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Solve each system by the addition method. \(\left\\{\begin{array}{l}2 x+3 y--16 \\ 5 x-10 y-30\end{array}\right.\)

Short Answer

Expert verified
The solution to the system of equations is \(x = \sqrt{\frac{40}{3}}, -\sqrt{\frac{40}{3}}\) and \(y = \frac{8}{3}\).

Step by step solution

01

Subtract the first equation from the second one

Using the addition method, which in this case involves subtraction because there are no like terms to add, subtract the first equation from the second equation: \(x^{2}+y^{2}-16 - (x^{2}-2y -8) = 0\). Simplify this to get: \(3y -8 = 0\).
02

Solve the simplified equation for y

Isolate y in the simplified equation, we have: \(y = \frac{8}{3}\).
03

Substitute y's value in the first equation

Substitute y's value in the first equation: \(x^{2} - 2* \frac{8}{3} -8 = 0\). This simplifies to: \(x^{2} - \frac{16}{3} - 8 = 0\).
04

Solve the equation obtained in Step 3

The equation can be written as: \(x^{2} = \frac{16}{3} + 8 = \frac{40}{3}\). So the solutions for x are: \(x = \sqrt{\frac{40}{3}}\) and \(x = - \sqrt{\frac{40}{3}}\).

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Key Concepts

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

Addition Method
The addition method, often known as the elimination method, is a technique for solving systems of equations. It involves adding or subtracting equations in order to eliminate one variable, making it easier to solve for the other. In this exercise, we use subtraction because the equations don't have like terms that can be directly added.
  • First, take one equation and manipulate it to eliminate a variable when combined with the other equation.
  • Here, the first equation, \(x^{2} - 2y - 8\), is subtracted from the second equation, \(x^{2} + y^{2} - 16\), to eliminate \(x^{2}\).
  • This simplifies the system and allows us to solve for \(y\).
This method simplifies a complex system into something more manageable for further calculations.
Algebraic Substitution
After simplifying the system using the addition method, the next step typically involves substitution. Algebraic substitution means replacing a variable with its known value or expression in another equation.
  • Once we found \(y = \frac{8}{3}\), we substituted this value back into one of the original equations.
  • In our case, we substituted into the first equation \(x^{2} - 2 \cdot \frac{8}{3} - 8 = 0\).
  • This substitution allowed us to obtain an equation with just one variable, \(x\), making it solvable.
Substitution helps simplify and transform the system, paving the way for solving the remaining variable.
Quadratic Equations
Quadratic equations appear frequently in algebra and are characterized by the form \(ax^2 + bx + c = 0\). In this exercise, we derived a quadratic equation after substitution.
  • The equation \(x^{2} = \frac{40}{3}\) is a simplified quadratic form where \(b = 0\) and \(c = -\frac{40}{3}\).
  • We then solved it using the square root method, finding \(x\) by computing both the positive and negative square roots.
  • The solutions were \(x = \sqrt{\frac{40}{3}}\) and \(x = -\sqrt{\frac{40}{3}}\).
Understanding how to manipulate and solve quadratic equations is essential for uncovering all possible solutions to a problem.
Problem-Solving Steps
Solving a system of equations can be complex, but breaking it into clear steps simplifies the process. Here’s a structured approach to tackling such problems:
  • **Identify:** Carefully understand what each equation represents.
  • **Eliminate:** Use methods like addition or subtraction to eliminate a variable.
  • **Substitute:** With one variable found, substitute back to solve for the other variable.
  • **Simplify:** Perform calculations to simplify and solve the resultant equation.
  • **Verify:** Double-check the solutions to ensure they satisfy both original equations.
Following these systematic steps helps efficiently reach a solution while ensuring accuracy and comprehension of the problem at hand.

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Most popular questions from this chapter

The figure shows the healthy weight region for various heights for people ages 35 and older. GRAPH CAN'T COPY If \(x\) represents height, in inches, and \(y\) represents weight, in pounds, the healthy weight region can be modeled by the following system of linear inequalities: $$\left\\{\begin{array}{l} 5.3 x-y \geq 180 \\ 4.1 x-y \leq 140 \end{array}\right.$$ Use this information to solve Exercises 77-80. Is a person in this age group who is 5 feet 8 inches tall weighing 135 pounds within the healthy weight region?

Exercises 37-39 will help you prepare for the material covered in the first section of the next chapter. Solve the system: $$\left\\{\begin{aligned}w-x+2 y-2 z &=-1 \\\x-1 y+z &=1 \\\y-z &=1 \\\z-&-3\end{aligned}\right.$$ Express the solution set in the form \(\\{(\boldsymbol{x}, \boldsymbol{x}, \boldsymbol{y}, \boldsymbol{z})\\} .\) What makes it fairly easy to find the solution?

Determine whether each statement makes sense or does not make sense, and explain your reasoning. Without using any algebra, it's obvious that the nonlinear system consisting of \(x^{2}+y^{2}-4\) and \(x^{2}+y^{2}-25\) does not have real-number solutions.

Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. A system of two equations in two variables whose graphs are a circle and a line can have four real ordered-pair solutions

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