Use the method of elimination to find the sol | vedclass

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

Question

(D) Given equations are:$5x + 6y = 14$ ............ $(1)$$3x - 2y = -14$ ............ $(2)$To eliminate $y$,multiply equation $(2)$ by $3$:$3(3x - 2y)

Vedclass · Accepted Answer

$(-2, 4)$

Vedclass · Answer

(D) Given equations are:$5x + 6y = 14$ ............ $(1)$$3x - 2y = -14$ ............ $(2)$To eliminate $y$,multiply equation $(2)$ by $3$:$3(3x - 2y) = 3(-14) \implies 9x - 6y = -42$ ............ $(3)$Now,add equation $(1)$ and equation $(3)$:$(5x + 6y) + (9x - 6y) = 14 + (-42)$$14x = -28$$x = \frac{-28}{14} = -2$Substitute $x = -2$ into equation $(1)$:$5(-2) + 6y = 14$$-10 + 6y = 14$$6y = 14 + 10$$6y = 24$$y = \frac{24}{6} = 4$Thus,the solution is $(x, y) = (-2, 4)$.

Use the method of elimination to find the solution of the following pair of linear equations: $5x + 6y = 14$ and $3x - 2y = -14$.

Similar Questions

One equation of a pair of dependent linear equations is $-5x + 7y = 2$. The second equation can be

If zero is inserted between the two digits of a two-digit number whose unit's place is $y$ and ten's place is $x$,then the new number is $\ldots \ldots \ldots \ldots$

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

If $\frac{3}{x}-\frac{2}{y}=5$ and $\frac{4}{x}-\frac{5}{y}=2,$ then the value of $\frac{1}{x}-\frac{1}{y}$ is ............ $(x, y \neq 0)$

The general form of a linear equation in two variables is .......... (where,$a^{2}+b^{2} \neq 0$).

Vedclass Test Series

Exam Paper Generator

Online Exam Module