Question
Solve the following systems of equations by using the method of cross multiplication:
$x + 2y + 1 = 0$
$2x - 3y - 12 = 0$
$x + 2y + 1 = 0$
$2x - 3y - 12 = 0$
✓
Answer
The given equations are:
$x + 2y + 1 = 0 ...(i)$
$2x - 3y - 12 = 0 ...(ii)$
Here, $a_1 = 1, b_1 = 2, c_1 = 1, a_2 = 2, b_2 = -3$ and $c_2 = -12$
By cross multiplication, we have
$\therefore\frac{\text{x}}{[2\times(-12)-1\times(-3)]}=\frac{{\text{y}}}{[1\times2-1\times(-12)]}=\frac{1}{[1\times(-3)-2\times2]}$
$\Rightarrow\frac{\text{x}}{(-24+3)}=\frac{\text{y}}{(2+12)}=\frac{1}{(-3-4)}$
$\Rightarrow\frac{\text{x}}{(-21)}=\frac{\text{y}}{(14)}=\frac{1}{(-7)}$
$\Rightarrow\text{x}=\frac{-21}{-7}=3,\ \text{y}=\frac{14}{-7}=-2$
Hence, $x = 3$ and $y = -2$ is the required solution.
$x + 2y + 1 = 0 ...(i)$
$2x - 3y - 12 = 0 ...(ii)$
Here, $a_1 = 1, b_1 = 2, c_1 = 1, a_2 = 2, b_2 = -3$ and $c_2 = -12$
By cross multiplication, we have
$\therefore\frac{\text{x}}{[2\times(-12)-1\times(-3)]}=\frac{{\text{y}}}{[1\times2-1\times(-12)]}=\frac{1}{[1\times(-3)-2\times2]}$
$\Rightarrow\frac{\text{x}}{(-24+3)}=\frac{\text{y}}{(2+12)}=\frac{1}{(-3-4)}$
$\Rightarrow\frac{\text{x}}{(-21)}=\frac{\text{y}}{(14)}=\frac{1}{(-7)}$
$\Rightarrow\text{x}=\frac{-21}{-7}=3,\ \text{y}=\frac{14}{-7}=-2$
Hence, $x = 3$ and $y = -2$ is the required solution.
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