Solve the following pair of linear equations | vedclass

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

Question

(A) Given equations are:$1) 5x - 3y = 1$$2) 2x + 5y = 19$From equation $(1)$,we get $5x = 1 + 3y$,so $x = \frac{1 + 3y}{5}$.Substitute this value of $

Vedclass · Accepted Answer

$x = 2, y = 3$

Vedclass · Answer

(A) Given equations are:$1) 5x - 3y = 1$$2) 2x + 5y = 19$From equation $(1)$,we get $5x = 1 + 3y$,so $x = \frac{1 + 3y}{5}$.Substitute this value of $x$ into equation $(2)$:$2(\frac{1 + 3y}{5}) + 5y = 19$Multiply by $5$ to clear the denominator:$2(1 + 3y) + 25y = 95$$2 + 6y + 25y = 95$$31y = 93$$y = 3$Now,substitute $y = 3$ back into the expression for $x$:$x = \frac{1 + 3(3)}{5} = \frac{1 + 9}{5} = \frac{10}{5} = 2$.Thus,the solution is $(x, y) = (2, 3)$.

Solve the following pair of linear equations by the method of substitution: $5x - 3y = 1$ and $2x + 5y = 19$.

Similar Questions

Solve the following pair of equations: $2x + y = \frac{7xy}{3}, x + 3y = \frac{11xy}{3}$

Solve the following pairs of equations by the cross-multiplication method:
$4x + 6y = 11$
$5x - 8y = 6$

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

Draw the graphs of the pair of linear equations $x-y+2=0$ and $4x-y-4=0$. Calculate the area of the triangle formed by the lines so drawn and the $x$-axis (in sq. units).

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module