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Chapter 20: Chapter 20

Problem 594

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Solve the system \(y=-x^{2}+7 x-5\) \(y-2 x=2\)

Short Answer

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The given system of equations has no real solution.
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TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Write the given equations

First, let's write down the given equations: 1) \( y = -x^2 + 7x - 5 \) 2) \( y = 2x + 2 \)

Step 2: Substitute y from equation (2) into equation (1)

To apply the substitution method, substitute the expression for y from equation (2) into equation (1): \( -x^2 + 7x - 5 = 2x + 2 \)

Step 3: Re-arrange and solve the equation for x

Now, we need to re-arrange the equation to get a quadratic equation in the form of \( Ax^2 + Bx + C = 0 \) and then solve for x: \( -x^2 + 7x - 5 - 2x - 2 = 0 \) Which simplifies to: \( -x^2 + 5x - 7 = 0 \) Now, solve the quadratic equation. In this case, it's not factorizable, so we will have to use the quadratic formula: \( x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \) Where \( A=-1, B=5, C=-7 \). So: \( x = \frac{-5 \pm \sqrt{5^2 - 4(-1)(-7)}}{-2} \) \( x = \frac{-5 \pm \sqrt{25 - 28}}{-2} \) \( x = \frac{-5 \pm \sqrt{-3}}{-2} \) Since we have a negative number inside the square root, the equation has no real solutions. Consequently, the given system of equations has no point of intersection and therefore no real solution.

Step 4: Final Answer

The given system of equations has no real solution.

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