Question

I need help with a couple questions that I need to take a picture of to do?

Answer 1

Hello!

Let's solve these questions to find the corresponding values.

First box:

$\begin{cases}y+12=x^2+x\text{ equation I} \\ x+y=\text{ }{3\text{ equation II}}\end{cases}$

Let's rewrite equation II (I'll just change the side of the values and their signal, to isolate one variable).

$\begin{gathered} x+y=3\rightarrow\boxed{y=3-x} \\ \end{gathered}$

Now, where's Y, we will replace it with (3 -x) in the second equation.

$\begin{gathered} y+12=x^2+x\rightarrow(3-x)+12=x^2+x \\ \\ 3-x+12=x^{2}+x \\ 0=x^2+x+x-12-3 \\ 0=x^2+2x-15 \end{gathered}$

Solving it, we'll obtain two values for x:

• x', = -5

,

• x", = 3

Now, let's calculate y' and y" by replacement:

$\begin{gathered} y^{\prime}=3-x^{\prime} \\ y^{\prime}=3-(-5) \\ y^{\prime}=3+5 \\ y^{\prime}=8 \\ \\ y"=3-x" \\ y"=3-3 \\ y"=0 \end{gathered}$

So, the solution for the first box is:

{ {-5, 8}; {3, 0} }

Third box:

$\begin{cases}y+5={x^2-3x} \\ 2x+y={1}\end{cases}$

Let's solve it in a similar way:

$y=1-2x$

So, we have:

$\begin{gathered} 1-2x+5=x^2-3x \\ 0=x^2-3x-1+2x-5 \\ 0=x^2-x-6 \end{gathered}$

Solving it, we'll obtain two values for x:

• x', = -2

,

• x", = 3

Finding y' and y" by replacement, we have:

$\begin{gathered} y^{\prime}=1-2x\~ \\ y^{\prime}=1-2(-2) \\ y^{\prime}=1-(-4) \\ y^{\prime}=1+4 \\ y^{\prime}=5 \\ \\ y"=1-2x" \\ y"=1-2(3) \\ y"=1-6 \\ y"=-5 \end{gathered}$

So, the solution for the third box is:

{ {-2, 5}; {3, -5} }