If $(2x-1,3y+1)$ and $(x+3,y-4)$ are equal ordered pairs. Find the values of $x$ and $y$.
Given:
$(2x-1,3y+1)$ and $(x+3,y-4)$ are equal ordered pairs.
To do:
We have to find the values of $x$ and $y$.
Solution:
We know that,
Two ordered pairs are equal if and only if the corresponding first components are equal and the second components are equal.
Therefore,
$2x-1 = x+3$ and $3y+1 = y-4$
$2x-x=3+1$ and $3y-y=-4-1$
$x=4$ and $2y=-5$
$x=4$ and $y=\frac{-5}{2}$
The values of $x$ and $y$ are $4$ and $\frac{-5}{2}$ respectively.
Related Articles
- Find the values of x and y from the following ordered pairs.$(2x-y, 2) = (5, y+1)$
- If \( 2 x+y=23 \) and \( 4 x-y=19 \), find the values of \( 5 y-2 x \) and \( \frac{y}{x}-2 \).
- If the points $(-2, -1), (1, 0), (x, 3)$ and $(1, y)$ form a parallelogram, find the values of $x$ and $y$.
- Find $25 x^{2}+16 y^{2}$, if $5 x+4 y=8$ and $x y=1$.
- If $(1, 2), (4, y), (x, 6)$ and $(3, 5)$ are the vertices of a parallelogram taken in order, find $x$ and $y$.
- If $x = 3$ and $y = -1$, find the values of each of the following using in identity:\( \left(9 y^{2}-4 x^{2}\right)\left(81 y^{4}+36 x^{2} y^{2}+16 x^{4}\right) \)
- If \( x=a, y=b \) is the solution of the equations \( x-y=2 \) and \( x+y=4 \), then the values of \( a \) and \( b \) are, respectively(A) 3 and 5(B) 5 and 3(C) 3 and 1,b>(D) \( -1 \) and \( -3 \)
- \Find $(x +y) \div (x - y)$. if,(i) \( x=\frac{2}{3}, y=\frac{3}{2} \)(ii) \( x=\frac{2}{5}, y=\frac{1}{2} \)(iii) \( x=\frac{5}{4}, y=\frac{-1}{3} \)(iv) \( x=\frac{2}{7}, y=\frac{4}{3} \)(v) \( x=\frac{1}{4}, y=\frac{3}{2} \)
- Simplify: $\frac{x^{-3}-y^{-3}}{x^{-3} y^{-1}+(x y)^{-2}+y^{-1} x^{-3}}$.
- Solve the following pairs of equations by reducing them to a pair of linear equations:(i) \( \frac{1}{2 x}+\frac{1}{3 y}=2 \)\( \frac{1}{3 x}+\frac{1}{2 y}=\frac{13}{6} \)(ii) \( \frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2 \)\( \frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1 \)(iii) \( \frac{4}{x}+3 y=14 \)\( \frac{3}{x}-4 y=23 \)(iv) \( \frac{5}{x-1}+\frac{1}{y-2}=2 \)\( \frac{6}{x-1}-\frac{3}{y-2}=1 \)(v) \( \frac{7 x-2 y}{x y}=5 \)\( \frac{8 x+7 y}{x y}=15 \),b>(vi) \( 6 x+3 y=6 x y \)\( 2 x+4 y=5 x y \)4(vii) \( \frac{10}{x+y}+\frac{2}{x-y}=4 \)\( \frac{15}{x+y}-\frac{5}{x-y}=-2 \)(viii) \( \frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4} \)\( \frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=\frac{-1}{8} \).
- If $x^2 + y^2 = 29$ and $xy = 2$, find the value of(i) $x + y$(ii) $x - y$(iii) $x^4 + y^4$
- If $x = 3$ and $y = -1$, find the values of each of the following using in identity:\( \left(\frac{x}{4}-\frac{y}{3}\right)\left(\frac{x^{2}}{16}+\frac{x y}{12}+\frac{y^{2}}{9}\right) \)
- 1. Factorize the expression \( 3 x y - 2 + 3 y - 2 x \)A) \( (x+1),(3 y-2) \)B) \( (x+1),(3 y+2) \)C) \( (x-1),(3 y-2) \)D) \( (x-1),(3 y+2) \)2. Factorize the expression \( \mathrm{xy}-\mathrm{x}-\mathrm{y}+1 \)A) \( (x-1),(y+1) \)B) \( (x+1),(y-1) \)C) \( (x-1),(y-1) \)D) \( (x+1),(y+1) \)
- If $x+y=5$ and $xy=6$ then find $x^3+y^3$.
- For $x=\frac{3}{4} $and $y=\frac{-9}{8}$, insert a rational number between:$(x+y)^{-1} and x^{-1}+y^{-1} $
Kickstart Your Career
Get certified by completing the course
Get Started
Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies