For the given pair of linear equations,
$a_{1} x + b_{1} y + c_{1} = 0$ and
$a_{2} x + b_{2} y + c_{2} = 0$ .

The values of x and y are ( , ). Fill in the blanks

\frac{(b_{1} c_{2} - b_{2} c_{1})}{(a_{1} b_{2} - a_{2} b_{1})}, \frac{(c_{1} a_{2} - c_{2} a_{1})}{(a_{1} b_{2} - a_{2} b_{1})}

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

\frac{(b_{2} c_{2} - b_{1} c_{1})}{(a_{1} b_{2} - a_{2} b_{1})}, \frac{(c_{2} a_{2} - c_{1} a_{1})}{(a_{1} b_{2} - a_{2} b_{1})}

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

\frac{(b_{2} c_{1} - b_{1} c_{2})}{(a_{1} b_{2} - a_{2} b_{1})}, \frac{(c_{1} a_{1} - c_{2} a_{2})}{(a_{1} b_{2} - a_{2} b_{1})}

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

None of the above

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

Open in App

Solution

The correct option is A $\frac{(b_{1} c_{2} - b_{2} c_{1})}{(a_{1} b_{2} - a_{2} b_{1})}, \frac{(c_{1} a_{2} - c_{2} a_{1})}{(a_{1} b_{2} - a_{2} b_{1})}$
The following table is useful to remember the solutions using the method of cross multiplication:

The solution is given by:

$\frac{x}{b_{1} c_{2} - b_{2} c_{1}} = \frac{y}{c_{1} a_{2} - c_{2} a_{1}} = \frac{1}{a_{1} b_{2} - a_{2} b_{1}}$

$\frac{x}{b_{1} c_{2} - b_{2} c_{1}} = \frac{y}{c_{1} a_{2} - c_{2} a_{1}} = \frac{1}{a_{1} b_{2} - a_{2} b_{1}}$

Therefore, $x = \frac{(b_{1} c_{2} - b_{2} c_{1})}{(a_{1} b_{2} - a_{2} b_{1})}, a n d y = \frac{(c_{1} a_{2} - c_{2} a_{1})}{(a_{1} b_{2} - a_{2} b_{1})}$ .

Suggest Corrections

Similar questions

The solution (x,y) for the given pair of linear equations is $a_{1}$ x + $b_{1}$ y + $c_{1}$ =0 and $a_{2}$ x + $b_{2}$ y + $c_{2}$ =0

For the given pair of linear equations $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0,$ the solution $(x, y)$ is

For the pair of linear equations $a_{1} x$ + $b_{1} y$ + $c_{1} = 0$ and $a_{2} x$ + $b_{2} y$ + $c_{2} = 0$ , the solution set are $x = \frac{(b_{1} c_{2} - b_{2} c_{1})}{(a_{1} b_{2} - a_{2} b_{1})}$ and $y = \frac{(c_{1} a_{2} - c_{2} a_{1})}{(a_{1} b_{2} - a_{2} b_{1})}$ . They can also be written as:

Solve the two equations using method of elimination.

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

The solution set x= $\frac{(b_{1} c_{2} - b_{2} c_{1})}{(a_{1} b_{2} - a_{2} b_{1})}$ and y= $\frac{(c_{1} a_{2} - c_{2} a_{1})}{(a_{1} b_{2} - a_{2} b_{1})}$ to the pair of linear equations $a_{1}$ x + $b_{1}$ y + $c_{1}$ =0 and $a_{2}$ x + $b_{2}$ y + $c_{2}$ =0 can be written as

Join BYJU'S Learning Program

For the given pair of linear equations, a1x+b1y+c1=0 and a2x+b2y+c2=0. The values of x and y are ( __ , __ ). Fill in the blanks

For the given pair of linear equations,
$a_{1} x + b_{1} y + c_{1} = 0$ and
$a_{2} x + b_{2} y + c_{2} = 0$ .

The values of x and y are ( , ). Fill in the blanks