Question

Calculate number of solutions of $(x,y)$

$4x-2y+3=8$ and $3x+6y=8y-x+5$

• A. $0$
• B. $1$
• C. $2$
• D. Infinitely many

Answer 1

Answer: (D)

Infinitely many

Given, $4x-2y+3=8$

$\Rightarrow 4x-2y=8-3$

$\Rightarrow 4x-2y=5$

Here $a_1=4,b_1=-2,c_1=5$

$\Rightarrow 3x+6y=8y-x+5$

$\Rightarrow 3x+x+6y-8y=5$

$\Rightarrow 4x-2y=5$

Here $a_1=4,b_1=-2,c_1=5$

If $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$ then equation has infinite number of solutions.

Here$\frac{a_1}{a_2}=\frac{4}{4}=1$

$\frac{b_1}{b_2}=\frac{-2}{-2}=1$

$\frac{c_1}{c_2}=\frac{5}{5}=1$

Then it has infinite number of solution