Question

Solve the following pair of linear equations by the substitution method.
(i) $x+y=14;x-y=4$
(ii) $s-t=3;\frac{s}{3}+\frac{t}{2}=6$
(iii) $3x-y=3;9x-3y=9$
(iv) $0.2x+0.3y=1.3;0.4x+0.5y=2.3$
(v) $\sqrt{2}x+\sqrt{3}y=0;\sqrt{3}x-\sqrt{8}y=0$
(vi) $\frac{3x}{2}-\frac{5y}{3}=-2;\frac{x}{3}+\frac{y}{2}=\frac{13}{6}$

Answer 1

(i) $x+y=14\Rightarrow y=14-x$

Substituting this value in the second equation, we get

$x-(14-x)=4$

$2x=18\Rightarrow x=9$

Substituting this value of x in the first equation, we get

$9+y=14\Rightarrow y=5$

(ii) $s-t=3\Rightarrow s=t+3$

Substituting in 2nd equation

$\frac{t+3}{3}+\frac{t}{2}=6$

$\frac{2t+6+3t}{6}=6$

$5t+6=36\Rightarrow t=6$

Substituting value of t in 1st equation

$s=t+3=9$

(iii) $\because \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

Hence all the points lying on the line $y=3x-3$ like $x=2,y=3;$ are a solution.

(iv) $0.2x+0.3y=1.3\Rightarrow x=\frac{1.3-0.3y}{0.2}$

Substituting this value in the second equation

$0.4\times \frac{1.3-0.3y}{0.2}+0.5y=2.3$

$2.6-0.6y+0.5y=2.3$

$2.6-2.3=0.6y-0.5y$

$0.1y=0.3\Rightarrow y=3$

Substituting this value of $y,$

$x=\frac{1.3-0.3y}{0.2}=\frac{1.3-0.3\times 3}{0.2}=\frac{0.4}{0.2}$

$x=2$

(v) $\sqrt{2}x+\sqrt{3}y=0\Rightarrow y=-\frac{\sqrt{2}}{\sqrt{3}}x$

Substituting this in 2nd equation

$\sqrt{3}x-\sqrt{8}\times (-\frac{\sqrt{2}}{\sqrt{3}}x)=0$

$3x+4x=0\Rightarrow x=0$

Substituting value of $x,$

$y=-\frac{\sqrt{2}}{\sqrt{3}}\times 0=0$

(vi) $\frac{3x}{2}-\frac{5y}{3}=-2$

$\frac{5y}{3}=\frac{3x}{2}+2\Rightarrow y=\frac{9x+12}{10}$

Substituting this in 2nd equation

$\frac{x}{3}+\frac{9x+12}{20}=\frac{13}{6}$

$\frac{47x+36}{60}=\frac{13}{6}$

$47x+36=130$

$47x=94\Rightarrow x=2$

Substituting this value of $x$ in 1st equation

$y=\frac{9x+12}{10}=\frac{9\times 2+12}{10}=3$