Question

Solve the following pair of equations by the elimination method and the substitution method:

$x+y=5$ and $2x-3y=4$

• A. $x=\frac{19}{5},y=\frac{6}{5}$
• B. $x=\frac{24}{7},y=\frac{6}{5}$
• C. $x=\frac{13}{2},y=\frac{6}{5}$
• D. $x=\frac{17}{7},y=\frac{6}{5}$

Answer 1

The correct option is A $x=\frac{19}{5},y=\frac{6}{5}$

Solution By Elimination Method.
$x+y=5$ (i)
$2x-3y=4$ (ii)
Multiplying (i) by 3 and (ii) by 1 and adding we get $3(x+y)+1(2x-3y)=3\times 5+1\times 4$
$\Rightarrow 3x+3y+2x-3y=19$
$5x=19\Rightarrow x=\frac{19}{5}$
From (i), substituting $x=\frac{19}{5}$, we get$\frac{19}{5}+y=5\Rightarrow y=5-\frac{19}{5}\Rightarrow y=\frac{6}{5}$
Hence, $x=\frac{19}{5},y=\frac{6}{5}$
Solution By Substitution Method :
$x+y=5$ (i)
$2x-3y=4$ (ii)
From (i), $y=5-x$ (iii)
Substituting y from (iii) in (ii), $2x-3(5-x)=4$
$\Rightarrow 2x-15+3x=4$
$\Rightarrow 5x=19\Rightarrow x=\frac{19}{5}$
Then from (iii), $y=5-\frac{19}{5}\Rightarrow y=\frac{6}{5}$
Hence, $x=\frac{19}{5},y=\frac{6}{5}$