Question

X + 2y² = 4
y=x-1
Solve system of equations step by step

Answer 1

(2, 1) (-0.5, -1.5)

Explanation:

`\begin{cases}x + 2y^2 = 4\\y = x - 1\end{cases}`

From here, we can substitute
`y = 2x - 1` into the first equation.

`x + 2(x - 1)^2 = 4\\x + 2(x^2 - 2x + 1) = 4\\x + 2x^2 - 4x + 2 = 4\\2x^2 - 3x + 2 = 4\\2x^2 - 3x - 2 = 0\\x = 2, -\frac12`

Substituting back into the second equation to find the corresponding y values:

`y(x = 2) = 2 - 1 = 1\\y(x = -\frac12) = -\frac12 - 1 = -1.5`

Answer 2

`\displaystyle x_1=2,\,x_2=-(1)/(2)`

`\displaystyle y_1=1,\,y_2=-(3)/(2)`

Explanation:

`\displaystyle x+2y^2=4\\x+2(x-1)^2=4\\x+2(x^2-2x+1)=4\\x+2x^2-4x+2=4\\2x^2-3x-2=0\\\\x=(-b\pm√(b^2-4ac))/(2a)\\\\x=(-(-3)\pm√((-3)^2-4(2)(-2)))/(2(2))\\ \\x=(3\pm√(9+16))/(4)\\ \\x=(3\pm√(25))/(4)\\ \\x=(3\pm5)/(4)\\\\x_1=(3+5)/(4)=(8)/(4)=2,\,x_2=(3-5)/(4)=(-2)/(4)=-(1)/(2)`

`\displaystyle y_1=2-1=1, y_2=-(1)/(2)-1=-(3)/(2)`