Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

If the ordered pairs \[({x^2} - 3x,{y^2} + 4y)\,and\,( - 2,5)\] are equal, then find x and y?

Answer
VerifiedVerified
509.4k+ views
Hint: Here in the given question we are given with two coordinates, and the x and y coordinates of both the coordinates are equal here, we need to solve for the values of x and y. to solve we need to equate the coordinate and thus solve with the equation form, here the equation form will be quadratic and need to be solved accordingly.

Complete step by step answer:
 Here in the given question we are provided with the coordinates, and it is given both the coordinate have same values of x and y, now on solving we get:
For x coordinate we have:
\[
   \Rightarrow {x^2} - 3x = - 2 \\
   \Rightarrow {x^2} - 3x + 2 = 0 \\
   \Rightarrow {x^2} - (2 + 1)x + 2 = 0 \\
   \Rightarrow {x^2} - 2x - 1x + 2 = 0 \\
   \Rightarrow x(x - 2) - 1(x - 2) = 0 \\
   \Rightarrow (x - 2)(x - 1) = 0 \\
   \Rightarrow x = 2,1 \\
   \Rightarrow for\,x = 2 \\
   \Rightarrow {x^2} - 3x \to {2^2} - 3(2) \to 4 - 6 = - 2 \\
   \Rightarrow for\,x = \\
   \Rightarrow {x^2} - 3x \to {1^2} - 3(1) \to 1 - 3 = - 2 \\
 \]
For y coordinate we have:
\[
   \Rightarrow {y^2} + 4y = 5 \\
   \Rightarrow {y^2} + 4y - 5 = 0 \\
   \Rightarrow {y^2} - (5 - 1)y - 5 = 0 \\
   \Rightarrow {y^2} - 5y + 1y - 5 = 0 \\
   \Rightarrow y(y - 5) + 1(y - 5) = 0 \\
   \Rightarrow (y - 5)(y + 1) = 0 \\
   \Rightarrow y = 5, - 1 \\
   \Rightarrow for\,y = 5 \\
   \Rightarrow {y^2} + 4y \to {5^2} - 4(5) \to 25 - 20 = 5 \\
 \]
\[
   \Rightarrow for\,y = - 1 \\
   \Rightarrow {y^2} + 4y \to {( - 1)^2} - 4( - 1) \to 1 + 4 = 5 \\
 \]
Here we get the values, of x and y as:
\[
   \Rightarrow x = 2,1 \\
   \Rightarrow y = 5, - 1 \\
 \]

Note: In the given question we need to find the values, after forming the equation between the coordinates, as here we know that the coordinates are equal and accordingly we create the equation, as per data given in the question.