SOLUTION: solve algebraically for each system given below x^2-4y^2=16 2y-x=2 show steps please

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Question 931587: solve algebraically for each system given below
x^2-4y^2=16
2y-x=2
show steps please
Found 2 solutions by ewatrrr, MathLover1:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
2y-x=2 0r x = 2y-2
.....
x^2-4y^2=16
(2y-2)^2 - 4y^2 = 16
4y^2 - 8y + 4 - 4y^2 = 16
-8y + 4 = 16
-8y = 12
y = -12/8 = -3/2
x = -5 2%28-3%2F2%29+-+2
(-5, -1.5) the solution set for this System

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2-4y%5E2=16.....eq.1
2y-x=2.....eq.2
____________________
start with
2y-x=2.....eq.2...solve for x
2y-2=x...substitute in eq.1

%282y-2%29%5E2-4y%5E2=16.....eq.1..solve for y
4y%5E2-8y%2B4-4y%5E2=16
cross%284y%5E2%29-8y%2B4-cross%284y%5E2%29=16
-8y%2B4=16
-16%2B4=8y
-12=8y
-12%2F8=y
highlight%28y=-3%2F2%29
go to eq.1 or eq.2, plug in value of y and solve for x

2%28-3%2F2%29-x=2.....eq.2
-3-x=2
-3-2=x
highlight%28x=-5%29
your solution is: highlight%28x=-5%29 and highlight%28y=-3%2F2%29