Chapter 7: Problem 11
Solve the system of nonlinear equations using elimination. $$ \begin{array}{l} 4 x^{2}-9 y^{2}=36 \\ 4 x^{2}+9 y^{2}=36 \end{array} $$
Short Answer
Expert verified
The solutions are \((3, 0)\) and \((-3, 0)\).
Step by step solution
01
Analyze the Given System
We have a system of two nonlinear equations: \(4x^2 - 9y^2 = 36\) and \(4x^2 + 9y^2 = 36\). Notice that both equations have terms with \(4x^2\) and they only differ in the sign of the \(9y^2\) term. This suggests using elimination to solve for one variable.
02
Eliminate \(y^2\) by Adding the Equations
Add the two equations together to eliminate the \(y^2\) term: \[ (4x^2 - 9y^2) + (4x^2 + 9y^2) = 36 + 36 \]which simplifies to \[ 8x^2 = 72. \]
03
Solve for \(x^2\)
Divide both sides of the equation \(8x^2 = 72\) by 8 to solve for \(x^2\): \[ x^2 = \frac{72}{8} = 9. \]Next, take the square root of both sides: \[ x = \pm 3. \]
04
Eliminate \(x^2\) by Subtracting the Equations
Subtract the second equation from the first to eliminate \(x^2\):\[ (4x^2 - 9y^2) - (4x^2 + 9y^2) = 36 - 36 \]which simplifies to \[ -18y^2 = 0. \]
05
Solve for \(y^2\)
Divide both sides of \(-18y^2 = 0\) by -18 to get: \[ y^2 = 0. \]Taking the square root gives: \[ y = 0. \]
06
Conclude the Solution
We found \(x = \pm 3\) from Step 3 and \(y = 0\) from Step 5. Therefore, the solutions to the system of equations are the ordered pairs \((3, 0)\) and \((-3, 0)\).
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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 popular technique for solving systems of equations, both linear and nonlinear. It allows you to reduce the number of variables by combining equations in a way that cancels out a particular variable. This can simplify the problem significantly.
Here's how it works:
Here's how it works:
- Identify pairs of equations, focusing on terms that can be added or subtracted to eliminate a variable.
- Combine these equations using addition or subtraction to cancel one of the variables.
- In this example, adding the equations resulted in the elimination of the variable \( y^2 \), while subtracting eliminated \( x^2 \).
System of Equations
A system of equations consists of multiple equations that share common variables. Solving a system typically involves finding values for the variables that satisfy every equation simultaneously.
In our example, we dealt with a nonlinear system defined by:
Systems can be linear or nonlinear, depending on whether the equations include terms higher than the first degree, such as squares (\(x^2\), \(y^2\)) or other non-linear expressions.
In our example, we dealt with a nonlinear system defined by:
- \( 4x^2 - 9y^2 = 36 \)
- \( 4x^2 + 9y^2 = 36 \)
Systems can be linear or nonlinear, depending on whether the equations include terms higher than the first degree, such as squares (\(x^2\), \(y^2\)) or other non-linear expressions.
Solving Quadratic Equations
Quadratic equations are polynomials that take the form \(ax^2 + bx + c = 0\). They can be solved using various methods such as factoring, completing the square, quadratic formula, or graphing.
In our example, one of the steps involved distilling a quadratic-like form \(8x^2 = 72\) to solve the system:
In the context of systems of equations, quadratic solutions are crucial, as they can uncover additional solutions that linear equations might miss.
In our example, one of the steps involved distilling a quadratic-like form \(8x^2 = 72\) to solve the system:
- We first simplified the expression by dividing each side by 8. This reduced it to \(x^2 = 9\).
- The next step involved taking the square root of both sides to solve for \(x\), resulting in possible solutions of \(x = \pm 3 \).
In the context of systems of equations, quadratic solutions are crucial, as they can uncover additional solutions that linear equations might miss.