Question
Solve the following systems of equations by using the method of cross multiplication:
$3x - 2y + 3 = 0$
$4x + 3y - 47 = 0$
$3x - 2y + 3 = 0$
$4x + 3y - 47 = 0$
✓
Answer
The given equations are: $3x - 2y + 3 = 0 ...(i) 4x + 3y - 47 = 0 ...(ii)$
Here, $a_1 = 3, b_1 = -2, c_1 = 3, a_2 = 4, b_2 = 3$ and $c_2 = -47$ By cross multiplication, we have:
$\therefore\frac{\text{x}}{[(-2)\times(-47)-3\times3]}=\frac{{\text{y}}}{[3\times4-(-47)\times3]}=\frac{1}{[3\times3-(-2)\times4]}$
$\Rightarrow\frac{\text{x}}{(94-9)}=\frac{\text{y}}{(12+141)}=\frac{1}{(9+8)}$
$\Rightarrow\frac{\text{x}}{85}=\frac{\text{y}}{153}=\frac{1}{17}$
$\Rightarrow\text{x}=\frac{85}{17}=5,\ \text{y}=\frac{153}{17}=9$
Hence, $x = 5$ and $y = 9$ is the required solution.
Here, $a_1 = 3, b_1 = -2, c_1 = 3, a_2 = 4, b_2 = 3$ and $c_2 = -47$ By cross multiplication, we have:
$\therefore\frac{\text{x}}{[(-2)\times(-47)-3\times3]}=\frac{{\text{y}}}{[3\times4-(-47)\times3]}=\frac{1}{[3\times3-(-2)\times4]}$
$\Rightarrow\frac{\text{x}}{(94-9)}=\frac{\text{y}}{(12+141)}=\frac{1}{(9+8)}$
$\Rightarrow\frac{\text{x}}{85}=\frac{\text{y}}{153}=\frac{1}{17}$
$\Rightarrow\text{x}=\frac{85}{17}=5,\ \text{y}=\frac{153}{17}=9$
Hence, $x = 5$ and $y = 9$ is the required solution.
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