Solve the following pairs of equations by cro | vedclass

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

Question

(C) Given equations are:$3x + y - 5 = 0$  --- $(1)$$5x + 3y - 3 = 0$  --- $(2)$Comparing with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we g

Vedclass · Accepted Answer

$(3, -4)$

Vedclass · Answer

(C) Given equations are:$3x + y - 5 = 0$  --- $(1)$$5x + 3y - 3 = 0$  --- $(2)$Comparing with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:$a_1 = 3, b_1 = 1, c_1 = -5$$a_2 = 5, b_2 = 3, c_2 = -3$Using the cross-multiplication formula:$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$Substituting the values:$\frac{x}{(1)(-3) - (3)(-5)} = \frac{y}{(-5)(5) - (-3)(3)} = \frac{1}{(3)(3) - (5)(1)}$$\frac{x}{-3 + 15} = \frac{y}{-25 + 9} = \frac{1}{9 - 5}$$\frac{x}{12} = \frac{y}{-16} = \frac{1}{4}$For $x$: $\frac{x}{12} = \frac{1}{4} \implies x = 3$For $y$: $\frac{y}{-16} = \frac{1}{4} \implies y = -4$Thus,the solution is $(3, -4)$.

Solve the following pairs of equations by cross-multiplication method:
$3x + y = 5$
$5x + 3y = 3$

Similar Questions

The graph of the equation $\ldots \ldots \ldots \ldots$ is a line perpendicular to the $Y-axis.$

If the sum of two numbers is $20$ and their difference is $8,$ then the larger number is ...........

The solution set of a pair of equations $3x + 3y - 3 = 0$ and $5x + 5y - 5 = 0$ is $\ldots \ldots \ldots \ldots$

Find the value of $k$ for the following pair of linear equations such that the system has no solution:
$2x - ky + 3 = 0$
$3x + 2y - 1 = 0$

By the graphical method,find whether the following pair of equations are consistent or not. If consistent,solve them.
$3x + y + 4 = 0$
$6x - 2y + 4 = 0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Solve the following pairs of equations by cross-multiplication method:$3x + y = 5$$5x + 3y = 3$