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Linear Equations In Two Variables — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsLinear Equations In Two Variables5 Marks
Question
Solve for x and y:
$\frac{35}{\text{x}+\text{y}}+\frac{14}{\text{x}-\text{y}}=19,$ $\frac{14}{\text{x}+\text{y}}+\frac{35}{\text{x}-\text{y}}=37$
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Answer

We have: $\frac{35}{\text{x+y}}+\frac{14}{\text{x}-\text{y}}=19$ and $\frac{14}{\text{x+y}}+\frac{35}{\text{x}-\text{y}}=37$ Taking $\frac{1}{\text{x+y}}=\text{u}$ and $\frac{1}{\text{x}-\text{y}}=\text{v}.$
$35 u+14 v-19=0.......(i) 14 u+35 v-37=0$
Here, $a_1=35, b_1=14, c_1=-19, a_2=14, b_2=35, c_2=-37$ By cross multiplying, we have:

$\therefore\frac{\text{u}}{[14\times(-37)-35\times(-19)]}=\frac{\text{v}}{[(-19)\times14-(-37)\times(35)]}=\frac{1}{[35\times35-14\times14]}$
$\Rightarrow\frac{\text{u}}{-518+665}=\frac{\text{v}}{-266+1295}=\frac{1}{1225-196}$
$\Rightarrow\frac{\text{u}}{147}=\frac{\text{v}}{1029}=\frac{1}{1029}$
$\Rightarrow\text{u}=\frac{147}{1029}=\frac{1}{7},\ \text{v}=\frac{1029}{1029}=1$
$\Rightarrow\frac{1}{\text{x+y}}=\frac{1}{7},\ \frac{1}{\text{x}-\text{y}}=1$
$\therefore( x + y )=7 \ldots$...iii) And, $( x - y )=1 \ldots$ (iv) Again, the equation (iii) and (iv) can be written as follows: $x + y -7=0 \ldots$ (v) x $-y-1=0 \ldots$ (vi) Here $a_1=1, b_1=1, c_1=-7, a_2=1, b_2=-1, c_2=-1$ By multiplication, we have:
$\Rightarrow\frac{\text{x}}{[1\times(-1)-(-1)\times(-7)]}=\frac{\text{y}}{[(-7)\times1-(-1)\times1]}=\frac{1}{[1\times(-1)-1\times1]}$ $\Rightarrow\frac{\text{x}}{-1-7}=\frac{\text{y}}{-7+1}=\frac{1}{-1-1}$
$\Rightarrow\frac{\text{x}}{-8}=\frac{\text{y}}{-6}=\frac{1}{-2}$
$\Rightarrow\text{x}=\frac{-8}{-2}=4,\ \text{y}=\frac{-6}{-2}=3$ Hence, x = 4 and y = 3 is the required solution.

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