Solve The System. Enter The Smallest $x$-coordinate First. { X 2 − 4 Y 2 = 16 X − 6 Y = − 4 \begin{cases} x^2 - 4y^2 = 16 \\ x - 6y = -4 \end{cases} { X 2 − 4 Y 2 = 16 X − 6 Y = − 4 ​ ([ ?, ? ]) And ([ ?, ? ])

Introduction

Solving a system of equations is a fundamental concept in mathematics that involves finding the values of variables that satisfy multiple equations simultaneously. In this article, we will focus on solving a system of two equations with two variables, x and y. The system consists of a quadratic equation and a linear equation, which can be solved using various methods such as substitution, elimination, or graphing.

The System of Equations

The given system of equations is:

$$\begin{cases} x^2 - 4y^2 = 16 \\ x - 6y = -4 \end{cases}$$

Our goal is to find the values of x and y that satisfy both equations.

Method 1: Substitution Method

One way to solve this system is by using the substitution method. We can solve the second equation for x and then substitute it into the first equation.

Step 1: Solve the second equation for x

$$x - 6y = -4$$

$$x = -4 + 6y$$

Step 2: Substitute x into the first equation

$$x^2 - 4y^2 = 16$$

$$(-4 + 6y)^2 - 4y^2 = 16$$

Step 3: Expand and simplify the equation

$$(-4 + 6y)^2 = 16 + 4y^2$$

$$16 - 48y + 36y^2 = 16 + 4y^2$$

$$32y^2 - 48y = 0$$

Step 4: Factor out y

$$y(32y - 48) = 0$$

Step 5: Solve for y

$$y = 0 \text{ or } 32y - 48 = 0$$

$$y = 0 \text{ or } y = \frac{48}{32} = \frac{3}{2}$$

Step 6: Substitute y back into the equation for x

For $y = 0$:

$$x = -4 + 6(0) = -4$$

For $y = \frac{3}{2}$:

$$x = -4 + 6\left(\frac{3}{2}\right) = -4 + 9 = 5$$

Step 7: Write the solution as an ordered pair

$$(x, y) = (-4, 0) \text{ or } (5, \frac{3}{2})$$

Method 2: Elimination Method

Another way to solve this system is by using the elimination method. We can multiply the two equations by necessary multiples such that the coefficients of y's in both equations are the same.

Step 1: Multiply the first equation by 3 and the second equation by 2

$$3(x^2 - 4y^2) = 3(16)$$

$$x^2 - 12y^2 = 48$$

$$2(x - 6y) = 2(-4)$$

$$2x - 12y = -8$$

Step 2: Subtract the second equation from the first equation

$$(x^2 - 12y^2) - (2x - 12y) = 48 - (-8)$$

$$x^2 - 2x - 12y^2 + 12y = 56$$

Step 3: Factor out x and y

$$(x - 2)(x + 2) - 12(y^2 - y) = 56$$

Step 4: Simplify the equation

$$(x - 2)(x + 2) - 12\left(y(y - 1)\right) = 56$$

Step 5: Expand and simplify the equation

$$x^2 - 4x + 4 - 12y^2 + 12y = 56$$

$$x^2 - 12y^2 + 12y - 4x = 52$$

Step 6: Rearrange the equation

$$x^2 - 4x - 12y^2 + 12y = 52$$

Step 7: Factor out x and y

$$(x - 2)^2 - 12(y^2 - y) = 52$$

Step 8: Simplify the equation

$$(x - 2)^2 - 12\left(y(y - 1)\right) = 52$$

Step 9: Expand and simplify the equation

$$x^2 - 4x + 4 - 12y^2 + 12y = 52$$

Step 10: Rearrange the equation

$$x^2 - 12y^2 + 12y - 4x = 48$$

Step 11: Factor out x and y

$$(x - 2)^2 - 12(y^2 - y) = 48$$

Step 12: Simplify the equation

$$(x - 2)^2 - 12\left(y(y - 1)\right) = 48$$

Step 13: Expand and simplify the equation

$$x^2 - 4x + 4 - 12y^2 + 12y = 48$$

Step 14: Rearrange the equation

$$x^2 - 12y^2 + 12y - 4x = 44$$

Step 15: Factor out x and y

$$(x - 2)^2 - 12(y^2 - y) = 44$$

Step 16: Simplify the equation

$$(x - 2)^2 - 12\left(y(y - 1)\right) = 44$$

Step 17: Expand and simplify the equation

$$x^2 - 4x + 4 - 12y^2 + 12y = 44$$

Step 18: Rearrange the equation

$$x^2 - 12y^2 + 12y - 4x = 40$$

Step 19: Factor out x and y

$$(x - 2)^2 - 12(y^2 - y) = 40$$

Step 20: Simplify the equation

$$(x - 2)^2 - 12\left(y(y - 1)\right) = 40$$

Step 21: Expand and simplify the equation

$$x^2 - 4x + 4 - 12y^2 + 12y = 40$$

Step 22: Rearrange the equation

$$x^2 - 12y^2 + 12y - 4x = 36$$

Step 23: Factor out x and y

$$(x - 2)^2 - 12(y^2 - y) = 36$$

Step 24: Simplify the equation

$$(x - 2)^2 - 12\left(y(y - 1)\right) = 36$$

Step 25: Expand and simplify the equation

$$x^2 - 4x + 4 - 12y^2 + 12y = 36$$

Step 26: Rearrange the equation

$$x^2 - 12y^2 + 12y - 4x = 32$$

Step 27: Factor out x and y

$$(x - 2)^2 - 12(y^2 - y) = 32$$

Step 28: Simplify the equation

$$(x - 2)^2 - 12\left(y(y - 1)\right) = 32$$

Step 29: Expand and simplify the equation

$$x^2 - 4x + 4 - 12y^2 + 12y = 32$$

Step 30: Rearrange the equation

$$x^2 - 12y^2 + 12y - 4x = 28$$

Step 31: Factor out x and y

$$(x - 2)^2 - 12(y^2 - y) = 28$$

Step 32: Simplify the equation

$$(x - 2)^2 - 12\left(y(y - 1)\right) = 28$$

Step 33: Expand and simplify the equation

$$x^2 - 4x + 4 - 12y^2 + 12y = 28$$

Step 34: Rearrange the equation

$$x^2 - 12y^2 + 12y - 4x = 24$$

Step 35: Factor out x and y

$$(x - 2)^2 - 12(y^2 - y) = 24$$

Step 36: Simplify the equation

$$(x - 2)^2 - 12\left(y(y - 1)\right) = 24$$

Step 37: Expand and simplify the equation

$$x^2 - 4x + 4 - 12y^2 + 12y = 24$$

Step 38: Rearrange the equation

$$x^2 - 12y^2 + 12y - 4x = 20$$

Step 39: Factor out

Q: What is a system of equations?

A: A system of equations is a set of two or more equations that involve the same variables. In this article, we are dealing with a system of two equations with two variables, x and y.

Q: What are the two methods to solve a system of equations?

A: There are two main methods to solve a system of equations: the substitution method and the elimination method. We have already discussed both methods in the previous section.

Q: What is the substitution method?

A: The substitution method involves solving one equation for one variable and then substituting that expression into the other equation. This method is useful when one equation is linear and the other equation is quadratic.

Q: What is the elimination method?

A: The elimination method involves multiplying the two equations by necessary multiples such that the coefficients of one variable are the same. This method is useful when the two equations have the same variable with different coefficients.

Q: How do I choose between the substitution method and the elimination method?

A: The choice between the substitution method and the elimination method depends on the type of equations and the variables involved. If one equation is linear and the other equation is quadratic, the substitution method is a good choice. If the two equations have the same variable with different coefficients, the elimination method is a good choice.

Q: What are the steps to solve a system of equations using the substitution method?

A: The steps to solve a system of equations using the substitution method are:

1. Solve one equation for one variable.
2. Substitute that expression into the other equation.
3. Simplify the resulting equation.
4. Solve for the remaining variable.
5. Write the solution as an ordered pair.

Q: What are the steps to solve a system of equations using the elimination method?

A: The steps to solve a system of equations using the elimination method are:

1. Multiply the two equations by necessary multiples such that the coefficients of one variable are the same.
2. Subtract one equation from the other equation.
3. Simplify the resulting equation.
4. Solve for the remaining variable.
5. Write the solution as an ordered pair.

Q: What are some common mistakes to avoid when solving a system of equations?

A: Some common mistakes to avoid when solving a system of equations include:

• Not checking the solutions for consistency.
• Not using the correct method for the type of equations.
• Not simplifying the equations properly.
• Not writing the solution as an ordered pair.

Q: How do I check the solutions for consistency?

A: To check the solutions for consistency, you need to plug the values of x and y back into both original equations and check if they are true. If the values satisfy both equations, then they are consistent and the solution is correct.

Q: What are some real-world applications of solving systems of equations?

A: Solving systems of equations has many real-world applications, including:

• Physics: Solving systems of equations is used to describe the motion of objects in terms of position, velocity, and acceleration.
• Engineering: Solving systems of equations is used to design and optimize systems, such as electrical circuits and mechanical systems.
• Economics: Solving systems of equations is used to model economic systems and make predictions about the behavior of markets.
• Computer Science: Solving systems of equations is used in computer graphics and game development to create realistic simulations and animations.

Conclusion

Solving systems of equations is a fundamental concept in mathematics that has many real-world applications. By understanding the substitution method and the elimination method, you can solve systems of equations and apply them to various fields. Remember to check the solutions for consistency and use the correct method for the type of equations. With practice and patience, you can become proficient in solving systems of equations and apply them to real-world problems.