Question

Y-x²=1

y + 2x=4
Use elimination or sent the two equations equal to one another and solve the quadratic

Answer 1

Ok simple. look in the step by step

Explanation:

-y + x^2 = 1

y + 2x = 4

combine.

x^2 + 2x =3

x^2 + 2x - 3

(x+3)(x-1), x = 1, -3. Quadratic in factored form!

y - x^2 = 1

y - 1 = 1

y = 2

y - 9 = 1

y = 10

(1, 2)(-3, 10)

Hoorah

Answer 2

(-3, 10) and (1, 2)

Explanation:

Given equations:

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

To solve the system of equations using the elimination method, we can eliminate one of the variables by adding or subtracting the equations to create a new equation. In this case, we can eliminate the variable y by subtracting the first equation from the second equation:

`\begin{aligned}(y+2x)-(y-x^2)&=4-1\\y+2x-y+x^2&=3\\x^2+2x&=3\end{aligned}`

Subtract 3 from both sides of the equation:

`\begin{aligned}x^2+2x-3&=3-3\\x^2+2x-3&=0\end{aligned}`

Now, we have a quadratic equation in terms of x.

Factor the left side of the equation:

`\begin{aligned}x^2+3x-x-3&=0\\x(x+3)-1(x+3)&=0\\(x+3)(x-1)&=0\end{aligned}`

Solve for x by setting each factor equal to zero:

`x+3=0 \implies x=-3`

`x-1=0 \implies x=1`

Now, we have found two values for x, which are x = -3 and x = 1.

To find the corresponding values of y, substitute the values of x into one of the original equations:

`\begin{aligned}x=-3 \implies y+2(-3)&=4\\y-6&=4\\y&=10\end{aligned}`

`\begin{aligned}x=1 \implies y+2(1)&=4\\y+2&=4\\y&=2\end{aligned}`

So, the solutions to the system of equations are (-3, 10) and (1, 2).