Question

If the same value of x and y satisfy the following equations; find the value of p :
$3x+7y+5=0$
$4x-3y-8=0$
$px+y-1=0$

• A. $1\frac{20}{47}$
• B. $1\frac{40}{41}$
• C. $1\frac{31}{45}$
• D. $1\frac{45}{21}$

Answer 1

The correct option is B $1\frac{40}{41}$

Let $3x+7y+5=0$ --- (1)

$4x-3y-8=0$ --- (2)

Multiplying equation $\left(1\right)$ with $4$ we get, $12x+28y+20=0$ ----- equation $\left(3\right)$

Multiplying equation $\left(2\right)$ with $3$ we get, $12x-9y-24=0$ ----- equation $\left(4\right)$

Subtracting equation $\left(4\right)$ from $\left(3\right)$, we get $37y=-44=>y=-\frac{44}{37}$

Substituting $y=-\frac{44}{37}$ in the equation $\left(2\right)$, we get

$4x-3(-\frac{44}{37})8=0=>4x+\frac{132}{37}=8=>4x=\frac{164}{37}=>x=\frac{41}{37}$

Putting the values of x and y in $px+y1=0$, we get

$p\left(\frac{41}{37}\right)-\frac{44}{37}-1=0$

$41p-44-37=0$
$41p=81$

$p=\frac{81}{41}=1\frac{40}{41}$