Solve for x and y: (2x + 3y -7)^2+ (x – y + 1)^2= 0
Answers
Answer:
y = 9/ 5 and x = 4/5
Explanation:
Here,
⇒ ( 2x + 3y - 7 )^2 + ( x - y + 1 )^2 = 0
Square on any real number can never be 0 if the number is not 0.
So here if sum of square of these numbers is 0, these number are 0 itself.
Case 1 : ( 2x + 3y - 7 ) = 0
⇒ 2x + 3y - 7 = 0
⇒ 2x + 3y = 7
Case 2 : ( x - y + 1 ) = 0
⇒ x - y + 1 = 0
⇒ x = y - 1
Substituting the value of x :
⇒ 2x + 3y = 7
⇒ 2( y - 1 ) + 3y = 7
⇒ 2y - 2 + 3y = 7
⇒ 5y = 9
⇒ y = 9/5
Hence,
x = 1 - (9/5) = 4/5
Answer:y = y = 9/ 5 and x = 4/5
Explanation:
Here,
⇒ ( 2x + 3y - 7 )^2 + ( x - y + 1 )^2 = 0
Square on any real number can never be 0 if the number is not 0.
So here if sum of square of these numbers is 0, these number are 0 itself.
Case 1 : ( 2x + 3y - 7 ) = 0
⇒ 2x + 3y - 7 = 0
⇒ 2x + 3y = 7
Case 2 : ( x - y + 1 ) = 0
⇒ x - y + 1 = 0
⇒ x = y - 1
Substituting the value of x :
⇒ 2x + 3y = 7
⇒ 2( y - 1 ) + 3y = 7
⇒ 2y - 2 + 3y = 7
⇒ 5y = 9
⇒ y = 9/5
Hence,
x = 1 - (9/5) = 4/5