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) Let $u = \frac{1}{x+y}$ and $v = \frac{1}{x-y}$.The equations become:$25u - 7v = -2$  --- $(1)$$15u - 7v = -4$  --- $(2)$Subtracting equation $(2)

Vedclass · Accepted Answer

$(3, 2)$

Vedclass · Answer

(C) Let $u = \frac{1}{x+y}$ and $v = \frac{1}{x-y}$.The equations become:$25u - 7v = -2$  --- $(1)$$15u - 7v = -4$  --- $(2)$Subtracting equation $(2)$ from equation $(1)$:$(25u - 7v) - (15u - 7v) = -2 - (-4)$$10u = 2 \implies u = \frac{2}{10} = \frac{1}{5}$.Substituting $u = \frac{1}{5}$ in equation $(1)$:$25(\frac{1}{5}) - 7v = -2$$5 - 7v = -2$$-7v = -7 \implies v = 1$.Now,$\frac{1}{x+y} = \frac{1}{5} \implies x+y = 5$ --- $(3)$And $\frac{1}{x-y} = 1 \implies x-y = 1$ --- $(4)$Adding $(3)$ and $(4)$:$2x = 6 \implies x = 3$.Substituting $x = 3$ in $(3)$:$3 + y = 5 \implies y = 2$.Thus,the solution is $(x, y) = (3, 2)$.

Solve the following pair of linear equations: $\frac{25}{x+y} - \frac{7}{x-y} = -2$ and $\frac{15}{x+y} - \frac{7}{x-y} = -4$.

Similar Questions

The ratio of the present ages of Rajan and Jay is $3:2$. After five years,the ratio of their ages will be $4:3$. Find the present ages of both.

Solve the following pair of linear equations by the substitution and cross-multiplication methods:
$8x + 5y = 9$
$3x + 2y = 4$

$A$ fraction becomes $\frac{1}{4}$ if $2$ is subtracted from the numerator and $3$ is added to the denominator. If $6$ is added to the numerator and the denominator is multiplied by $3$,it becomes $\frac{2}{3}$. Find the fraction.

If $3x + 5y = 8$ and $5x + 3y = 24$,then $x - y = \ldots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module