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Linear Equation In Two Variable — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsLinear Equation In Two Variable5 Marks
Question
Solve the following systems of equations by using the method of cross multiplication:
$\frac{5}{(\text{x}+\text{y})}-\frac{2}{(\text{x}-\text{y})}+1=0,$
$\frac{15}{(\text{x}+\text{y})}+\frac{7}{(\text{x}-\text{y})}-10=0$ $(\text{x}\neq\text{y},\ \text{x}\neq-\text{y}).$
✓

Answer

Taking $\frac{1}{\text{x}+\text{y}}=\text{u}$ and $\frac{1}{\text{x}-\text{y}}=\text{v},$ the given equations become:
$5u - 2v + 1 = 0 ...(i) 15u + 7v - 10 = 0 ...(ii)$
Here, $a_1 = 5, b_1 = -2, c_1 = 1, a_2 = 15, b_2 = -7$ and $c_2 = -10$ By cross multiplication, we have:



$\therefore\frac{\text{u}}{[-2\times(-10)-1\times7]}=\frac{\text{v}}{[1\times15-(-10)\times5]}=\frac{1}{[35+30]}$
$\Rightarrow\frac{\text{u}}{20-7}=\frac{\text{v}}{15+50}=\frac{1}{65}$
$\Rightarrow\frac{\text{u}}{13}=\frac{\text{v}}{65}=\frac{1}{65}$
$\Rightarrow\text{u}=\frac{13}{65}=\frac{1}{5},\ \text{v}=\frac{65}{65}=1$
$\Rightarrow\frac{1}{\text{x}+\text{y}}=\frac{1}{5},\ \frac{1}{\text{x}-\text{y}}=1$
So, $(x + y) = 5 ...(iii)$ and $(x - y) = 1 ...(iv)$
Again, the above equations (iii) and (iv) may be written as: $x + y - 5 = 0 ...(v) x - y - 1 = 0 ...(vi)$
Here, $a_1 = 1, b_1 = 1, c_1 = -5, a_2 = 1, b_2 = -1$ and $c_2 = -1$ By cross multiplication, we have:

$\therefore\frac{\text{x}}{[1\times(-1)-(-5)\times(-1)]}=\frac{\text{y}}{(-5)\times1-(-1)\times1}=\frac{1}{[1\times(-1)-1\times1]}$
$\Rightarrow\frac{\text{x}}{(-1-5)}=\frac{\text{y}}{(-5+1)}=\frac{1}{(-1-1)}$
$\Rightarrow\frac{\text{x}}{-6}=\frac{\text{y}}{-4}=\frac{1}{-2}$
$\Rightarrow\text{u}=\frac{-6}{-2}=3,\ \text{y}=\frac{-4}{-2}=2$
Hence, $x = 3$ and $y = 2$ is the required solution.

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