Solve the following pairs of equations:43x + | vedclass

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

Question

(C) Given pair of linear equations is:$43x + 67y = -24 .....(i)$$67x + 43y = 24 .....(ii)$Adding equations $(i)$ and $(ii)$:$(43 + 67)x + (67 + 43)y =

Vedclass · Accepted Answer

$1, -1$

Vedclass · Answer

(C) Given pair of linear equations is:$43x + 67y = -24 .....(i)$$67x + 43y = 24 .....(ii)$Adding equations $(i)$ and $(ii)$:$(43 + 67)x + (67 + 43)y = -24 + 24$$110x + 110y = 0$$110(x + y) = 0$$x + y = 0 \Rightarrow x = -y .....(iii)$Subtracting equation $(i)$ from equation $(ii)$:$(67 - 43)x + (43 - 67)y = 24 - (-24)$$24x - 24y = 48$$24(x - y) = 48$$x - y = 2 .....(iv)$Substituting $x = -y$ from $(iii)$ into $(iv)$:$-y - y = 2$$-2y = 2$$y = -1$Substituting $y = -1$ into $(iii)$:$x = -(-1) = 1$Thus,the solution is $x = 1$ and $y = -1$.

Solve the following pairs of equations:
$43x + 67y = -24$
$67x + 43y = 24$

Similar Questions

For which values of $p$ and $q$ will the following pair of linear equations have infinitely many solutions?
$4x + 5y = 2$
$(2p + 7q)x + (p + 8q)y = 2q - p + 1$

If $2x + y = 35$ and $3x + 4y = 65$,then find $\frac{x}{y}$.

Solve the following pairs of equations:
$\frac{1}{2x} - \frac{1}{y} = -1$
$\frac{1}{x} + \frac{1}{2y} = 8, \quad x, y \neq 0$

The pair of equations $2x + y = 3$ and $x + 2y = 4$ is:

After $x$ years,the sum of the ages of Sachin and Sehwag will be $y$ years,then the sum of the ages before $3$ years was $\ldots \ldots \ldots \ldots$ years.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Solve the following pairs of equations:$43x + 67y = -24$$67x + 43y = 24$