Question

x + y =2 and y = -1/4x^2 + 3

explain how to solve the system of equations algebraically.

Solve the system of equations.

Answer 1

\[x+y=2 \]
\[y=-\frac{ 1 }{ 4x^2 +3 }\]
Since "y" is already solved for in the equation number 2 you can just plug in the value on equation 1:
\[(2) \rightarrow (1) \iff x+(-\frac{ 1 }{ 4x^2 +3 } )=2 \]
multiplying both sides by the denominator of the second term:
\[x(4x^2 + 3 ) -1 =2(4x^2 +3)\]
Applying the distributive property:
\[4x^3+3x-1=8x^2 + 6 \]
\[4x^3 -8x^2 +3x-7=0 \]
Now, all you have to do is solve the third degree polynomial as you normally would.

Answer 2

The second equation is written wrong here. It's -1/4*x^2 + 3

Answer 3

That was a slip up on my part.

Answer 4

Well, the process is analogical, you take (2) and replace in (1):
\[x+(-\frac{ 1 }{ 4 }x^2+3)=2\]
Multiply both sides by "4" in order to get rid of the fraction:
\[4x-x^2 +12=8\]
And then forming a quadratic equation:
\[-x^2 -4x+4=0 \]

Answer 5

Does this include the first equation of x + y = 2?

Answer 6

Solve the first equation for y by eye ball.
y = 2-x
Replace y in the second equation using the above.
2-x = -1/4x^2 + 3
Solve the above for the two values of x.
Plug each x value into the y = 2-x equation to fine the cooresponding y value.

Answer 7

Based on my assignment, I have to solve the quadratic equation by either Factoring, using the Square Root Method, Completing the Square, or by used the Quadratic Formula. The linear equation would be solved as any other one of course. Then I simply explain how to solve the system which would be telling what form I used to solve the Quadratic Equation, and how I solved the Linear Equation.