Question

( 5 pts) Use substitution to solve the systern of quotations below. \[ \begin{array}{l} y=-x^{2}-2 x+9 \\ y=3 x+20 \end{array} \]

Ask by Bernard Rowe. in the United States
Mar 19,2025

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Tutor-Verified Answer

Answer

There are no real solutions to the system of equations.

Solution

Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}y=-x^{2}-2x+9\\y=3x+20\end{array}\right.\) - step1: Substitute the value of \(y:\) \(-x^{2}-2x+9=3x+20\) - step2: Move the expression to the left side: \(-x^{2}-2x+9-\left(3x+20\right)=0\) - step3: Calculate: \(-x^{2}-5x-11=0\) - step4: Multiply both sides: \(x^{2}+5x+11=0\) - step5: Solve using the quadratic formula: \(x=\frac{-5\pm \sqrt{5^{2}-4\times 11}}{2}\) - step6: Simplify the expression: \(x=\frac{-5\pm \sqrt{-19}}{2}\) - step7: The expression is undefined: \(x \notin \mathbb{R}\) - step8: The system of equations has no solution in the set of real numbers: \(\left(x,y\right) \notin \mathbb{R}^{2}\) - step9: Alternative Form: \(\textrm{No real solution}\) The system of equations given is: 1. \( y = -x^{2} - 2x + 9 \) 2. \( y = 3x + 20 \) To solve this system using substitution, we can set the two expressions for \( y \) equal to each other: \[ -x^{2} - 2x + 9 = 3x + 20 \] Now, let's rearrange this equation to find \( x \): 1. Move all terms to one side of the equation: \[ -x^{2} - 2x - 3x + 9 - 20 = 0 \] 2. Combine like terms: \[ -x^{2} - 5x - 11 = 0 \] 3. Multiply through by -1 to make the leading coefficient positive: \[ x^{2} + 5x + 11 = 0 \] Next, we can use the discriminant to determine if there are real solutions. The discriminant \( D \) is given by: \[ D = b^{2} - 4ac \] where \( a = 1 \), \( b = 5 \), and \( c = 11 \): \[ D = 5^{2} - 4 \cdot 1 \cdot 11 = 25 - 44 = -19 \] Since the discriminant is negative (\( D < 0 \)), this indicates that there are no real solutions to the system of equations. Thus, the conclusion is that the system has **no real solution**.

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To solve the system of equations using substitution, we first set the two equations equal to each other since they both equal \(y\): \[ -x^2 - 2x + 9 = 3x + 20 \] Now, let's rearrange the equation: \[ -x^2 - 2x - 3x + 9 - 20 = 0 \] This simplifies to: \[ -x^2 - 5x - 11 = 0 \] Multiplying through by -1 gives us: \[ x^2 + 5x + 11 = 0 \] Next, we can use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1, b = 5, c = 11\): Calculating the discriminant: \[ b^2 - 4ac = 5^2 - 4(1)(11) = 25 - 44 = -19 \] Since the discriminant is negative, this indicates that there are no real solutions for this system of equations. Thus, the system has no intersection points (real solutions).

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