Question

How many solutions does the pair of linear equations will have if the equation are -2x+8y=23 and -8x+32y-4=0?

Answer 1

Given :-

The pair of linear equations are:

• -2x + 8y = 23
• -8x + 32y - 4 = 0

To Find :-

• Number of solutions that the pair of linear equations have.

Solution :-

Given equations are:-

• -2x + 8y = 23
• -8x + 32y = 4

Here,

$\implies\rm{\dfrac{a_1}{a_2}=\dfrac{-2x}{-8x}=\dfrac{1}{4}}$

$\implies\rm{\dfrac{b_1}{b_2}=\dfrac{8y}{32y}=\dfrac{1}{4}}$

$\implies\rm{\dfrac{c_1}{c_2}=\dfrac{23}{4}}$

Therefore,

$\implies\rm{\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\neq\dfrac{c_1}{c_2}}$

Hence, the given pair of linear equations have no solution.

$\hrulefill$

Additional Information :

If,

• $\rm{\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}}$

The pair of linear equations will have infinite number of solutions.

If,

• $\rm{\dfrac{a_1}{a_2}\neq\dfrac{b_1}{b_2}}$

The pair of linear equations has a unique solution.

Answer 2

Aɴsᴡᴇʀ :-

Gɪᴠᴇɴ :-

ᴛʜᴇ ᴘᴀɪʀ ᴏғ ʟɪɴᴇᴀʀ ᴇǫᴜᴀᴛɪᴏɴs

$=>$-2x + 8y = 23 => -2x + 8y - 23 = 0

$=>$ -8x + 32y - 4 = 0

ʜᴇʀᴇ,

$\frac{a_1}{a_2} => \frac{-2x}{-8x} => \frac{1}{4}$

$\frac{b_1}{b_2} => \frac{8y}{32y} => \frac {1}{4}$

$\frac{c_1}{c_2} => \frac{-23}{-4} => \frac{23}{4}$

Tʜᴇʀᴇғᴏʀᴇ ,

$\frac{a_1}{a_2} = \frac{b_1}{b_2} ≠ \frac{c_1}{c_2 }$

Hᴇɴᴄᴇ ,

ᴛʜᴇsᴇ ᴘᴀɪʀ ᴏғ ʟɪɴᴇᴀʀ ᴇǫᴜᴀᴛɪᴏɴs ʜᴀᴠᴇ ɴᴏ sᴏʟᴜᴛɪᴏɴ .

Hᴏᴘᴇ ɪᴛ ʜᴇʟᴘs..!