Question
Solve the following simultaneous equations by the substitution method$:\ 2x + 3y = 31,5x - 4 = 3y$
✓
Answer
The given equations are
$ 2 x+3 y=31 ....(i)$
$5 x-4=3 y ....(ii) $
Now, consider equation
$ 2 x+3 y=31$
$\Rightarrow 2 x=31-3 y$
$\Rightarrow x=\frac{31-3 y}{2} ....(iii) $
Substituting the value of $x$ in eqn. $(ii),$ we get
$ 5\left(\frac{31-3 y}{2}\right)-4=3 y$
$\Rightarrow \frac{155-15 y}{2}-4=3 y$
$\Rightarrow \frac{155-15 y-8}{2}=3 y$
$\Rightarrow 147-15 y=6 y$
$\Rightarrow 21 y=147$
$\Rightarrow y=\frac{147}{21}=7 $
Putting the value of $y$ in eqn. $(iii),$ we get
$ x=\frac{31-3(7)}{2}$
$=\frac{31-21}{2}$
$=\frac{10}{2}$
$=5 $
Thus, the solution set is $(5,7)$.
$ 2 x+3 y=31 ....(i)$
$5 x-4=3 y ....(ii) $
Now, consider equation
$ 2 x+3 y=31$
$\Rightarrow 2 x=31-3 y$
$\Rightarrow x=\frac{31-3 y}{2} ....(iii) $
Substituting the value of $x$ in eqn. $(ii),$ we get
$ 5\left(\frac{31-3 y}{2}\right)-4=3 y$
$\Rightarrow \frac{155-15 y}{2}-4=3 y$
$\Rightarrow \frac{155-15 y-8}{2}=3 y$
$\Rightarrow 147-15 y=6 y$
$\Rightarrow 21 y=147$
$\Rightarrow y=\frac{147}{21}=7 $
Putting the value of $y$ in eqn. $(iii),$ we get
$ x=\frac{31-3(7)}{2}$
$=\frac{31-21}{2}$
$=\frac{10}{2}$
$=5 $
Thus, the solution set is $(5,7)$.
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