Question

What is the solution set of y=2x^2+8x and x-2.23y+10.34=0

Answer 1

First, let's put the second equation, x-2.23y+10.34=0, in terms of y:
x - 2.23y +10.34 = 0   
2.23y = x + 10.34
y =  .45x + 4.64
Now we can substitute the right side of this equation for y in the first equation
y=2x^2+8x
.45x + 4.64 = 2x^2 + 8x
Turn it into a quadratic by getting 0 on one side:
2x^2 + 8x - .45x - 4.64 = 0
2x^2 + 7.55x - 4.64 = 0 Divide both sides by 2
x^2 + 3.76x - 2.32 = 0
x =( -b +/- √(b² - 4ac) ) / 2a
x =( -3.76 +/- √(14.14 + 9.28)) ÷ 2
x = .54 or -4.31

Plug the x values into y = .45x + 4.64
y = .45 (.54) + 4.64
y = 4.88 when x= .54

y = .45 (-4.31) + 4.64
y = 2.70 when x= -4.31

Solution set:
{ (0.54, 4.88) , (-4.31, 2.70) }