2x+3y=9 and 3x+4y=5. Solve by substitution Method [ICSE Class 9 ML Aggarwal Chapter 5 Exercise 5.1 Simultaneous Linear Equations Solution Question Number 1(iii)]
2x+3y=9 —(i)
3x+4y=5 —(ii)
From equation (i) we get,
2x= 9-3y —-(iii)
$\Rightarrow x$ = $\frac{9 – 3y}{2}$
Substituting the value of x in equation (ii) we get,
$3\left(\frac{9 – 3y}{2}\right) + 4y$ = 5
$\Rightarrow \frac{27 – 9y}{2} + 4y $= 5
$\Rightarrow \frac{27 – 9y + 8y}{2}$ = 5
$\Rightarrow 27 – y$ = 10
$\Rightarrow y$ = 27 – 10
$\Rightarrow y$ = 17
From equation (iii) we get,
2x = 9-3(17)
2x=9-51
⟹ 2x=-42
⟹ x=$\frac{ – 42}{2}$
⟹ x= -21
Therefore Required Solutions are x =-21 and y = 17.