3x – y = 5 and 4x – 3y = – 1 Hence, find p, if y = px – 3.Solve the simultaneous equations [ICSE Class 9 ML Aggarwal Latest Solution in English of Chapter 5 Exercise 5.1 Simultaneous Linear Equations ]
3x – y = 5 —(i)
4x – 3y = – 1 —-(ii)
From equation (i) we get,
3x-5=y —(iii)
Substituting the value of y in equation (ii) we get,
4x – 3(3x-5) = -1
⟹ 4x -9x+15= -1
⟹ -5x +15=-1
⟹ -5x = -15-1
⟹ -5x = -16
⟹ x = $\frac{16}{5}$
From equation (iii) we get,
$3\left(\frac{16}{5}\right) – 5$ = y
$\Rightarrow \frac{48}{5} – 5$ = y
$\Rightarrow \frac{48 – 25}{5}$ = y
$\Rightarrow y$ = $\frac{23}{5}$
Given that,
y = px – 3
$\frac{23}{5}$ = $p\left(\frac{16}{5}\right) – 3$
$\Rightarrow \frac{23}{5}$ + 3 = $\frac{16p}{5}$
$\Rightarrow \frac{23 + 15}{5}$ = $\frac{16{\mathrm{p}}}{5}$
$\Rightarrow 16{\mathrm{p}}$ = 38
$\Rightarrow {\mathrm{p}}$ = $\frac{38}{16}$
$\Rightarrow {\mathrm{p}}$ = $\frac{19}{8}$
Required Solutions are x =$\frac{16}{5}$ , y=$\frac{23}{5}$ and p = $\frac{19}{8}$.