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Linear Equations In Two Variables — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsLinear Equations In Two Variables5 Marks
Question
Solve the following systems of equations by using the method of cross multiplication:
$\frac{\text{x}}{6}+\frac{\text{y}}{15}=4,$
$\frac{\text{x}}{3}-\frac{\text{y}}{15}=\frac{\text{19}}{4}$
✓

Answer

The given equations may be written as: $\frac{\text{x}}{6}+\frac{\text{y}}{15}-4=0\ \dots(\text{i})$ $\frac{\text{x}}{3}-\frac{\text{y}}{15}-\frac{\text{19}}{4}=0\ \dots(\text{ii})$ Here, $\text{a}_1=\frac{1}{6},\ \text{b}_1=\frac{1}{15},\ \text{c}_1=-4,$ $\text{a}_2=\frac{1}{3},\ \text{b}_2=-\frac{1}{12}$ and $\text{c}_2=-\frac{19}{4}$ By cross multiplication, we have:
$\therefore\frac{\text{x}}{\big[\frac{1}{15}\times\big(-\frac{19}{4}\big)-\big(-\frac{1}{12}\big)\times(-4)\big]}=\frac{{\text{y}}}{\big[(-4)\times\frac{1}{3}-\big(\frac{1}{6}\big)\times\big(-\frac{19}{4}\big)\big]}\\=\frac{1}{\big[\frac{1}{{6}}\times\big(\frac{-1}{12}\big)-\frac{1}{3}\times\frac{1}{15}\big]}$ $\Rightarrow\frac{\text{x}}{\big(-\frac{19}{60}-\frac{1}{3}\big)}=\frac{\text{y}}{\big(-\frac{4}{3}+\frac{19}{24}\big)}=\frac{1}{\big(-\frac{1}{72}-\frac{1}{45}\big)}$ $\Rightarrow\frac{\text{x}}{\big(-\frac{39}{60}\big)}=\frac{\text{y}}{\big(-\frac{13}{24}\big)}=\frac{1}{\big(\frac{13}{360}\big)}$ $\Rightarrow\text{x}=\Big[\Big(-\frac{39}{60}\Big)\times\Big(-\frac{360}{13}\Big)\Big]=18,$ $\text{y}=\Big[\Big(-\frac{13}{24}\Big)\times\Big(-\frac{360}{13}\Big)\Big]=15$ Hence, x = 18 and y = 15 is the required solution.

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