Solve: 3x+7y+5=0 4x-3y = 8 and px+y=1 if the following equations hold simultaneously for x and y, find p.
3x+7y+5=0 —(i)
4x-3y = 8 —(ii)
px+y=1 —-(iii)
From equation (i) we get,
3x+7y+5=0
⟹ 3x = -5-7y
⟹ 3x = -(5+7y)
x = $\frac{ – (5 + 7y)}{3}$ —(iv)
Substituting the value of x in equation (ii) we get,
4x-3y = 8
$\Rightarrow 4\left[\frac{ – (5 + 7y)}{3}\right] – 3y = 8$
$\Rightarrow \frac{ – 20 – 28y}{3} – 3y = 8$
$\Rightarrow \frac{ – 20 – 28y – 9y}{3} = 8$
$\Rightarrow – 20 – 37y = 24$
$\Rightarrow – 20 – 24 = 37y$
$\Rightarrow – 44 = 37y$
$\Rightarrow y = \frac{ – 44}{37}$
From equation (iv) we get,
$x = \frac{ – (5 + 7y)}{3}$
$\Rightarrow x = \frac{ – 5 – 7y}{3}$
$\Rightarrow x = \frac{ – 5 – 7\left(\frac{ – 44}{37}\right)}{3}$
$\Rightarrow x = \frac{ – 5 + \frac{308}{37}}{3}$
$\Rightarrow x = \frac{ – 185 + 308}{3 \times 37}$
$\Rightarrow x = \frac{ 123}{3 \times 37}$
$\Rightarrow x = \frac{ 41}{37}$
Since the given equations hold simultaneously then we can substitute the value of x and y in equation (iii)
$\therefore {\mathrm{px}} + {\mathrm{y}} = 1$
$\Rightarrow {\mathrm{p}}\left(\frac{41}{37}\right) + \left(\frac{ – 44}{37}\right) = 1$
$\Rightarrow \frac{41{\mathrm{p}} – 44}{37} = 1$
$\Rightarrow 41{\mathrm{p}} – 44 = 37$
$\Rightarrow 41{\mathrm{p}} = 44 + 37$
$\Rightarrow 41{\mathrm{p}} = 81$
$\Rightarrow {\mathrm{p}} = \frac{81}{41}$
$\therefore x = \frac{41}{37},y = – \frac{44}{37} and p = \frac{81}{41}$