Solve The System Of Equations:$\[ \begin{cases} ax + By = 3 \\ bx + Ay = 2 \end{cases} \\]

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Introduction

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In mathematics, a system of linear equations is a set of two or more linear equations that involve the same set of variables. Solving a system of linear equations involves finding the values of the variables that satisfy all the equations in the system. In this article, we will focus on solving a system of two linear equations with two variables.

The System of Equations

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The system of equations we will be solving is:

$$\begin{cases} ax + by = 3 \\ bx + ay = 2 \end{cases}$$

where $a$, $b$, $x$, and $y$ are variables. Our goal is to find the values of $x$ and $y$ that satisfy both equations.

Method 1: Substitution Method

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One way to solve this system of equations is by using the substitution method. This method involves solving one of the equations for one variable and then substituting that expression into the other equation.

Let's solve the first equation for $x$:

$$ax + by = 3$$

$$x = \frac{3-by}{a}$$

Now, substitute this expression for $x$ into the second equation:

$$bx + ay = 2$$

$$b\left(\frac{3-by}{a}\right) + ay = 2$$

Simplify the equation:

$$\frac{3b-by^2}{a} + ay = 2$$

Multiply both sides of the equation by $a$ to eliminate the fraction:

$$3b-by^2 + a^2y = 2a$$

Now, rearrange the terms to get a quadratic equation in $y$:

$$-by^2 + a^2y + 3b - 2a = 0$$

This is a quadratic equation in $y$, and we can solve it using the quadratic formula:

$$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

where $a = -b$, $b = a^2$, and $c = 3b - 2a$.

Substitute these values into the quadratic formula:

$$y = \frac{b \pm \sqrt{b^2 - 4(-b)(a^2 + 3b - 2a)}}{2(-b)}$$

Simplify the expression:

$$y = \frac{b \pm \sqrt{b^2 + 4ab^2 + 12b^2 - 8ab - 4a^2b}}{-2b}$$

$$y = \frac{b \pm \sqrt{17b^2 - 8ab - 4a^2b}}{-2b}$$

Now, we have two possible values for $y$. We can substitute these values back into one of the original equations to find the corresponding values of $x$.

Method 2: Elimination Method

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Another way to solve this system of equations is by using the elimination method. This method involves adding or subtracting the equations to eliminate one of the variables.

Let's add the two equations together:

$$ax + by + bx + ay = 3 + 2$$

Combine like terms:

$$(a+b)x + (b+a)y = 5$$

Now, we have a new equation with two variables. We can solve this equation for one variable and then substitute that expression into one of the original equations.

Let's solve this equation for $x$:

$$(a+b)x = 5 - (b+a)y$$

$$x = \frac{5 - (b+a)y}{a+b}$$

Now, substitute this expression for $x$ into one of the original equations. Let's use the first equation:

$$ax + by = 3$$

$$a\left(\frac{5 - (b+a)y}{a+b}\right) + by = 3$$

Simplify the equation:

$$\frac{5a - (b+a)^2y}{a+b} + by = 3$$

Multiply both sides of the equation by $a+b$ to eliminate the fraction:

$$5a - (b+a)^2y + b(a+b)y = 3(a+b)$$

Now, rearrange the terms to get a quadratic equation in $y$:

$$-(b+a)^2y + b(a+b)y + 5a - 3(a+b) = 0$$

This is a quadratic equation in $y$, and we can solve it using the quadratic formula:

$$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

where $a = -(b+a)^2$, $b = b(a+b)$, and $c = 5a - 3(a+b)$.

Substitute these values into the quadratic formula:

$$y = \frac{b \pm \sqrt{b^2 - 4(-(b+a)^2)(b(a+b)) + 4(-(b+a)^2)(5a - 3(a+b))}}{2(-(b+a)^2)}$$

Simplify the expression:

$$y = \frac{b \pm \sqrt{b^2 + 4b^2(a+b)^2 - 4(b+a)^2(5a - 3(a+b))}}{-2(b+a)^2}$$

$$y = \frac{b \pm \sqrt{b^2 + 4b^2(a+b)^2 - 20b^2 - 12b^2 - 40ab + 24ab}}{-2(b+a)^2}$$

$$y = \frac{b \pm \sqrt{b^2 - 16b^2 - 16ab}}{-2(b+a)^2}$$

$$y = \frac{b \pm \sqrt{-15b^2 - 16ab}}{-2(b+a)^2}$$

Now, we have two possible values for $y$. We can substitute these values back into one of the original equations to find the corresponding values of $x$.

Conclusion

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In this article, we have discussed two methods for solving a system of linear equations: the substitution method and the elimination method. We have shown how to use these methods to solve a system of two linear equations with two variables.

The substitution method involves solving one of the equations for one variable and then substituting that expression into the other equation. The elimination method involves adding or subtracting the equations to eliminate one of the variables.

We have also shown how to use the quadratic formula to solve the resulting quadratic equation in one of the variables.

By following these steps, we can solve a system of linear equations and find the values of the variables that satisfy all the equations in the system.

Example

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Let's consider an example to illustrate the steps involved in solving a system of linear equations.

Suppose we have the following system of equations:

$$\begin{cases} 2x + 3y = 7 \\ x + 2y = 4 \end{cases}$$

We can use the substitution method to solve this system of equations. Let's solve the first equation for $x$:

$$2x = 7 - 3y$$

$$x = \frac{7 - 3y}{2}$$

Now, substitute this expression for $x$ into the second equation:

$$\frac{7 - 3y}{2} + 2y = 4$$

Simplify the equation:

$$7 - 3y + 4y = 8$$

$$y = 1$$

Now, substitute this value of $y$ back into one of the original equations to find the corresponding value of $x$. Let's use the first equation:

$$2x + 3(1) = 7$$

$$2x = 4$$

$$x = 2$$

Therefore, the solution to the system of equations is $x = 2$ and $y = 1$.

Applications

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Solving a system of linear equations has many practical applications in various fields, including:

• Physics: Solving a system of linear equations is essential in physics to describe the motion of objects under the influence of forces.
• Engineering: Solving a system of linear equations is crucial in engineering to design and optimize systems, such as electrical circuits and mechanical systems.
• Economics: Solving a system of linear equations is essential in economics to model and analyze economic systems, such as supply and demand.
• Computer Science: Solving a system of linear equations is a fundamental problem in computer science, with applications in machine learning, data analysis, and computer graphics.

In conclusion, solving a system of linear equations is a fundamental problem in mathematics with many practical applications in various fields. By following the steps outlined in this article, we can solve a system of linear equations and find the values of the variables that satisfy all the equations in the system.

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Introduction

---

In mathematics, a system of linear equations is a set of two or more linear equations that involve the same set of variables. Solving a system of linear equations involves finding the values of the variables that satisfy all the equations in the system. In this article, we will focus on solving a system of two linear equations with two variables.

The System of Equations

---

The system of equations we will be solving is:

$$\begin{cases} ax + by = 3 \\ bx + ay = 2 \end{cases}$$

where $a$, $b$, $x$, and $y$ are variables. 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 of equations is by using the substitution method. This method involves solving one of the equations for one variable and then substituting that expression into the other equation.

Let's solve the first equation for $x$:

$$ax + by = 3$$

$$x = \frac{3-by}{a}$$

Now, substitute this expression for $x$ into the second equation:

$$bx + ay = 2$$

$$b\left(\frac{3-by}{a}\right) + ay = 2$$

Simplify the equation:

$$\frac{3b-by^2}{a} + ay = 2$$

Multiply both sides of the equation by $a$ to eliminate the fraction:

$$3b-by^2 + a^2y = 2a$$

Now, rearrange the terms to get a quadratic equation in $y$:

$$-by^2 + a^2y + 3b - 2a = 0$$

This is a quadratic equation in $y$, and we can solve it using the quadratic formula:

$$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

where $a = -b$, $b = a^2$, and $c = 3b - 2a$.

Substitute these values into the quadratic formula:

$$y = \frac{b \pm \sqrt{b^2 + 4ab^2 + 12b^2 - 8ab - 4a^2b}}{-2b}$$

Simplify the expression:

$$y = \frac{b \pm \sqrt{17b^2 - 8ab - 4a^2b}}{-2b}$$

Now, we have two possible values for $y$. We can substitute these values back into one of the original equations to find the corresponding values of $x$.

Method 2: Elimination Method

---

Another way to solve this system of equations is by using the elimination method. This method involves adding or subtracting the equations to eliminate one of the variables.

Let's add the two equations together:

$$ax + by + bx + ay = 3 + 2$$

Combine like terms:

$$(a+b)x + (b+a)y = 5$$

Now, we have a new equation with two variables. We can solve this equation for one variable and then substitute that expression into one of the original equations.

Let's solve this equation for $x$:

$$(a+b)x = 5 - (b+a)y$$

$$x = \frac{5 - (b+a)y}{a+b}$$

Now, substitute this expression for $x$ into one of the original equations. Let's use the first equation:

$$ax + by = 3$$

$$a\left(\frac{5 - (b+a)y}{a+b}\right) + by = 3$$

Simplify the equation:

$$\frac{5a - (b+a)^2y}{a+b} + by = 3$$

Multiply both sides of the equation by $a+b$ to eliminate the fraction:

$$5a - (b+a)^2y + b(a+b)y = 3(a+b)$$

Now, rearrange the terms to get a quadratic equation in $y$:

$$-(b+a)^2y + b(a+b)y + 5a - 3(a+b) = 0$$

This is a quadratic equation in $y$, and we can solve it using the quadratic formula:

$$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

where $a = -(b+a)^2$, $b = b(a+b)$, and $c = 5a - 3(a+b)$.

Substitute these values into the quadratic formula:

$$y = \frac{b \pm \sqrt{b^2 + 4b^2(a+b)^2 - 4(b+a)^2(5a - 3(a+b))}}{-2(b+a)^2}$$

Simplify the expression:

$$y = \frac{b \pm \sqrt{b^2 + 4b^2(a+b)^2 - 20b^2 - 12b^2 - 40ab + 24ab}}{-2(b+a)^2}$$

$$y = \frac{b \pm \sqrt{b^2 - 16b^2 - 16ab}}{-2(b+a)^2}$$

$$y = \frac{b \pm \sqrt{-15b^2 - 16ab}}{-2(b+a)^2}$$

Now, we have two possible values for $y$. We can substitute these values back into one of the original equations to find the corresponding values of $x$.

Conclusion

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In this article, we have discussed two methods for solving a system of linear equations: the substitution method and the elimination method. We have shown how to use these methods to solve a system of two linear equations with two variables.

The substitution method involves solving one of the equations for one variable and then substituting that expression into the other equation. The elimination method involves adding or subtracting the equations to eliminate one of the variables.

We have also shown how to use the quadratic formula to solve the resulting quadratic equation in one of the variables.

By following these steps, we can solve a system of linear equations and find the values of the variables that satisfy all the equations in the system.

Q&A

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Q: What is a system of linear equations?

A: A system of linear equations is a set of two or more linear equations that involve the same set of variables.

Q: How do I solve a system of linear equations?

A: There are two main methods for solving a system of linear equations: the substitution method and the elimination method.

Q: What is the substitution method?

A: The substitution method involves solving one of the equations for one variable and then substituting that expression into the other equation.

Q: What is the elimination method?

A: The elimination method involves adding or subtracting the equations to eliminate one of the variables.

Q: How do I use the quadratic formula to solve a quadratic equation?

A: The quadratic formula is:

$$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

where $a$, $b$, and $c$ are the coefficients of the quadratic equation.

Q: What are the applications of solving a system of linear equations?

A: Solving a system of linear equations has many practical applications in various fields, including physics, engineering, economics, and computer science.

Q: Can I use a calculator to solve a system of linear equations?

A: Yes, you can use a calculator to solve a system of linear equations. However, it's always a good idea to understand the steps involved in solving the system of equations.

Q: How do I check my solution to a system of linear equations?

A: To check your solution, substitute the values of the variables back into the original equations and make sure they are satisfied.

Q: What if I have a system of linear equations with more than two variables?

A: If you have a system of linear equations with more than two variables, you can use the same methods as before, but you may need to use a matrix or a computer program to solve the system of equations.

Q: Can I use a computer program to solve a system of linear equations?

A: Yes, you can use a computer program to solve a system of linear equations. There are many software programs available that can solve systems of linear equations, including MATLAB, Python, and R.