Question

Solve algebraically the simultaneous equations
x^2– y^2 = 24
x= 3 + 2y
Write your two pairs of answers in this form:
when x =... y =...
When x=... y=...

Answer 1

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)