Solve the following pair of linear equations | vedclass

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

Question

(C) Given equations are:$1$) $\frac{x}{a} + \frac{y}{b} = 0 \implies bx + ay = 0$$2$) $(a+b)x + (a-b)y = a^2 + b^2$Rewriting in standard form $a_1x +

Vedclass · Accepted Answer

$(a, -b)$

Vedclass · Answer

(C) Given equations are:$1$) $\frac{x}{a} + \frac{y}{b} = 0 \implies bx + ay = 0$$2$) $(a+b)x + (a-b)y = a^2 + b^2$Rewriting in standard form $a_1x + b_1y + c_1 = 0$:$1$) $bx + ay + 0 = 0$$2$) $(a+b)x + (a-b)y - (a^2 + b^2) = 0$Using cross-multiplication method $\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$:$\frac{x}{a(-(a^2+b^2)) - (a-b)(0)} = \frac{y}{0(a+b) - (- (a^2+b^2))b} = \frac{1}{b(a-b) - a(a+b)}$$\frac{x}{-a(a^2+b^2)} = \frac{y}{b(a^2+b^2)} = \frac{1}{ab - b^2 - a^2 - ab}$$\frac{x}{-a(a^2+b^2)} = \frac{y}{b(a^2+b^2)} = \frac{1}{-(a^2+b^2)}$For $x$: $x = \frac{-a(a^2+b^2)}{-(a^2+b^2)} = a$For $y$: $y = \frac{b(a^2+b^2)}{-(a^2+b^2)} = -b$Thus,the solution is $(a, -b)$.

Solve the following pair of linear equations by the cross-multiplication method:
$\frac{x}{a} + \frac{y}{b} = 0$
$(a+b)x + (a-b)y = a^2 + b^2$

Similar Questions

$x=-1$ and $y=-2$ is one of the solutions of the equation $\ldots \ldots \ldots \ldots$

For the pair of equations $\frac{2}{x} + \frac{3}{y} = 3$ and $\frac{3}{x} + \frac{2}{y} = 7$,find the value of $\frac{3}{x} + \frac{3}{y}$.

The sum of a two-digit number and the number obtained by interchanging its digits is $110$. If the number obtained by subtracting $10$ from the original number exceeds five times the sum of the digits of the original number by $4$,then find the original number.

Do the equations $4x + 3y - 1 = 5$ and $12x + 9y = 15$ represent a pair of coincident lines? Justify your answer.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Solve the following pair of linear equations by the cross-multiplication method:$\frac{x}{a} + \frac{y}{b} = 0$$(a+b)x + (a-b)y = a^2 + b^2$