Question

Solve:(by substitution method)
99x+101y=499
101x+99y=501

Answer 1

$$\begin{gathered} 99\mathrm{x}+101\mathrm{y}=499...\left(1\right) \\ 101\mathrm{x}+99\mathrm{y}=501...\left(2\right) \\ \mathrm{From}\mathrm{eqn}\left(1\right),\mathrm{we}\mathrm{get} \\ 99\mathrm{x}=499-101\mathrm{y} \\ \mathrm{x}=\frac{499-101\mathrm{y}}{99} \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{x}\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ 101\left(\frac{499-101\mathrm{y}}{99}\right)+99\mathrm{y}=501 \\ \frac{101\left(499\right)-{\left(101\right)}^2\mathrm{y}+{\left(99\right)}^2\mathrm{y}}{99}=501 \\ 101\left(499\right)-{\left(101\right)}^2\mathrm{y}+{\left(99\right)}^2\mathrm{y}=501\left(99\right) \\ \mathrm{y}\left({\left(99\right)}^2-{\left(101\right)}^2\right)=\left(500+1\right)\left(100-1\right)-\left(100+1\right)\left(500-1\right) \\ \mathrm{y}\left(99+101\right)\left(99-101\right)=\left(50000-500+100-1\right)-\left(50000-100+500-1\right) \\ \mathrm{y}\left(200\right)\left(-2\right)=-1000+200 \\ \mathrm{y}=\frac{-800}{-400}=2 \\ \mathrm{So}, \\ \mathrm{x}=\frac{499-101\left(2\right)}{99}=\frac{499-204}{99}=\frac{295}{99}=2.97 \\ \mathrm{So},\mathrm{value}\mathrm{of}\mathrm{x}=2.97\mathrm{and}\mathrm{y}=2. \end{gathered}$$