Question

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

$3x-5y-4=0$ and $9x=2y+7$

• A. $x=\frac{1}{18}$ and $y=-\frac{7}{18}$
• B. $x=\frac{5}{6}$ and $y=-\frac{3}{11}$
• C. $x=\frac{2}{11}$ and $y=-\frac{2}{19}$
• D. $x=\frac{9}{13}$ and $y=-\frac{5}{13}$

Answer 1

Answer: (D)

$x=\frac{9}{13}$ and $y=-\frac{5}{13}$

Given equations are:
$3x-5y-4=0...\left(1\right)$
$9x=2y+7...\left(2\right)$
From eq$\left(1\right)$
$3x-5y=4...\left(3\right)$
From eq$\left(2\right)$
$9x-2y=7...\left(4\right)$

Elimination Method
Multiplying eq$\left(3\right)$ by $3$ and subtracting eq$\left(4\right)$, we get:
$\Rightarrow 3(3x-5y)-1(9x-2y)=3\times 4-1\times 7$
$\Rightarrow 9x-15y-9x+2y=12-7$
$\Rightarrow -13y=5$
$y=-\frac{5}{13}$

Now putting $y$ in eq$\left(3\right)$ will get
$\Rightarrow 3x-5(-\frac{13}{5})=4$
$15x+25=52$
$65x=52-25$
$x=\frac{27}{65}$
$x=\frac{9}{13}$
Hence $x=\frac{9}{13},y=-\frac{5}{13}$
Substitution Method
From eq$\left(3\right)$ will get
$\Rightarrow -5y=4-3x$
$\Rightarrow y=\frac{-4+3x}{5}$
Now substituting $y$ in eq$\left(4\right)$
$\Rightarrow 9x-2\left(\frac{-4+3x}{5}\right)=7$
$\Rightarrow 45x+8-6x=35$
$39x=27$
$x=\frac{9}{13}$
Now substituting $x$ in eq$\left(3\right)$
$\Rightarrow 3\left(\frac{9}{13}\right)-5y=4$
$\Rightarrow 27-65y=52$
$\Rightarrow -65y=52-27$
$\Rightarrow -65y=25$
$\Rightarrow y=-\frac{25}{65}$
$\Rightarrow y=-\frac{5}{13}$
Hence $x=\frac{9}{13},y=-\frac{5}{13}$