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Chapter 3: Problem 29

Solve the system by elimination. \(-3 x^2+y=-18 x+29\) \(-3 x^2-y=18 x-25\)

Short Answer

Expert verified
The solutions to the system of equations are \(x = \pm \sqrt{2/3}\) and the corresponding values for 'y' can be attained by substituting these 'x' values into equation 'b' or 'c'.

Step by step solution

01

Add the equations

Adding the two equations together:\(-3x^2 + y + (-3x^2 - y) = -18x + 29 + 18x - 25\), simplifying the left side will cause the 'y' variable to cancel out, while the right side simplifies and results in a quadratic equation 'a'. Add the right side of the equations individually to get 'a'
02

Solve the quadratic equation

After the simplification, we obtain \(a: -6x^2=-4\), which is a simple quadratic equation. To solve for 'x', you have to first remove the negative sign on the left side of the equation by multiplying the entire equation by -1. Next, divide the equation by the constant in front of \(x^2\), which is -6, to isolate \(x^2\). Then take a square root of both sides to solve for 'x'. After finding the value of 'x', choose any equation provided in the exercise to solve for 'y'.
03

Find the y-value

After finding 'x', you substitute the value of 'x' into equation 'b: y = -3x^2 + 18x - 29' or 'c: y = 3x^2 - 18x + 25', and this gives you the possible values for 'y'.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Elimination Method
The elimination method is a powerful tool for solving systems of equations. It is particularly useful when you want to simplify a system by removing one of the variables. In the given problem, we begin with two equations:
  • \(-3x^2 + y = -18x + 29\)
  • \(-3x^2 - y = 18x - 25\)
The key idea here is to add these equations together. Why? Because it allows us to cancel out the \(y\) terms. Doing so, the equation becomes simplified as follows:
  • The left side: \(-3x^2 + y + (-3x^2 - y) = -6x^2\) (\(y\) terms cancel out).
  • The right side: \(-18x + 29 + 18x - 25 = -4\).
When you add both sides together, you are left with a quadratic equation: \(-6x^2 = -4\). By using elimination, we have managed to transform two equations into one, making it easier to solve the system.
Solving Quadratic Equations
After the elimination method leaves us with the quadratic equation \(-6x^2 = -4\), the next step is to solve for \(x\). Let's break down this process:
First, you eliminate the negative signs by multiplying the entire equation by -1, giving us:
  • \(6x^2 = 4\)
Next, to isolate \(x^2\), divide both sides by 6:
  • \(x^2 = \frac{4}{6}\)
  • \(x^2 = \frac{2}{3}\)
To solve for \(x\), take the square root of both sides:
  • \(x = \pm \sqrt{\frac{2}{3}}\)
This result means there are two possible values for \(x\), which stems from the nature of quadratic equations.
Variable Isolation
Variable isolation refers to the process of rearranging equations to solve for one particular variable. After solving for \(x\), it鈥檚 time to determine the corresponding \(y\) values in our original system of equations. You can substitute the values of \(x\) back into one of the original equations:
  • For instance, using equation \(b\): \(y = -3x^2 + 18x - 29\)
Substitute each \(x\) value calculated earlier:
  • Replace \(x\) with \(\sqrt{\frac{2}{3}}\) and calculate \(y\).
  • Replace \(x\) with \(-\sqrt{\frac{2}{3}}\) and calculate \(y\).
This step allows you to find the corresponding \(y\) values, completing the solution of the system by giving us the points \((x, y)\) that satisfy both equations. Remember, isolating variables effectively turns complex systems into simpler, manageable parts.

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Most popular questions from this chapter

Suppose the quadratic equation \(a x^2+5 x+c=0\) has one real solution. Is it possible for \(a\) and \(c\) to be integers? rational numbers? Explain your reasoning. Then describe the possible values of \(a\) and \(c\).

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