Question
Solve, using cross-multiplication $:4x - 3y = 0,2x + 3y = 18$
✓
Answer
Given equation are $4 x-3 y=0$ and
$2 x+3 y=18$
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
$a_1=4, b_1=-3, c_1=0$ and $a_2=2, b_2=3, c_2=-18$
Now, $x=\frac{b_1 c_2-b_2 c_1}{a_1 b_2-a_2 b_1}$ and $y=\frac{c_1 a_2-c_2 a_1}{a_1 b_2-a_2 b_1}$
$\Rightarrow x= \frac{-3 \times(-18)-3 \times 0}{4 \times 3-2 \times(-3)}$ and $y=\frac{0 \times 2-(-18) \times 4}{4 \times 3-2 \times(-3)}$
$\Rightarrow x=\frac{54-0}{12+6} \quad$ and $y=\frac{0+72}{12+6}$
$\Rightarrow \mathrm{x}=\frac{54}{18}$ and $y=\frac{72}{18}$
$\Rightarrow x=3$ and $y=4$.
$2 x+3 y=18$
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
$a_1=4, b_1=-3, c_1=0$ and $a_2=2, b_2=3, c_2=-18$
Now, $x=\frac{b_1 c_2-b_2 c_1}{a_1 b_2-a_2 b_1}$ and $y=\frac{c_1 a_2-c_2 a_1}{a_1 b_2-a_2 b_1}$
$\Rightarrow x= \frac{-3 \times(-18)-3 \times 0}{4 \times 3-2 \times(-3)}$ and $y=\frac{0 \times 2-(-18) \times 4}{4 \times 3-2 \times(-3)}$
$\Rightarrow x=\frac{54-0}{12+6} \quad$ and $y=\frac{0+72}{12+6}$
$\Rightarrow \mathrm{x}=\frac{54}{18}$ and $y=\frac{72}{18}$
$\Rightarrow x=3$ and $y=4$.
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