Question

Based on equations reducible to linear equations
Solve for x and y: $\frac{24}{2x+y}-\frac{13}{3x+2y}=2;\frac{26}{3x+2y}+\frac{8}{2x+y}=3$

• A. $x=-2,y=6$
• B. $x=-5,y=3$
• C. $x=3,y=2$
• D. $x=1,y=0$

Answer 1

The correct option is C $x=3,y=2$

The given equations are
$\Rightarrow \frac{24}{2x+y}-\frac{13}{3x+2y}=\mathrm{2...}eq\left(1\right)$
$\Rightarrow \frac{26}{3x+2y}+\frac{8}{2x+y}=\mathrm{3...}eq\left(1\right)$
$Put\frac{1}{2x+y}=uand\frac{1}{3x+y}=vineq\left(1\right)and\left(2\right)$
$\Rightarrow 24u-13v=\mathrm{2.....}\left(3\right)$
$\Rightarrow 8u+26v=\mathrm{3......}\left(4\right)$
$Multiplyeq\left(3\right)by26andeq\left(4\right)by13andsubtractboth$
$\Rightarrow (624u-338v=52)-(104u+338v=39)$
$\Rightarrow 728u=91\Rightarrow u=\frac{1}{8}$
$Putu=\frac{1}{8}ineq\left(4\right)$
$\Rightarrow 8\times \frac{1}{8}+26v=3\Rightarrow 26v=2\Rightarrow v=\frac{1}{13}$
$Hence,u=\frac{1}{2x+y}=\frac{1}{8}\Rightarrow 2x+y=\mathrm{8.....}eq\left(5\right)$
$v=\frac{1}{3x+2y}=\frac{1}{13}\Rightarrow 3x+2y=\mathrm{13...}eq\left(6\right)$
$Multiplyeq\left(5\right)by2andsubtractfromeq\left(6\right)$
$\Rightarrow (4x+2y=16)-(3x+3y=13)$
$\Rightarrow x=3$
$putx=3ineq\left(5\right)$
$\Rightarrow 2\times 3+y=8\Rightarrow y=2$