MATHEMATICS
JAMB 2012 - Question 13
Mathematics 2012 JAMB Past Questions - Question 13: Solve for x and y in the equation belowx² - y² = 4 , x +y =2
A:
B:
C:
D:
Correct Answer
C
Explanation
x = 2 – y (2 –y)² - y² = 44 – 2y + y² - y² =4 4 – 2y = 4 4 – 4 = 2y 2y = 0 i.e y = 0 X + y = 2X + 0 =2 X = 2, y = 0To solve the system of equations x² - y² = 4 and x + y = 2, we can use the method of substitution.From the second equation, we can express x in terms of y as x = 2 - y.Substituting this value of x into the first equation, we have:(2 - y)² - y² = 4Expanding the equation:4 - 4y + y² - y² = 4Combining like terms:-4y = 0Dividing both sides by -4:y = 0Now, substituting the value of y back into the second equation:x + 0 = 2x = 2Therefore, the solution to the system of equations x² - y² = 4 and x + y = 2 is x = 2 and y = 0.
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