Solving Systems Of Equations Algebraically Find Solutions For Y=-x^2+6x+16 And Y=-4x+37

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When faced with a system of equations like $y = -x^2 + 6x + 16$ and $y = -4x + 37$, our goal is to find the point(s) where these equations intersect. In simpler terms, we're looking for the x and y values that satisfy both equations simultaneously. This is a fundamental concept in algebra with numerous applications in various fields. This article will walk you through a step-by-step approach to solve this system algebraically, understand the underlying concepts, and interpret the results. We will explore how to solve this specific problem, discuss the different methods available, and highlight common pitfalls to avoid. By the end of this guide, you'll have a solid understanding of how to tackle similar problems with confidence.

To begin, let's restate the core question: Which of the given options (A. (3,25), B. (-3,49), C. (3,25) and (7,9), D. (-3,49) and (-7,65)) represents the solution(s) of the system of equations $y = -x^2 + 6x + 16$ and $y = -4x + 37$? We will also determine the solution set algebraically, providing a clear and methodical approach that can be applied to other systems of equations. The solutions to a system of equations are the points where the graphs of the equations intersect. In this case, one equation is a parabola (quadratic) and the other is a straight line (linear). The intersection points represent the (x, y) pairs that satisfy both equations. Understanding this graphical interpretation can often provide a visual check on our algebraic solutions.

The given equations are:

  1. y = -x^2 + 6x + 16$ (a quadratic equation representing a parabola)

  2. y = -4x + 37$ (a linear equation representing a straight line)

To solve this system algebraically, we will use the substitution method. This involves setting the two expressions for y equal to each other and solving for x. Once we find the x values, we can substitute them back into either equation to find the corresponding y values. This process will give us the coordinates of the intersection points, which are the solutions to the system. Remember, the key to solving systems of equations is to find values that work for all equations in the system. This means that each solution (x, y) pair must satisfy both the quadratic and the linear equations. This rigorous approach ensures that our solutions are accurate and reliable. Let's dive into the step-by-step solution process.

Step-by-Step Algebraic Solution

1. Set the Equations Equal

Since both equations are expressed in terms of y, we can set them equal to each other:

−x2+6x+16=−4x+37-x^2 + 6x + 16 = -4x + 37

This step is crucial because it allows us to eliminate y and create a single equation in terms of x. By equating the two expressions, we're essentially finding the x values where the y values of both equations are the same. This is the algebraic equivalent of finding the intersection points of the two graphs. The resulting equation is a quadratic equation, which we can solve using several methods, such as factoring, completing the square, or the quadratic formula. Each method has its advantages and disadvantages, but the goal remains the same: to find the values of x that satisfy the equation.

2. Rearrange into a Quadratic Equation

To solve for x, we need to rearrange the equation into the standard quadratic form, which is $ax^2 + bx + c = 0$. To do this, we'll add $x^2$ to both sides, subtract $6x$ from both sides, and subtract 16 from both sides:

0=x2−10x+210 = x^2 - 10x + 21

This rearrangement is a key step in solving the quadratic equation. By bringing all the terms to one side, we set the stage for factoring or using the quadratic formula. The standard form of a quadratic equation is essential for applying these methods effectively. It's also important to double-check the signs and coefficients during this rearrangement to avoid errors. A small mistake in this step can lead to incorrect solutions. Once the equation is in the standard form, we can proceed to find the roots, which are the solutions for x.

3. Solve the Quadratic Equation

Now we have a quadratic equation $x^2 - 10x + 21 = 0$. We can solve this by factoring. We are looking for two numbers that multiply to 21 and add up to -10. Those numbers are -3 and -7. Therefore, we can factor the quadratic as:

(x−3)(x−7)=0(x - 3)(x - 7) = 0

Factoring is a powerful technique for solving quadratic equations, especially when the roots are integers. It involves breaking down the quadratic expression into a product of two linear expressions. This method relies on the principle that if the product of two factors is zero, then at least one of the factors must be zero. In this case, we found the factors (x - 3) and (x - 7). Setting each factor equal to zero gives us the solutions for x. Factoring is often quicker than other methods like the quadratic formula, but it requires some practice to identify the correct factors. If factoring proves difficult, the quadratic formula is a reliable alternative.

Setting each factor equal to zero gives us two possible solutions for x:

x−3=0extorx−7=0x - 3 = 0 ext{ or } x - 7 = 0

Solving these equations, we get:

x=3extorx=7x = 3 ext{ or } x = 7

These x values are the x-coordinates of the intersection points of the parabola and the line. They represent the x values where the y values of both equations are the same. To find the corresponding y values, we will substitute these x values back into either of the original equations. Choosing the simpler equation (the linear one) can make the calculation easier. This step is crucial for completing the solution and finding the coordinates of the intersection points.

4. Find the Corresponding y Values

We have found two values for x: $x = 3$ and $x = 7$. Now we need to find the corresponding y values. We can use either of the original equations, but the linear equation $y = -4x + 37$ is simpler. Let's substitute each x value into this equation.

For $x = 3$:

y=−4(3)+37=−12+37=25y = -4(3) + 37 = -12 + 37 = 25

For $x = 7$:

y=−4(7)+37=−28+37=9y = -4(7) + 37 = -28 + 37 = 9

This step is essential for finding the complete solution to the system of equations. By substituting the x values into one of the original equations, we determine the y values that correspond to those x values. This gives us the coordinates of the points where the two graphs intersect. Choosing the simpler equation can reduce the chance of making a calculation error. The resulting (x, y) pairs are the solutions to the system, and they represent the points that satisfy both equations simultaneously.

5. Write the Solutions as Ordered Pairs

So, the solutions are the ordered pairs (3, 25) and (7, 9). These pairs represent the points where the parabola and the line intersect. They are the only points that satisfy both equations in the system. This is the final step in solving the system of equations algebraically. The ordered pairs represent the coordinates of the intersection points, and they provide a clear and concise representation of the solution. It's always a good idea to check these solutions by substituting them back into both original equations to ensure they are correct. This verification step helps to catch any errors made during the solution process.

Final Answer and Conclusion

Therefore, the correct answer is C. (3,25) and (7,9). We have successfully solved the system of equations algebraically by using the substitution method, factoring the resulting quadratic equation, and finding the corresponding y values for each x value. This step-by-step approach demonstrates a clear and methodical way to solve systems of equations. The solutions represent the points where the parabola and the line intersect, and they are the only points that satisfy both equations simultaneously. This process can be applied to other systems of equations, making it a valuable tool for solving mathematical problems.

This exercise demonstrates the power of algebraic methods in solving systems of equations. By understanding the underlying principles and applying a systematic approach, we can find solutions accurately and efficiently. Remember to always check your solutions to ensure they satisfy the original equations. This problem also highlights the connection between algebraic solutions and graphical interpretations. The solutions we found algebraically correspond to the intersection points of the graphs of the two equations.

Which ordered pair(s) represent the solution(s) to the system of equations $y=-x^2+6x+16$ and $y=-4x+37$? Determine the solution set algebraically.

Solving Systems of Equations Algebraically Find Solutions for y=-x^2+6x+16 and y=-4x+37