If the following pair of equations is consist | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The given equations are:$x - ky - 2 = 0$ ...... $(1)$$3x + 2y + 5 = 0$ ...... $(2)$Comparing these with the general form $a_1x + b_1y + c_1 = 0$ a

Vedclass · Accepted Answer

$-\frac{2}{3}$

Vedclass · Answer

(D) The given equations are:$x - ky - 2 = 0$ ...... $(1)$$3x + 2y + 5 = 0$ ...... $(2)$Comparing these with the general form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:$a_1 = 1, b_1 = -k, c_1 = -2$$a_2 = 3, b_2 = 2, c_2 = 5$For a unique solution,the condition is $\frac{a_1}{a_2} 
eq \frac{b_1}{b_2}$.Substituting the values:$\frac{1}{3} 
eq \frac{-k}{2}$Multiplying both sides by $6$:$2 
eq -3k$$k 
eq -\frac{2}{3}$Thus,the system has a unique solution for all real values of $k$ except $k = -\frac{2}{3}$.

If the following pair of equations is consistent (unique solution),then find the value of $k$:
$x - ky = 2$
$3x + 2y = -5$

Similar Questions

The numerator of a fraction is $4$ less than the denominator. If the numerator is decreased by $2$ and the denominator is increased by $1,$ then the denominator becomes eight times the numerator. Find the fraction.

If $5x + 3y = 10xy$ $(x \neq 0, y \neq 0)$,then $\frac{6}{x} + \frac{10}{y} = ...........$

Solve the following pairs of equations:
$\frac{x}{3} + \frac{y}{4} = 4$
$\frac{5x}{6} - \frac{y}{8} = 4$

Raj got $\frac{3}{4}$ times the marks as obtained by Nil in the annual examination of Mathematics. The sum of their marks is $140$. Represent this as a pair of linear equations in two variables.

Find the value$(s)$ of $p$ for the pair of equations $-x + py = 1$ and $px - y = 1$ if the system has no solution.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

If the following pair of equations is consistent (unique solution),then find the value of $k$:$x - ky = 2$$3x + 2y = -5$