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:$21x + 47y = 110$  --- $(1)$$47x + 21y = 162$  --- $(2)$Adding $(1)$ and $(2)$:$(21x + 47x) + (47y + 21y) = 110 + 162$$68x + 6

Vedclass · Accepted Answer

$x = 3, y = 1$

Vedclass · Answer

(A) Given equations are:$21x + 47y = 110$  --- $(1)$$47x + 21y = 162$  --- $(2)$Adding $(1)$ and $(2)$:$(21x + 47x) + (47y + 21y) = 110 + 162$$68x + 68y = 272$Dividing by $68$:$x + y = 4$  --- $(3)$Subtracting $(1)$ from $(2)$:$(47x - 21x) + (21y - 47y) = 162 - 110$$26x - 26y = 52$Dividing by $26$:$x - y = 2$  --- $(4)$Adding $(3)$ and $(4)$:$(x + y) + (x - y) = 4 + 2$$2x = 6$$x = 3$Substituting $x = 3$ in $(3)$:$3 + y = 4$$y = 1$Thus,the solution is $x = 3, y = 1$.

Solve the following pair of linear equations:
$21x + 47y = 110$
$47x + 21y = 162$

Similar Questions

Solve the following pair of linear equations by the method of substitution: $8x - 3y = 1$ and $34x - 3y = 14$.

If $\frac{5}{x}+\frac{3}{y}=4$ and $\frac{3}{x}+\frac{5}{y}=2,$ then $\frac{1}{x}-\frac{1}{y}=\ldots \ldots \ldots \ldots .(x, y \neq 0)$

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

Solve the following pairs of equations:
$\frac{1}{2x} - \frac{1}{y} = -1$
$\frac{1}{x} + \frac{1}{2y} = 8, \quad x, y \neq 0$

Draw the graphs of the equations $x=3, x=5$ and $2x-y-4=0$. Also,find the area of the quadrilateral formed by these lines and the $x$-axis.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Solve the following pair of linear equations:$21x + 47y = 110$$47x + 21y = 162$