Question

$\frac{4-3x}{5}+\frac{7-x}{3}+4\frac{1}{3}=0$

Hence, find the value of $p$, if $3p-2x+1=0.$

• A. $x=2$ and $p=2$
• B. $x=8$ and $p=5$
• C. $x=6$ and $p=3$
• D. $x=4$ and $p=7$

Answer 1

Answer: (B)

$x=8$ and $p=5$

$\frac{4-3x}{5}+\frac{7-x}{3}+4\frac{1}{3}=0$
$\Rightarrow \frac{12-9x+35-5x+65}{15}=0$
$\Rightarrow 12-9x+35-5x+65=0$
$\Rightarrow 14x=112$
$\Rightarrow x=8$
Now,
$\Rightarrow 3p-2x+1=0$
$\Rightarrow 3p-2\left(8\right)+1=0$
$\Rightarrow 3p=15$
$\Rightarrow p=5$