Question
Solve the following simultaneous equations by the substitution method$:\ 0.5x + 0.7y = 0.74,0.3x + 0.5y = 0.5$
✓
Answer
The given equations are
$ 0.5 x+0.7 y=0.74 ....(i)$
$0.3 x+0.5 y=0.5 ....(ii) $
Now, consider equation
$ 0.5 x+0.7 y=0.74$
$\Rightarrow 0.5 x=0.74-0.7 y$
$\Rightarrow x=\frac{0.74-0.7 y}{0.5} ....(iii) $
Substituting the value of $x$ in eqn. $(ii),$ we get
$ 0.3\left(\frac{0.74-0.7 y}{0.5}\right)+0.5 y=0.5$
$\Rightarrow \frac{0.222-0.21}{0.5}+0.5 y=0.5$
$\Rightarrow \frac{0.222-0.2 y+0.25}{0.5}=0.5$
$\Rightarrow 0.222+0.04 y=0.25$
$\Rightarrow 0.04 y=0.028$
$\Rightarrow y=\frac{0.028}{0.04}$
$=\frac{28}{40}$
$=\frac{7}{10}$
$=0.7 $
Putting the value of $y$ in eqn. $(iii),$ we get
$ x=\frac{0.74-0.7(0.7)}{0.5}$
$=\frac{0.74-0.49}{0.5}$
$=\frac{0.25}{0.5}$
$=\frac{25}{50}$
$=\frac{1}{2}$
$=0.5 $
Thus, the solution set is $(0.5,0.7)$.
$ 0.5 x+0.7 y=0.74 ....(i)$
$0.3 x+0.5 y=0.5 ....(ii) $
Now, consider equation
$ 0.5 x+0.7 y=0.74$
$\Rightarrow 0.5 x=0.74-0.7 y$
$\Rightarrow x=\frac{0.74-0.7 y}{0.5} ....(iii) $
Substituting the value of $x$ in eqn. $(ii),$ we get
$ 0.3\left(\frac{0.74-0.7 y}{0.5}\right)+0.5 y=0.5$
$\Rightarrow \frac{0.222-0.21}{0.5}+0.5 y=0.5$
$\Rightarrow \frac{0.222-0.2 y+0.25}{0.5}=0.5$
$\Rightarrow 0.222+0.04 y=0.25$
$\Rightarrow 0.04 y=0.028$
$\Rightarrow y=\frac{0.028}{0.04}$
$=\frac{28}{40}$
$=\frac{7}{10}$
$=0.7 $
Putting the value of $y$ in eqn. $(iii),$ we get
$ x=\frac{0.74-0.7(0.7)}{0.5}$
$=\frac{0.74-0.49}{0.5}$
$=\frac{0.25}{0.5}$
$=\frac{25}{50}$
$=\frac{1}{2}$
$=0.5 $
Thus, the solution set is $(0.5,0.7)$.
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