \frac{y}{20} = \frac{1}{- 1} = - 1$

$\frac{x}{- 5} = - 1$

$∴$ $x = 5$

$\frac{y}{20} = - 1 ⟹ y = - 20$

Thus $x = 5, y = - 20$

Suggest Corrections

Similar questions

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

$3 x + 2 y + 25 = 0$ ,

$2 x + y + 10 = 0.$

x + 2 y + 1 = 0

2 x - 3 y - 12 = 0

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

3x+2y+25=0

2x+y+10=0

\frac{5}{x + y} - \frac{2}{x - y} = - 1 \frac{15}{x + y} + \frac{7}{x - y} 10,

Join BYJU'S Learning Program