Question
Solve the following simultaneous equations by the substitution method$:\ 7(y + 3) - 2(x + 2) = 14;4(y - 2) + 3(x - 3) = 2$
✓
Answer
The given equations are
$ 7(y+3)-2(x+2)=14 ....(i)$
$4(y-2)+3(x-3)=2 ....(ii) $
Consider
$ 7(y+3)-2(x+2)=14$
$\Rightarrow 7 y+21-2 x-4=14$
$\Rightarrow-2 x+7 y=-3$
$\Rightarrow 2 x-7 y=3$
$\Rightarrow 2 x=7 y+3$
$\Rightarrow x=\frac{7 y+3}{2} ....(iii) $
Now, consider equation
$ 4(y-2)+3(x-3)=2$
$\Rightarrow 4 y-8+3 x-9=2$
$\Rightarrow 3 x+4 y=19$
$\Rightarrow 3\left(\frac{7 y+3}{2}\right)+4 y=19 \ldots .[$ From $(iii)]$
$\Rightarrow \frac{21 y+9+8 y}{2}=19$
$\Rightarrow 29 y+9=38$
$\Rightarrow 29 y=29$
$\Rightarrow y=1 $
Substituting value of $y$ in eqn. $(iii),$ we get
$ x=\frac{7(1)+3}{2}$
$=\frac{10}{2}$
$=5 $
Thus, the solution set is $(5,1)$.
$ 7(y+3)-2(x+2)=14 ....(i)$
$4(y-2)+3(x-3)=2 ....(ii) $
Consider
$ 7(y+3)-2(x+2)=14$
$\Rightarrow 7 y+21-2 x-4=14$
$\Rightarrow-2 x+7 y=-3$
$\Rightarrow 2 x-7 y=3$
$\Rightarrow 2 x=7 y+3$
$\Rightarrow x=\frac{7 y+3}{2} ....(iii) $
Now, consider equation
$ 4(y-2)+3(x-3)=2$
$\Rightarrow 4 y-8+3 x-9=2$
$\Rightarrow 3 x+4 y=19$
$\Rightarrow 3\left(\frac{7 y+3}{2}\right)+4 y=19 \ldots .[$ From $(iii)]$
$\Rightarrow \frac{21 y+9+8 y}{2}=19$
$\Rightarrow 29 y+9=38$
$\Rightarrow 29 y=29$
$\Rightarrow y=1 $
Substituting value of $y$ in eqn. $(iii),$ we get
$ x=\frac{7(1)+3}{2}$
$=\frac{10}{2}$
$=5 $
Thus, the solution set is $(5,1)$.
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