STATEMENT -1 : $3 x + 2 y = 5, 2 x - 3 y = 7$ are consistent pair of equation.
STATEMENT - 2: If the pair of linear equations is given by $a_{1} x + b_{1} y + c_{1} = 0, a_{2} x + b_{2} y + c_{2} = 0$ and $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$ .
In this case, the pair of linear equations is consistent.

Statement - 1 is True, Statement - 2 is True, Statement - 2 is a correct explanation for Statement - 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Statement - 1 is True, Statement - 2 is True : Statement 2 is NOT a correct explanation for Statement - 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Statement - 1 is True, Statement - 2 is False

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Statement - 1 is False, Statement - 2 is True

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C Statement - 1 is True, Statement - 2 is False
For equations $S_{1} : \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$ and hence the pair of lines will be consistent.
If $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$ , then there will not be any solution and hence the system of equations will be inconsistent.

Suggest Corrections

Similar questions

\sqrt{2} x + \sqrt{3} y = 0, \sqrt{3} x - \sqrt{8} y = 0

a_{1} x + b_{1} y + c_{1} = 0, a_{2} x + b_{2} y + c_{2} = 0

\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}

- 1 : {log}_{10} x < {log}_{π} x < {log}_{e} x < {log}_{2} x \forall x > 1

- 2 : x < y \Rightarrow {log}_{a} x > {log}_{a} y

0 < a < 1

- 1 :

0 < x < y \Rightarrow {log}_{a} x > {log}_{a} y

a > 1

- 2 :

a > 1, {log}_{a} x

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} = (n + 2) 2^{n - 1} .

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} x^{r} = (1 + x)^{n} + n x (1 + x)^{n - 1} .

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} = (n + 2) 2^{n - 1} .

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} x^{r} = (1 + x)^{n} + n x (1 + x)^{n - 1} .

Join BYJU'S Learning Program