Question

Solve the following system of equations. Show all work and solutions.

y = 3x^2 + 6x + 4
y = −3x^2 + 4

Answer 1

Answer:

y = 4, x = 0

y = 1, x = -1

Step-by-step explanation:

$y = 3 {x}^{2} + 6x + 4$

$y = - 3 {x}^{2} + 4$

$\:$

Since both of them are equal to y, we can just equate them and solve.

$\:$

$3 {x}^{2} + 6x + 4 = - 3 {x}^{2} + 4$

$3 {x}^{2} + 6x + 4 + 3 {x}^{2} - 4 = 0$

$6 {x}^{2} + 6x = 0$

$6x(x + 1) = 0$

$6x = 0$

$x = 0$

$\:$

$x + 1 = 0$

$x = - 1$

$\:$

$for \: x = 0$

$y = 3( {0})^{2} + 6(0) + 4$

$y = 4$

$\:$

$for \: x = - 1$

$y = 3( { - 1})^{2} + 6( - 1) + 4$

$y = 1$

$\:$

$values \: of \: x \: and \: y \: for \: the \: system \: of \: equations \: are$

y = 4, x = 0

y = 1, x = -1

Answer 2

The system of equations has been solved and the solution set of the equation are (-1,1) and (0,4).

What is a system of equation?

A system of equation is the set of equation in which the finite set of equation is present for which the common solution is sought.

The first equation given as,

y = 3x² + 6x + 4

The second equation given as,

y = -3x² + 4

Equate the value of y from both equation as,

3x² + 6x + 4=-3x²+4

3x²+3x²+6x+4-4=0

6x²+6x=0

6x(x+1)=0

x=0:x=-1

Put these values of x, in first equation,

y = 3(-1)^2 + 6(-1) + 4

y = 3 - 6 + 4

y=1

Put 0 on the place of x,

y = 3(0)^2 + 6(0) + 4

y = 0 +0+ 4

y=4

Thus, the system of equations has been solved and the solution set of the equation are (-1,1) and (0,4).

Learn more about the system of equations here;

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