Do the following pair of linear equations have no solution? Justify your answer.
$y + 6x = 6$ and $y = 2x$

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(B) The condition for a pair of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ to have no solution is $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.
Given equations are:
$6x + y - 6 = 0$ --- $(i)$
$2x - y = 0$ --- $(ii)$
Here,$a_1 = 6, b_1 = 1, c_1 = -6$ and $a_2 = 2, b_2 = -1, c_2 = 0$.
Calculating the ratios:
$\frac{a_1}{a_2} = \frac{6}{2} = 3$
$\frac{b_1}{b_2} = \frac{1}{-1} = -1$
Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ $(3 \neq -1)$,the lines intersect at a single point.
Therefore,the given pair of linear equations has a unique solution,not 'no solution'.

Explore More

Similar Questions

$A$ fraction becomes $\frac{4}{5}$ if $1$ is added to both the numerator and the denominator. If $5$ is subtracted from both the numerator and the denominator,it becomes $\frac{1}{2}$. Find the fraction.

The solution set of the pair of equations $x+y+1=0$ and $3x+3y+k=0$ is an infinite set. Then $k = \dots$

If $4x - 12y = 20$,then $5x - 15y = \ldots$

Solve the following pair of linear equations: $\frac{25}{x+y} - \frac{7}{x-y} = -2$ and $\frac{15}{x+y} - \frac{7}{x-y} = -4$.

Difficult
View Solution

The solution set of the pair of equations $2x + 3y = 4$ and $6x + 9y = 17$ is ............

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo