QuestionJuly 28, 2025

Instructions: Solve the following systems of equations algebraically. If there are no real solutions, type "none" in both blanks.If there is only one, type "none" in the other blank. I(I/eft{}2+5x-2vert end{array) Iright.D 1(10 square 160 square 100 square 1(1) square lov I(I/eftil/begin(array)fl]y=-2 x-7||y=x 2 +4x-2|end{array)lright.V) (1) square square 10 (1) square 16-5 I(IIeft{1begin(array)fl]y=2 x-7||y=x~2+3x-1 (end(array) (right.) 1(1) square 16 square 10 (1) square ICD square lov

Instructions: Solve the following systems of equations algebraically. If there are no real solutions, type "none" in both blanks.If there is only one, type "none" in the other blank. I(I/eft{}2+5x-2vert end{array) Iright.D 1(10 square 160 square 100 square 1(1) square lov I(I/eftil/begin(array)fl]y=-2 x-7||y=x 2 +4x-2|end{array)lright.V) (1) square square 10 (1) square 16-5 I(IIeft{1begin(array)fl]y=2 x-7||y=x~2+3x-1 (end(array) (right.) 1(1) square 16 square 10 (1) square ICD square lov
Instructions: Solve the following systems of equations algebraically. If there are no real solutions, type "none" in both blanks.If there is only one, type "none" in the other blank. I(I/eft{}2+5x-2vert end{array) Iright.D 1(10 square 160 square 100 square 1(1) square lov I(I/eftil/begin(array)fl]y=-2 x-7||y=x 2 +4x-2|end{array)lright.V) (1) square square 10 (1) square 16-5 I(IIeft{1begin(array)fl]y=2 x-7||y=x~2+3x-1 (end(array) (right.) 1(1) square 16 square 10 (1) square ICD square lov

Solution
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Answer

First System: (-2+\sqrt{5}, -3+\sqrt{5}) and (-2-\sqrt{5}, -3-\sqrt{5}) ### Second System: (-5, 3) and (-1, -5) ### Third System: none Explanation 1. Solve the first system of equations Given 3y = 3x - 3 and y = x^2 + 5x - 2. Simplify the first equation to y = x - 1. Set x - 1 = x^2 + 5x - 2. Rearrange to x^2 + 4x - 1 = 0. Use the quadratic formula x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} with a=1, b=4, c=-1. Calculate x = \frac{-4 \pm \sqrt{16 + 4}}{2} = \frac{-4 \pm \sqrt{20}}{2} = \frac{-4 \pm 2\sqrt{5}}{2} = -2 \pm \sqrt{5}. For each x, find y: y = (-2 + \sqrt{5}) - 1 and y = (-2 - \sqrt{5}) - 1. 2. Solve the second system of equations Given y = -2x - 7 and y = x^2 + 4x - 2. Set -2x - 7 = x^2 + 4x - 2. Rearrange to x^2 + 6x + 5 = 0. Factor to (x+5)(x+1) = 0. Solutions are x = -5 and x = -1. For each x, find y: y = -2(-5) - 7 = 3 and y = -2(-1) - 7 = -5. 3. Solve the third system of equations Given y = 2x - 7 and y = x^2 + 3x - 1. Set 2x - 7 = x^2 + 3x - 1. Rearrange to x^2 + x + 6 = 0. The discriminant is 1^2 - 4(1)(6) = 1 - 24 = -23. Since it's negative, there are no real solutions.

Explanation

1. Solve the first system of equations<br /> Given $3y = 3x - 3$ and $y = x^2 + 5x - 2$. Simplify the first equation to $y = x - 1$. Set $x - 1 = x^2 + 5x - 2$. Rearrange to $x^2 + 4x - 1 = 0$. Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=1, b=4, c=-1$. Calculate $x = \frac{-4 \pm \sqrt{16 + 4}}{2} = \frac{-4 \pm \sqrt{20}}{2} = \frac{-4 \pm 2\sqrt{5}}{2} = -2 \pm \sqrt{5}$. For each $x$, find $y$: $y = (-2 + \sqrt{5}) - 1$ and $y = (-2 - \sqrt{5}) - 1$.<br /><br />2. Solve the second system of equations<br /> Given $y = -2x - 7$ and $y = x^2 + 4x - 2$. Set $-2x - 7 = x^2 + 4x - 2$. Rearrange to $x^2 + 6x + 5 = 0$. Factor to $(x+5)(x+1) = 0$. Solutions are $x = -5$ and $x = -1$. For each $x$, find $y$: $y = -2(-5) - 7 = 3$ and $y = -2(-1) - 7 = -5$.<br /><br />3. Solve the third system of equations<br /> Given $y = 2x - 7$ and $y = x^2 + 3x - 1$. Set $2x - 7 = x^2 + 3x - 1$. Rearrange to $x^2 + x + 6 = 0$. The discriminant is $1^2 - 4(1)(6) = 1 - 24 = -23$. Since it's negative, there are no real solutions.
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