Question
Solve the following systems of equations by using the method of cross multiplication:
$2ax + 3by = (a + 2b),$
$3ax + 2by = (2a + b).$
$2ax + 3by = (a + 2b),$
$3ax + 2by = (2a + b).$
✓
Answer
The given equations may be written as: $2ax + 3by = (a + 2b) ...(i) 3ax + 2by = (2a + b) ...(ii)$
Here, $a_1 = 2a, b_1 = 3b, c_1 = -(a + 2b),$
$a_2 = 3a, b_2 = 2b$ and $c_2 = -(2a + b) $By cross multiplication, we have:
$\therefore\frac{\text{x}}{[\text{3b}\times(-(\text{2a}+\text{b}))-2\text{b}\times(-(\text{a}+2\text{b}))]}=\frac{\text{y}}{[-(\text{a}+2\text{b})\times3\text{a}-\text{2a}\times(-(\text{2a}+\text{b}))]}=\frac{1}{[\text{2a}\times2\text{b}-3\text{a}\times\text{3b}]}$ $\Rightarrow\frac{\text{x}}{\big(-\text{6ab}-\text{3b}^2+\text{2ab}+\text{4b}^2\big)}=\frac{\text{x}}{\big(-\text{3a}^2-\text{6ab}+\text{4a}^2+\text{2ab}\big)}=\frac{1}{\text{4ab}-\text{9ab}}$
$\Rightarrow\frac{\text{x}}{\text{b}^2-\text{4ab}}=\frac{\text{y}}{\text{a}^2-\text{4ab}}=\frac{1}{-\text{5ab}}$
$\Rightarrow\frac{\text{x}}{-\text{b}(\text{4a}-\text{b})}=\frac{\text{y}}{-\text{a}(4\text{b}-\text{a})}=\frac{1}{-5\text{ab}}$
$\Rightarrow\text{x}=\frac{-\text{b}(\text{4a}-\text{b})}{-5\text{ab}}=\frac{(\text{4a}-\text{b})}{\text{5a}},$ $\text{y}=\frac{-\text{a}(\text{4b}-\text{a})}{-\text{5ab}}=\frac{(\text{4b}-\text{a})}{\text{5b}}$
Hence, $\text{x}=\frac{(\text{4a}-\text{b})}{\text{5a}}$ and $\text{y}=\frac{(\text{4a}-\text{b})}{\text{5b}}$ is the required solution.
Here, $a_1 = 2a, b_1 = 3b, c_1 = -(a + 2b),$
$a_2 = 3a, b_2 = 2b$ and $c_2 = -(2a + b) $By cross multiplication, we have:
$\therefore\frac{\text{x}}{[\text{3b}\times(-(\text{2a}+\text{b}))-2\text{b}\times(-(\text{a}+2\text{b}))]}=\frac{\text{y}}{[-(\text{a}+2\text{b})\times3\text{a}-\text{2a}\times(-(\text{2a}+\text{b}))]}=\frac{1}{[\text{2a}\times2\text{b}-3\text{a}\times\text{3b}]}$ $\Rightarrow\frac{\text{x}}{\big(-\text{6ab}-\text{3b}^2+\text{2ab}+\text{4b}^2\big)}=\frac{\text{x}}{\big(-\text{3a}^2-\text{6ab}+\text{4a}^2+\text{2ab}\big)}=\frac{1}{\text{4ab}-\text{9ab}}$
$\Rightarrow\frac{\text{x}}{\text{b}^2-\text{4ab}}=\frac{\text{y}}{\text{a}^2-\text{4ab}}=\frac{1}{-\text{5ab}}$
$\Rightarrow\frac{\text{x}}{-\text{b}(\text{4a}-\text{b})}=\frac{\text{y}}{-\text{a}(4\text{b}-\text{a})}=\frac{1}{-5\text{ab}}$
$\Rightarrow\text{x}=\frac{-\text{b}(\text{4a}-\text{b})}{-5\text{ab}}=\frac{(\text{4a}-\text{b})}{\text{5a}},$ $\text{y}=\frac{-\text{a}(\text{4b}-\text{a})}{-\text{5ab}}=\frac{(\text{4b}-\text{a})}{\text{5b}}$
Hence, $\text{x}=\frac{(\text{4a}-\text{b})}{\text{5a}}$ and $\text{y}=\frac{(\text{4a}-\text{b})}{\text{5b}}$ is the required solution.
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