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

If $x+1$ is a factor of $2x^{3}+ax^{2}+2bx+1$,then find the values of $a$ and $b$ given that $2a-3b=4$.

Solve the following pair of equations by the method of cross-multiplication:
$2x + y = 35$
$3x + 4y = 65$

Solve the following pair of equations: $(a \neq 0, b \neq 0)$
$\frac{2}{a} = -1 - \frac{3}{b}$
$\frac{9}{b} = \frac{4}{a} + 7$

If $\frac{x+y}{x y}=2$ and $\frac{x-y}{x y}=6,$ then $x=\ldots \ldots \ldots \ldots .$

The cost of $2\,kg$ of apples and $1\,kg$ of grapes on a day was found to be Rs. $160$. After a month,the cost of $4\,kg$ of apples and $2\,kg$ of grapes is Rs. $300$. Represent this situation as a pair of linear equations in two variables.

Vedclass Test Series

Exam Paper Generator

Online Exam Module