Nathaniel G. answered • 03/23/14
Tutor
New to Wyzant
Math, English, and Physics
You solve these just like any other system of linear equations, only for a quadratic, you will have two solutions to both x and y.
3x+2y=5 and 2x^2+xy+y^2=14
3x+2y=5 and 2x^2+xy+y^2=14
Solve the first equation for x (or y, it doesn't matter!) and plug it into the second equation.
x=5/3-2/3 y
x=5/3-2/3 y
2(5/3-2/3 y)^2+(5/3-2/3 y)y+y^2=14
This simplifies down to,
1/9 (11y^2-25y-34)=0
1/9 (11y^2-25y-34)=0
Now either 1/9 equals zero of the y term down. Since 1/9 cannot be zero, all we need to do is factor the quadratic and solve for y. Using the formula
(-b±√(b^2-4ac))/2a
(-b±√(b^2-4ac))/2a
We get y=3.23, -0.957 rounded.
We then go back to our x express and plug in both values for y to get the values for x.
x=5/3-(2/3)(3.23) x=5/3-(2/3)(-0.957)
We get x= -0.487, 2.305 rounded.
We get x= -0.487, 2.305 rounded.
