Math Problem Statement

Question content area top Part 1 Solve the system by substitution. Check the​ answer(s). left brace Start 2 By 1 Matrix 1st Row 1st Column y equals 2 x squared minus 27 x minus 1 2nd Row 1st Column y equals x minus 99 EndMatrix

Question content area bottom Part 1 Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice. A. The​ solution(s) is/are

enter your response here. ​(Type an ordered pair. Use a comma to separate answers as​ needed.) B. There is no solution. left brace Start 2 By 1 Matrix 1st Row 1st Column y equals 2 x squared minus 27 x minus 1 2nd Row 1st Column y equals x minus 99 EndMatrixSelect the correct choice below and, if necessary, fill in the answer box to complete your choice. Help me solve this View an example Get more help

Solution

To solve the system of equations by substitution, let's proceed step by step. The system is:

1) & \quad y = 2x^2 - 27x - 1 \\ 2) & \quad y = x - 99 \end{aligned}$$ ### Step 1: Substitute the second equation into the first. Since both equations equal $$y$$, we can set them equal to each other: $$2x^2 - 27x - 1 = x - 99$$ ### Step 2: Simplify and solve for $$x$$. Move all terms to one side to set the equation equal to zero: $$2x^2 - 27x - 1 - x + 99 = 0$$ Simplify the equation: $$2x^2 - 28x + 98 = 0$$ Now, divide the entire equation by 2 to make it simpler: $$x^2 - 14x + 49 = 0$$ ### Step 3: Factor the quadratic equation. The quadratic equation can be factored as: $$(x - 7)^2 = 0$$ So, $$x - 7 = 0$$ leads to: $$x = 7$$ ### Step 4: Solve for $$y$$ using the value of $$x$$. Substitute $$x = 7$$ back into the second equation to find $$y$$: $$y = 7 - 99 = -92$$ ### Step 5: Check the solution. Substitute $$x = 7$$ and $$y = -92$$ into the first equation: $$y = 2(7)^2 - 27(7) - 1 = 2(49) - 189 - 1 = 98 - 190 = -92$$ Since both equations are satisfied, the solution is correct. ### Final Answer: The solution is $$(7, -92)$$. Would you like to explore the steps in more detail, or do you have any questions? --- Here are five related questions you might consider: 1. How would the solution change if the second equation was $$y = x - 50$$? 2. What if the first equation was $$y = 2x^2 - 27x + 1$$ instead? 3. How do you solve the system if both equations were linear? 4. Can the quadratic equation $$x^2 - 14x + 49 = 0$$ have any other solutions? 5. How do you determine if a system of equations has no solution? **Tip:** When solving by substitution, always ensure that after substituting, the equation is simplified properly before solving for the variable.

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Math Problem Analysis

Mathematical Concepts

Quadratic Equations
Substitution Method

Formulas

Quadratic equation ax^2 + bx + c = 0

Theorems

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Suitable Grade Level

Grades 9-12

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