Answer:
(5, 1) and (-5, -7)
Step-by-step explanation:
Given the simultaneous equations
x^2– y^2 = 24 .... 1
x= 3 + 2y .... 2
Substitute 2 into 1
(3-2y)^2-y^2 = 24
Expand
9 - 12y + 4y^2 - y^2 = 24
9 - 12y + 3y^2 = 24
3y^2 - 12y + 9-24 = 0
3y^2 -12y -15 = 0
y^2-4y-5 = 0
y^2+5y - y - 5 = 0
y(y+5) - 1 (y+5) = 0
y-1 = 0 and y+5 = 0
y = 1 and -5
Substitute the values of y into 2
Recall that x = 3+3y
when y = 1
x = 3 + 2(1)
x = 5
when x = -5
x = 3 + 2(-5)
x = 3 - 10
x = -7
Hence the solutions are (5, 1) and (-5, -7)
