Solve: 3x+7y+5=0 4x-3y = 8 and px+y=1 if the following equations hold simultaneously for x and y, find p.

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)

$\therefore {\mathrm{px}} + {\mathrm{y}} = 1$

$\Rightarrow {\mathrm{p}}\left(\frac{41}{37}\right) + \left(\frac{ – 44}{37}\right) = 1$

$\Rightarrow \frac{41{\mathrm{p}} – 44}{37} = 1$

$\Rightarrow 41{\mathrm{p}} – 44 = 37$

$\Rightarrow 41{\mathrm{p}} = 44 + 37$

$\Rightarrow 41{\mathrm{p}} = 81$

$\Rightarrow {\mathrm{p}} = \frac{81}{41}$

$\therefore x = \frac{41}{37},y = – \frac{44}{37} and p = \frac{81}{41}$

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