We want to solve the following system of equations:
$$
\left\{\begin{array}{l}
2 x+y=-4 \
y=(x+1)^{2}-2
\end{array}\right.
$$

The line given by $2 x+y=-4$ is graphed.
1) Draw the parabola given by $y=(x+1)^{2}-2$ in the interactive graph.
2) Select all solutions to the system.
Choose all answers that apply:
A $(-3,2)$
B $(-2,0)$
c. $(-1,-2)$
D $(0,-4)$

Question

We want to solve the following system of equations:
$$
\left\{\begin{array}{l}
2 x+y=-4 \
y=(x+1)^{2}-2
\end{array}\right.
$$

The line given by $2 x+y=-4$ is graphed.
1) Draw the parabola given by $y=(x+1)^{2}-2$ in the interactive graph.
2) Select all solutions to the system.
Choose all answers that apply:
A $(-3,2)$
B $(-2,0)$
c. $(-1,-2)$
D $(0,-4)$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ We are given a system of equations: $2x + y = -4$ and $y = (x + 1)^2 - 2$. We need to find the values of $x$ and $y$ that satisfy both equations simultaneously. ‖ ‖ step 2 ⋮ Substitute the expression for $y$ from the second equation into the first equation: $2x + ((x + 1)^2 - 2) = -4$. ‖ ‖ step 3 ⋮ Expand the quadratic expression and simplify: $2x + (x^2 + 2x + 1 - 2) = -4$ becomes $x^2 + 4x - 1 = -4$. ‖ ‖ step 4 ⋮ Further simplify the equation to find $x$: $x^2 + 4x + 3 = 0$. ‖ ‖ step 5 ⋮ Factor the quadratic equation: $(x + 3)(x + 1) = 0$. ‖ ‖ step 6 ⋮ Solve for $x$: $x = -3$ or $x = -1$. ‖ ‖ step 7 ⋮ Substitute $x = -3$ into the second equation to find $y$: $y = ((-3) + 1)^2 - 2 = 2$. ‖ ‖ step 8 ⋮ Substitute $x = -1$ into the second equation to find $y$: $y = ((-1) + 1)^2 - 2 = -2$. ‖ ‖ step 9 ⋮ The solutions to the system of equations are $x = -3, y = 2$ and $x = -1, y = -2$. ‖ ***A, C*** ∻Key Concept∻ ⚹Solving systems of equations⚹ ∻Explanation∻ ⚹To solve a system of equations, one can substitute one equation into the other and solve for the variables. The solutions are the points where the graphs of the equations intersect.⚹◊What method would you use to determine which points are the solutions to the system of equations given in the problem?⍭ Generate me a similar question◊

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