Question
Solve, using cross-multiplication :$0.4 x-1.5 y=6.5; 0.3 x+0.2 y=0.9$
✓
Answer
Given equation are $0.4 x-1.5 y=6.5$ and $0.3 x+0.2 y=0.9$
Comparing with $a_1 x+b_1 y+c_1=0$ and $a_2 x+b_2 y+c_2=0$, We have
$a_1=0.4, b_1=-1.5, c_1=-6.5$ and $a_2=0.3, b_2=0.2, c_2=0.9$
Now, $x=\frac{b_1 c_2-b_2 c_1}{a_1 b_2-a_2 b_1}$ and $y=\frac{c_1 a_2-c_2 a_1}{a_1 b_2-a_2 b_1}$
$\Rightarrow x=\frac{(-1.5) \times(-0.9)-(0.2) \times(-6.5)}{0.4 \times(0.2)-(0.3) \times(-1.5)}$ and $y=\frac{(-6.5) \times(0.3)-(-0.9) \times(0.4)}{0.4 \times(0.2)-(0.3) \times(-1.5)}$
$ \Rightarrow x=\frac{1.35+1.3}{0.08+0.45}$ and $y=\frac{-1.95+0.36}{0.08+0.45}$
$ \Rightarrow x=\frac{2.65}{0.53}$ and $y=\frac{-1.59}{0.53}$
$ \Rightarrow x=5$ and $y=-3$
Comparing with $a_1 x+b_1 y+c_1=0$ and $a_2 x+b_2 y+c_2=0$, We have
$a_1=0.4, b_1=-1.5, c_1=-6.5$ and $a_2=0.3, b_2=0.2, c_2=0.9$
Now, $x=\frac{b_1 c_2-b_2 c_1}{a_1 b_2-a_2 b_1}$ and $y=\frac{c_1 a_2-c_2 a_1}{a_1 b_2-a_2 b_1}$
$\Rightarrow x=\frac{(-1.5) \times(-0.9)-(0.2) \times(-6.5)}{0.4 \times(0.2)-(0.3) \times(-1.5)}$ and $y=\frac{(-6.5) \times(0.3)-(-0.9) \times(0.4)}{0.4 \times(0.2)-(0.3) \times(-1.5)}$
$ \Rightarrow x=\frac{1.35+1.3}{0.08+0.45}$ and $y=\frac{-1.95+0.36}{0.08+0.45}$
$ \Rightarrow x=\frac{2.65}{0.53}$ and $y=\frac{-1.59}{0.53}$
$ \Rightarrow x=5$ and $y=-3$
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