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$We want to find the intersection points of the graphs given by the following system of equations: {[x-y=-4],[y=5(x+1)^(2)-3]:} One of the intersection points is (-2,2). Find the other intersection point. Your answer must be exact.$

We want to find the intersection points of the graphs given by the following system of equations: $\newline$ $\left\{\begin{array}{l} x-y=-4 \\ y=5(x+1)^{2}-3 \end{array}\right.$ $\newline$ One of the intersection points is $(-2,2)$ . $\newline$ Find the other intersection point. Your answer must be exact.

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Grade 7

Quantities that combine to zero: word problems

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Q. We want to find the intersection points of the graphs given by the following system of equations:

\newline

\left\{\begin{array}{l} x-y=-4 \\ y=5(x+1)^{2}-3 \end{array}\right.

\newline

One of the intersection points is

(-2,2)

\newline

Find the other intersection point. Your answer must be exact.

Write down equations: First, let's write down the system of equations we need to solve: $\newline$ $1$ ) $x - y = -4$ $\newline$ $2$ ) $y = 5(x + 1)^2 - 3$
Substitute and solve for $x$ : We can substitute the expression for $y$ from the second equation into the first equation to solve for $x$ : $\newline$ $x - ($ $5$ (x + $1$ )^ $2$ - $3$ ) = $-4$
Expand and simplify: Now, let's expand the equation and simplify: $\newline$ $x - 5(x^2 + 2x + 1) + 3 = -4$ $\newline$ $x - 5x^2 - 10x - 5 + 3 = -4$ $\newline$ $-5x^2 - 9x - 2 = -4$
Set equation to zero: Next, we add $4$ to both sides of the equation to set it to zero: $\newline$ $-5x^2 - 9x - 2 + 4 = 0$ $\newline$ $-5x^2 - 9x + 2 = 0$
Calculate discriminant: We now have a quadratic equation. Let's solve for x using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = -5$ , $b = -9$ , and $c = 2$ .
Solve for x: First, calculate the discriminant $b^2 - 4ac$ : $\text{Discriminant} = (-9)^2 - 4(-5)(2) = 81 + 40 = 121$
Find y-coordinate: Since the discriminant is positive, we have two real solutions. Now, calculate the two possible values for x: $\newline$ $x_1 = \frac{-(-9) + \sqrt{121}}{2 \cdot -5} = \frac{9 + 11}{-10} = \frac{20}{-10} = -2$ $\newline$ $x_2 = \frac{-(-9) - \sqrt{121}}{2 \cdot -5} = \frac{9 - 11}{-10} = \frac{-2}{-10} = -\frac{1}{5}$
Simplify expression: We already know that one of the intersection points is $(-2, 2)$ , which corresponds to $x_1 = -2$ . Therefore, the other intersection point must correspond to $x_2 = -\frac{1}{5}$ .
Calculate y: Now, let's find the y-coordinate for the intersection point when $x = -\frac{1}{5}$ by substituting $x$ into the second equation: $\newline$ $y = 5(-\frac{1}{5} + 1)^2 - 3$
Calculate y: Now, let's find the y-coordinate for the intersection point when $x = -\frac{1}{5}$ by substituting $x$ into the second equation: $\newline$ $y = 5\left(-\frac{1}{5} + 1\right)^2 - 3$ Simplify the expression inside the parentheses: $\newline$ $y = 5\left(\frac{4}{5}\right)^2 - 3$
Calculate y: Now, let's find the y-coordinate for the intersection point when $x = -\frac{1}{5}$ by substituting $x$ into the second equation: $\newline$ $y = 5\left(-\frac{1}{5} + 1\right)^2 - 3$ Simplify the expression inside the parentheses: $\newline$ $y = 5\left(\frac{4}{5}\right)^2 - 3$ Calculate the square and multiply by $5$ : $\newline$ $y = 5\left(\frac{16}{25}\right) - 3$ $\newline$ $y = 5 \times \frac{16}{25} - 3$ $\newline$ $y = \frac{16}{5} - 3$
Calculate y: Now, let's find the y-coordinate for the intersection point when $x = -\frac{1}{5}$ by substituting $x$ into the second equation: $\newline$ $y = 5\left(-\frac{1}{5} + 1\right)^2 - 3$ Simplify the expression inside the parentheses: $\newline$ $y = 5\left(\frac{4}{5}\right)^2 - 3$ Calculate the square and multiply by $5$ : $\newline$ $y = 5\left(\frac{16}{25}\right) - 3$ $\newline$ $y = 5 \times \frac{16}{25} - 3$ $\newline$ $y = \frac{16}{5} - 3$ Convert $3$ to a fraction with a denominator of $5$ to combine the terms: $\newline$ $y = \frac{16}{5} - \frac{15}{5}$ $\newline$ $y = \frac{(16 - 15)}{5}$ $\newline$ $y = \frac{1}{5}$