Question
Solve, using cross$-$multiplication $:4x + 3y = 17,3x - 4y + 6 = 0$
✓
Answer
Given equation are $4 x+3 y=17$ and $3 x-4 y+6=0$
Comparing with $\mathrm{a}_ 1 \mathrm{x}+\mathrm{b} _1 \mathrm{y}+\mathrm{c} _1=0$ and $\mathrm{a} _2 \mathrm{x}+\mathrm{b}_ 2 \mathrm{y}+\mathrm{c} _2=0$,
We have
$\mathrm{a} _1=4, \mathrm{~b} _1=3, \mathrm{c} _1=-17$ and $\mathrm{a} _2=3, \mathrm{~b} _2=-4, \mathrm{c} _2=6$
Now, $x=\frac{b_1 c_2-b_2 c_1}{a_1 b_2-a_2 b_1}$ and $y=\frac{c_1 a_2-c_2 a_1}{a_1 b_2-a_2 b_1}$
$\Rightarrow \mathrm{x}= \frac{3 \times 6-(-4) \times(-17)}{4 \times(-4)-3 \times 3}$ and $y=\frac{-17 \times 3-6 \times 4}{4 \times(-4)-3 \times 3}$
$ \Rightarrow \mathrm{x}=\frac{18-68}{-16-9}$ and $y=\frac{-51-24}{-16-9}$
$ \Rightarrow x=\frac{-50}{-25}$ and $y=\frac{-75}{-25}$
$\Rightarrow x=2$ and $y=3$.
Comparing with $\mathrm{a}_ 1 \mathrm{x}+\mathrm{b} _1 \mathrm{y}+\mathrm{c} _1=0$ and $\mathrm{a} _2 \mathrm{x}+\mathrm{b}_ 2 \mathrm{y}+\mathrm{c} _2=0$,
We have
$\mathrm{a} _1=4, \mathrm{~b} _1=3, \mathrm{c} _1=-17$ and $\mathrm{a} _2=3, \mathrm{~b} _2=-4, \mathrm{c} _2=6$
Now, $x=\frac{b_1 c_2-b_2 c_1}{a_1 b_2-a_2 b_1}$ and $y=\frac{c_1 a_2-c_2 a_1}{a_1 b_2-a_2 b_1}$
$\Rightarrow \mathrm{x}= \frac{3 \times 6-(-4) \times(-17)}{4 \times(-4)-3 \times 3}$ and $y=\frac{-17 \times 3-6 \times 4}{4 \times(-4)-3 \times 3}$
$ \Rightarrow \mathrm{x}=\frac{18-68}{-16-9}$ and $y=\frac{-51-24}{-16-9}$
$ \Rightarrow x=\frac{-50}{-25}$ and $y=\frac{-75}{-25}$
$\Rightarrow x=2$ and $y=3$.
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