Find the values of x and y from the following ordered pairs.$(2x-y, 2) = (5, y+1)$
Given :
The given expression is $(2x-y, 2) = (5, y+1)$.
To do :
We have to find the value of x and y.
Solution :
$(2x-y, 2) = (5, y+1)$
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,
$2x-y = 5$ and $2 = y+1$
$y+1 = 2$
$y=2-1$
$y=1$
Substituting $y=1$ in $2x-y = 5$
$2x-(1) = 5$
$2x = 5+1$
$2x = 6$
$x = \frac{6}{2}$
$x = 3$.
The values of $x$ and $y$ are 3 and 1 respectively.
Related Articles
- If $(2x-1,3y+1)$ and $(x+3,y-4)$ are equal ordered pairs. Find the values of $x$ and $y$.
- If \( 2 x+y=23 \) and \( 4 x-y=19 \), find the values of \( 5 y-2 x \) and \( \frac{y}{x}-2 \).
- Simplify the following:$5+[x-[2y-(6x+y-4)+2x]-[x-(y-2)]]$
- Solve the following pairs of equations:\( \frac{1}{2 x}-\frac{1}{y}=-1 \)\( \frac{1}{x}+\frac{1}{2 y}=8, x, y ≠ 0 \)
- Find $25 x^{2}+16 y^{2}$, if $5 x+4 y=8$ and $x y=1$.
- Solve the following pairs of equations:\( \frac{2 x y}{x+y}=\frac{3}{2} \)\( \frac{x y}{2 x-y}=\frac{-3}{10}, x+y ≠ 0,2 x-y ≠ 0 \)
- If $x=1,\ y=2$ and $z=5$, find the value of $x^{2}+y^{2}+z^{2}$.
- Find the value of y : $\frac{y^2 + 1}{y^2 - 1} = \frac{5}{4}$
- Solve the following pairs of equations:\( x+y=3.3 \)\( \frac{0.6}{3 x-2 y}=-1,3 x-2 y ≠ 0 \)
- Solve the following system of equations:$\frac{5}{x+1} -\frac{2}{y-1}=\frac{1}{2}$$\frac{10}{x+1}+\frac{2}{y-1}=\frac{5}{2}$, where $x≠-1$ and $y≠1$
- Find the following products and verify the result for $x = -1, y = -2$:\( \left(\frac{1}{3} x-\frac{y^{2}}{5}\right)\left(\frac{1}{3} x+\frac{y^{2}}{5}\right) \)
- 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) \( (x+1),(y+1) \)
- Solve the following system of equations:$\frac{5}{x+y} -\frac{2}{x-y}=-1$$\frac{15}{x+y}+\frac{7}{x-y}=10$
- Simplify each of the following:\( (2 x-5 y)^{3}-(2 x+5 y)^{3} \)
- Find the sum of the following arithmetic progressions:\( \frac{x-y}{x+y}, \frac{3 x-2 y}{x+y}, \frac{5 x-3 y}{x+y}, \ldots \) to \( n \) terms
Kickstart Your Career
Get certified by completing the course
Get Started