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${:[y=3(2x^(2)-x-1)],[y=4x+2]:} If (x_(1),y_(1)) and (x_(2),y_(2)) are distinct solutions to the system of equations shown, what is the value of y_(1)+y_(2) ?$

$\begin{array}{l} y=3\left(2 x^{2}-x-1\right) \\ y=4 x+2 \end{array}$ $\newline$ If $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ are distinct solutions to the system of equations shown, what is the value of $y_{1}+y_{2}$ ?

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Algebra 1

Compare linear, exponential, and quadratic growth

Full solution

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

\newline

\left(x_{1}, y_{1}\right)

and

\left(x_{2}, y_{2}\right)

are distinct solutions to the system of equations shown, what is the value of

y_{1}+y_{2}

Set Equations Equal: We have the system of equations: $\newline$ $y = 3(2x^2 - x - 1)$ $\newline$ $y = 4x + 2$ $\newline$ To find the solutions $(x_1, y_1)$ and $(x_2, y_2)$ , we need to set the two equations equal to each other and solve for $x$ . $\newline$ $3(2x^2 - x - 1) = 4x + 2$
Distribute and Simplify: Distribute the $3$ on the left side of the equation: $6x^2 - 3x - 3 = 4x + 2$
Quadratic Formula: Move all terms to one side to set the equation to zero: $\newline$ $6x^2 - 3x - 3 - 4x - 2 = 0$ $\newline$ $6x^2 - 7x - 5 = 0$
Substitute Values: Now we need to solve the quadratic equation $6x^2 - 7x - 5 = 0$ . This can be done by factoring, completing the square, or using the quadratic formula. The equation does not factor easily, so we will use the quadratic formula: $\newline$ $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $\newline$ where $a = 6$ , $b = -7$ , and $c = -5$ .
Solve for x: Substitute the values of $a$ , $b$ , and $c$ into the quadratic formula: $\newline$ $x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(6)(-5)}}{2(6)}$ $\newline$ $x = \frac{7 \pm \sqrt{49 + 120}}{12}$ $\newline$ $x = \frac{7 \pm \sqrt{169}}{12}$ $\newline$ $x = \frac{7 \pm 13}{12}$
Find y Values: Solve for the two possible values of x: $\newline$ $x_1 = \frac{7 + 13}{12} = \frac{20}{12} = \frac{5}{3}$ $\newline$ $x_2 = \frac{7 - 13}{12} = \frac{-6}{12} = -\frac{1}{2}$
Calculate $y_1 + y_2$ : Now we need to find the corresponding $y$ values for each $x$ by substituting $x_1$ and $x_2$ into either of the original equations. We'll use the simpler equation $y = 4x + 2$ . For $x_1 = \frac{5}{3}$ : $y_1 = 4\left(\frac{5}{3}\right) + 2$ $y_1 = \frac{20}{3} + 2$ $y_1 = \frac{20}{3} + \frac{6}{3}$ $y$ $0$
Calculate $y_1 + y_2$ : Now we need to find the corresponding $y$ values for each $x$ by substituting $x_1$ and $x_2$ into either of the original equations. We'll use the simpler equation $y = 4x + 2$ . For $x_1 = \frac{5}{3}$ : $y_1 = 4\left(\frac{5}{3}\right) + 2$ $y_1 = \frac{20}{3} + 2$ $y_1 = \frac{20}{3} + \frac{6}{3}$ $y$ $0$ For $y$ $1$ : $y$ $2$ $y$ $3$ $y$ $4$
Calculate $y_1 + y_2$ : Now we need to find the corresponding $y$ values for each $x$ by substituting $x_1$ and $x_2$ into either of the original equations. We'll use the simpler equation $y = 4x + 2$ . For $x_1 = \frac{5}{3}$ : $y_1 = 4(\frac{5}{3}) + 2$ $y_1 = \frac{20}{3} + 2$ $y_1 = \frac{20}{3} + \frac{6}{3}$ $y$ $0$ For $y$ $1$ : $y$ $2$ $y$ $3$ $y$ $4$ Now we can find $y_1 + y_2$ : $y$ $6$ $y$ $7$