Question 79579
Solve this system of equations: I would use substitution.
1) {{{x^2-4y^2 = 16}}}
2) {{{2y-x = 2}}} Write this as:
2a) {{{2y = x+2}}} Now square both sides. {{{4y^2 = x^2+4x+4}}} and substitute into equation 1).
{{{x^2-(x^2+4x+4) = 16}}} Remove the parentheses (Remember to change the signs of the terms inside the parentheses!)
{{{x^2-x^2-4x-4 = 16}}} Simplify.
{{{-4x-4 = 16}}} Add 4 to both sides.
{{{-4x = 20}}} Divide both sides by -4.
{{{x = -5}}} Now substitute this into equation 2a) and solve for y.
2a) {{{2y = -5+2}}}
{{{2y = -3}}} Divide both sides by 2.
{{{y = -3/2}}}
The solution is: (-5, -3/2)
Check:
{{{(-5)^2-4(-3/2)^2 = 25-4(9/4)}}}={{{25-9 = 16}}}