Respuesta :

Answer:

(0, 4) and (-1, 0)

Step-by-step explanation:

Given system of equations:

[tex]\begin{cases}y=2x^2+6x+4\\y=-4x^2+4\end{cases}[/tex]

Solve by substitution

Substitute the first equation into the second:

[tex]\implies 2x^2+6x+4=-4x^2+4[/tex]

Add 4x² to both sides:

[tex]\implies 2x^2+6x+4+4x^2=-4x^2+4+4x^2[/tex]

[tex]\implies 6x^2+6x+4=4[/tex]

Subtract 4 from both sides:

[tex]\implies 6x^2+6x+4-4=4-4[/tex]

[tex]\implies 6x^2+6x=0[/tex]

Factor out 6x from the left side:

[tex]\implies 6x(x+1)=0[/tex]

Therefore:

[tex]\implies 6x=0 \implies x=0[/tex]

[tex]\implies x+1=0 \implies x=-1[/tex]

To find the y-coordinates of the found x-values, substitute the found values of x into one of the equations:

[tex]x=0 \implies -4(0)^2+4=4 \implies (0,4)[/tex]

[tex]x=-1\implies -4(-1)^2+4=0\implies (-1,0)[/tex]

Therefore, the solutions to the system of equations are:

(0, 4) and (-1, 0)

Esther

Answer:

Solutions:

a) x = 0, y = 4 ⇒ (0, 4)

b) x = -1, y = 0 ⇒ (-1, 0)

Step-by-step explanation:

Given system of equations:

a) y = 2x² + 6x + 4

b) y = -4x² + 4

1. Substitute the value of y in the second equation into the first equation:

⇒ -4x² + 4 = 2x² + 6x + 4

2. Solve for x:

⇒ -4x² + 4 = 2x² + 6x + 4 [subtract 4 from both sides]

⇒ -4x² + 4 - 4 = 2x² + 6x + 4 - 4

⇒ -4x² = 2x² + 6x [subtract 2x² from both sides]

⇒ -4x² - 2x² = 2x² - 2x² + 6x

⇒ -6x² = 6x [subtract 6x from both sides]

⇒ -6x² - 6x = 6x - 6x

⇒ -6x² - 6x = 0 [factor out -6x from the equation]

⇒ -6x(x + 1) = 0

Two cases:

a)

⇒ -6x = 0 [divide both sides by -6]        

⇒ -6x ÷ -6 = 0 ÷ -6

x = 0

b)

⇒ x + 1 = 0 [subtract 1 from both sides]

⇒ x + 1 - 1 = 0 - 1

x = -1

3. Find the value of y by substituting the found x values into one of the given equations:

a) x = 0:

⇒ y = -4x² + 4

⇒ y = -4(0)² + 4

⇒ y = -4(0) + 4

⇒ y = = 0 + 4

⇒ y = 4

coordinate: (0, 4)

b) x = -1:

⇒ y = -4x² + 4

⇒ y = -4(-1)² + 4

⇒ y = -4(1) + 4

⇒ y = -4 + 4

⇒ y = 0

coordinate: (-1, 0)

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