Solve the given two equations and select the | vedclass

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

Question

(D) Given equations are:$4x + 7y = 209$  --- $(1)$$12x - 14y = -38$ --- $(2)$To eliminate $y$,multiply equation $(1)$ by $2$:$8x + 14y = 418$ --- $(3)

Vedclass · Accepted Answer

if $x = y$ or the relationship cannot be established.

Vedclass · Answer

(D) Given equations are:$4x + 7y = 209$  --- $(1)$$12x - 14y = -38$ --- $(2)$To eliminate $y$,multiply equation $(1)$ by $2$:$8x + 14y = 418$ --- $(3)$Now,add equation $(2)$ and equation $(3)$:$(12x - 14y) + (8x + 14y) = -38 + 418$$20x = 380$$x = 380 / 20 = 19$Substitute $x = 19$ into equation $(1)$:$4(19) + 7y = 209$$76 + 7y = 209$$7y = 209 - 76$$7y = 133$$y = 133 / 7 = 19$Since $x = 19$ and $y = 19$,we have $x = y$.

Solve the given two equations and select the correct answer from the given options.
$I.$ $4x + 7y = 209$
$II.$ $12x - 14y = -38$

Similar Questions

If $a, b, c \in \mathbb{Q}$,then the roots of the equation $(b + c - 2a)x^2 + (c + a - 2b)x + (a + b - 2c) = 0$ are

The sum of all real values of $x$ satisfying the equation $(x^2 - 5x + 5)^{x^2 + 4x - 60} = 1$ is:

If $\alpha$ and $\beta$ are the roots of the equation $ax^2 + bx + c = 0$,then the equation whose roots are $\alpha + \frac{1}{\beta}$ and $\beta + \frac{1}{\alpha}$ is:

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2}+5x-6=0$
$II.$ $2y^{2}-11y+15=0$

If $8, 2$ are the roots of ${x^2} + ax + \beta = 0$ and $3, 3$ are the roots of ${x^2} + \alpha x + b = 0$,then the roots of ${x^2} + ax + b = 0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Solve the given two equations and select the correct answer from the given options.$I.$ $4x + 7y = 209$$II.$ $12x - 14y = -38$