Solve the following pair of equations by the | vedclass

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

Question

(B) The given equations are:$ax + by + (-a + b) = 0$$bx - ay + (-a - b) = 0$Comparing with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y

Vedclass · Accepted Answer

$(1, -1)$

Vedclass · Answer

(B) The given equations are:$ax + by + (-a + b) = 0$$bx - ay + (-a - b) = 0$Comparing with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:$a_1 = a, b_1 = b, c_1 = -a + b$$a_2 = b, b_2 = -a, c_2 = -a - b$Using the cross-multiplication method:$\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}{b(-a - b) - (-a)(-a + b)} = \frac{y}{(-a + b)(b) - (-a - b)(a)} = \frac{1}{a(-a) - b(b)}$$\frac{x}{-ab - b^2 - (a^2 - ab)} = \frac{y}{-ab + b^2 - (-a^2 - ab)} = \frac{1}{-a^2 - b^2}$$\frac{x}{-ab - b^2 - a^2 + ab} = \frac{y}{-ab + b^2 + a^2 + ab} = \frac{1}{-a^2 - b^2}$$\frac{x}{-(a^2 + b^2)} = \frac{y}{a^2 + b^2} = \frac{1}{-(a^2 + b^2)}$Now,solving for $x$ and $y$:$x = \frac{-(a^2 + b^2)}{-(a^2 + b^2)} = 1$$y = \frac{a^2 + b^2}{-(a^2 + b^2)} = -1$Thus,the solution is $(x, y) = (1, -1)$.

Solve the following pair of equations by the method of cross multiplication:
$ax + by - a + b = 0$
$bx - ay - a - b = 0$

Similar Questions

Do the following equations represent a pair of coincident lines? Justify your answer.
$-2x - 3y = 1$
$6y + 4x = -2$

Solve the following pair of equations: $(a \neq 0, b \neq 0)$
$\frac{5}{a} + \frac{3}{2b} = 1$
$\frac{1}{2a} - \frac{3}{b} = 1$

The solution set of $x + 2y - 3 = 0$ and $5x + ky + 7 = 0$ is an empty set. The value of $k$ is .......

Solve the following pairs of equations by the method of elimination:
$\frac{x}{2} + \frac{3y}{5} + 1 = 0$
$\frac{x}{2} + \frac{y}{3} - \frac{1}{3} = 0$

Two years ago,the sum of the ages of a mother,father,and two daughters was $40$ years. After three years,what will be the sum of their ages?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Solve the following pair of equations by the method of cross multiplication:$ax + by - a + b = 0$$bx - ay - a - b = 0$