Question

Two straight lines have equations y= px + 4 and py = qx-7, where p and q are constants.
The two lines meet at the point (3,1)
What is the value of q

Answer 1

Given : Two straight lines have equations y= px + 4 and py = qx-7, where p and q are constants.

The two lines meet at the point (3,1)

To Find : Value of q

Solution:

y= px + 4 and py = qx-7,  meet at ( 3 , 1)

Hence x = 3 , y = 1 is solution of both equations

y= px + 4

=> 1 = 3p + 4

=> 3p = - 3

=> p = - 1

py = qx-7

=> -1(1) = q(3) - 7

=> - 1 = 3q - 7

=> 6 = 3q

=> 2 = q

Value of q = 2

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Answer 2

$\huge\mathtt\colorbox{lime}{{\color{white}{ANSWER࿐}}}$

$\huge\color{lime}{\mathbb{\mathtt{GIVEN :}}}$

$\red\bigstar$ Two straight line have equations y = px + 4 & py = qx - 7, where p & q are constants.

$\red\bigstar$ The two lines meet at the point (3, 1).

$\huge\color{lime}{\mathbb{\mathtt{TO \: FIND :}}}$

$\red\bigstar$ What is the value of q ?

$\huge\color{lime}{\mathbb{\mathtt{SOLUTION :}}}$

$\red\bigstar$ Let x = 3 & y = 1

Now put the values in equation y = px +4,

$y = px + 4$

$➞ \: 1 = 3p + 4$

$➞ \: 3p = - 4 + 1$

$➞ \: 3p = - 3$

$➞ \: p = \frac{ - 3}{3}$

$➞ \: p = - 1$

Now put the values in equation py = qx - 7,

$py = qx - 7$

$➞ \: - 1(1) = q(3) - 7$

$➞ \: - 1 = 3q - 7$

$➞ \: 3q = 7 - 1$

$➞ \: 3q = 6$

$➞ \: q = \frac{6}{3}$

$➞ \: q = \small\boxed{{\sf\red{ \: 2 \: }}}$

$\huge\boxed{{\sf\red{Hence, \: the \: value \: of \: q \: is \: 2.}}}$

Hope it Helps Buddy ♥️