Question
Solve the following systems of equations by using the method of cross multiplication:
$\text{7x}-\text{2y}=3,$
$\text{11x}-\frac{3}{2}\text{y}=8$
$\text{7x}-\text{2y}=3,$
$\text{11x}-\frac{3}{2}\text{y}=8$
✓
Answer
The given equations may be written as: $\text{7x}-\text{2y}-3=0\ \dots(\text{i})$
$\text{11x}-\frac{3}{2}\text{y}-8=0\ \dots(\text{ii})$
Here, $a_1 = 7, b_1 = -2, c_1 = -3,$
$a_2 = 11, b_2 =$ $-\frac{3}{2}$ and $c_2 = -8$ By cross multiplication,
we have:
$\therefore\frac{\text{x}}{\big[(-2)\times(-8)-\big(\frac{3}{2}\big)\times(-3)\big]}=\frac{{\text{y}}}{[(-3)\times11-(-8)\times7]}=\frac{1}{\big[7\times\big(\frac{-3}{2}\big)-11\times(-2)\big]}$
$\Rightarrow\frac{\text{x}}{\big(16-\frac{9}{2}\big)}=\frac{\text{y}}{(-33+56)}=\frac{1}{\big(-\frac{21}{2}+22\big)}$
$\Rightarrow\frac{\text{x}}{\big(\frac{23}2{}\big)}=\frac{\text{y}}{23}=\frac{1}{\big(\frac{23}{2}\big)}$
$\Rightarrow\text{x}=\frac{\frac{23}{2}}{\frac{23}{2}}=1,\ \text{y}=\frac{23}{\frac{23}{2}}=2$
Hence, x = 1 and y = 2 is the required solution.
$\text{11x}-\frac{3}{2}\text{y}-8=0\ \dots(\text{ii})$
Here, $a_1 = 7, b_1 = -2, c_1 = -3,$
$a_2 = 11, b_2 =$ $-\frac{3}{2}$ and $c_2 = -8$ By cross multiplication,
we have:
$\therefore\frac{\text{x}}{\big[(-2)\times(-8)-\big(\frac{3}{2}\big)\times(-3)\big]}=\frac{{\text{y}}}{[(-3)\times11-(-8)\times7]}=\frac{1}{\big[7\times\big(\frac{-3}{2}\big)-11\times(-2)\big]}$
$\Rightarrow\frac{\text{x}}{\big(16-\frac{9}{2}\big)}=\frac{\text{y}}{(-33+56)}=\frac{1}{\big(-\frac{21}{2}+22\big)}$
$\Rightarrow\frac{\text{x}}{\big(\frac{23}2{}\big)}=\frac{\text{y}}{23}=\frac{1}{\big(\frac{23}{2}\big)}$
$\Rightarrow\text{x}=\frac{\frac{23}{2}}{\frac{23}{2}}=1,\ \text{y}=\frac{23}{\frac{23}{2}}=2$
Hence, x = 1 and y = 2 is the required solution.
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