Solved Problems

Tag: solving equations

Solving Systems of Equations Using Cramer’s Rule
Cramer’s Rule is a powerful method for solving systems of linear equations using determinants. It is particularly effective for small systems where calculating determinants manually is feasible. In this post, we will explain the method step-by-step and provide examples to help you master the technique.

What is Cramer’s Rule?

Cramer’s Rule is applicable to systems of linear equations where the number of equations matches the number of variables. The system:

$\begin{cases}a_{11}x_1 + a_{12}x_2 + \dots + a_{1n}x_n = b_1 \\a_{21}x_1 + a_{22}x_2 + \dots + a_{2n}x_n = b_2 \\\vdots \\a_{n1}x_1 + a_{n2}x_2 + \dots + a_{nn}x_n = b_n\end{cases}$

can be expressed in matrix form as:

$A \vec{x} = \vec{b},$

where $A$ is the coefficient matrix, $\vec{x}$ is the vector of variables, and $\vec{b}$ is the constants vector.

To solve for $x_1, x_2, \dots, x_n$ using Cramer’s Rule:
1. Compute the determinant of the coefficient matrix, $\det(A)$ .
2. Replace the $i$ -th column of $A$ with $\vec{b}$ to create a new matrix $A_i$ .
3. Solve for each variable using the formula:
$x_i = \frac{\det(A_i)}{\det(A)},$

provided $\det(A) \neq 0$ .

Steps to Solve Using Cramer’s Rule
1. Write the coefficient matrix $A$ and constants vector $\vec{b}$ .
2. Compute $\det(A)$ .
3. For each variable $x_i$ :
- Replace the $i$ -th column of $A$ with\ ( \vec{b} \) to form $A_i$ .
- Compute $\det(A_i)$ .
- Solve for $x_i$ using $x_i = \frac{\det(A_i)}{\det(A)}$ .
Example 1: Solving a 2×2 System

Consider the system:

$\begin{cases}x + 2y = 5 \\3x - y = 4\end{cases}$

Step 1: Write the Coefficient Matrix and Constants Vector

$A = \begin{bmatrix}1 & 2 \\3 & -1\end{bmatrix}, \quad \vec{b} = \begin{bmatrix} 5 \ 4 \end{bmatrix}.$

Step 2: Compute $\det(A)$

$\det(A) = \begin{vmatrix}1 & 2 \\3 & -1\end{vmatrix}$

$\det(A) = (1)(-1) - (2)(3) = -1 - 6 = -7.$

Step 3: Solve for Each Variable
- Replace the 1st column of $A$ with $\vec{b}$ to get $A_1$ :
$A_1 = \begin{bmatrix}5 & 2 \\4 & -1\end{bmatrix}.$

Compute $\det(A_1)$ :

$\det(A_1) = \begin{vmatrix}5 & 2 \\4 & -1\end{vmatrix}$

$\det(A_1) = (5)(-1) - (2)(4) = -5 - 8 = -13.$

Solve for $x$ :

$x = \frac{\det(A_1)}{\det(A)} = \frac{-13}{-7} = \frac{13}{7}.$
- Replace the 2nd column of $A$ with $\vec{b}$ to get $A_2$ :
$A_2 = \begin{bmatrix}1 & 5 \\3 & 4\end{bmatrix}.$

Compute $\det(A_2)$ :

$\det(A_2) = \begin{vmatrix}1 & 5 \\3 & 4\end{vmatrix}.$

$\det(A_2) = (1)(4) - (5)(3) = 4 - 15 = -11.$

Solve for $y$ :

$y = \frac{\det(A_2)}{\det(A)} = \frac{-11}{-7} = \frac{11}{7}.$

Solution:

$x = \frac{13}{7}, \quad y = \frac{11}{7}.$

Below, you can see the lines defined by this system of equations and their intersection point, which represents the solution to the system.

Example 2: Solving a 3×3 System

Solve the system:
$\begin{cases}2x + y - z &= 1, \\3x - 2y + 4z &= 7, \\x + 3y - 2z &= -3.\end{cases}$

Step 1: Write the coefficient matrix $A$ and constants vector $\vec{b}$ :

$A = \begin{bmatrix}2 & 1 & -1 \\3 & -2 & 4 \\1 & 3 & -2\end{bmatrix}, \quad\vec{b} = \begin{bmatrix}1 \ 7 \ -3\end{bmatrix}.$

Step 2: Compute $\det(A)$ :

$\begin{split}\det(A) = &2(-2 \times -2 - 4 \times 3) \\&- 1(3 \times -2 - 4 \times 1) \\&- 1(3 \times 3 - (-2 \times 1)).\end{split}$

$\begin{split}\det(A) &= 2(4 - 12) - 1(-6 - 4) - 1(9 + 2) \\&= 2(-8) + 10 - 11 \\ &= -16 + 10 - 11 \\&= -17.\end{split}$

Step 3: Solve for $x_1$ , $x_2$ , and $x_3$ :

Solve for $x_1$ :

Replace the first column of $A$ with $\vec{b}$ :
$A_1 = \begin{bmatrix}1 & 1 & -1 \\7 & -2 & 4 \\-3 & 3 & -2\end{bmatrix}.$

$\begin{split}\det(A_1) = &1(-2 \times -2 - 4 \times 3) \\&- 1(7 \times -2 - 4 \times -3) \\&+ (-1)(7 \times 3 - (-2 \times -3)).\end{split}$

$\begin{split}\det(A_1) &= 1(4 - 12) - 1(-14 + 12) - (21 - 6) \\&= -8 + 2 - 15 \\&= -21.\end{split}$

$x_1 = \frac{\det(A_1)}{\det(A)} = \frac{-21}{-17} = \frac{21}{17}.$

Solve for $x_2$ :

Replace the second column of $A$ with $\vec{b}$ :
$A_2 = \begin{bmatrix}2 & 1 & -1 \\3 & 7 & 4 \\1 & -3 & -2\end{bmatrix}.$

$\begin{split}\det(A_2) = &2(7 \times -2 - 4 \times -3) \\&- 1(3 \times -2 - 4 \times 1) \\&+ (-1)(3 \times -3 - 7 \times 1).\end{split}$

$\begin{split}\det(A_2) &= 2(-14 +12) - 1(-6 - 4) - 1(-9 - 7) \\&= 2(-2) + 10 +16 \\&= -4 + 10 + 16 \\&= 22.\end{split}$

$x_2 = \frac{\det(A_2)}{\det(A)} = \frac{22}{-17} = -\frac{22}{17}.$

Solve for $x_3$ :

Replace the third column of $A$ with $\vec{b}$ :
$A_3 = \begin{bmatrix}2 & 1 & 1 \\3 & -2 & 7 \\1 & 3 & -3\end{bmatrix}.$

$\begin{split}\det(A_3) = &2(-2 \times -3 - 7 \times 3) \\&- 1(3 \times -3 - 7 \times 1) \\&+ 1(3 \times 3 - (-2 \times 1)).\end{split}$

$\begin{split}\det(A_3) &= 2(6 - 21) - 1(-9 - 7) + 1(9 + 2) \\&= 2(-15) + 16 + 11 \\&= -30 + 16 + 11 \\&= -3.\end{split}$

$x_3 = \frac{\det(A_3)}{\det(A)} = \frac{-3}{-17} = \frac{3}{17}.$

Solution:

$x_1 = \frac{21}{17}, \quad x_2 = -\frac{22}{17}, \quad x_3 = \frac{3}{17}.$

Below is an interactive plot displaying three planes defined by the system of equations, along with their intersection point, which represents the solution to the system. You can change the viewpoint by dragging the plot to explore the relationship between the planes and their intersection more effectively. Enjoy visualizing the solution!

Example 3: Special Case – No Solution

Solve the system:
$\begin{cases}x + y + z &= 2, \\2x + 2y + 2z &= 5, \\x - y + z &= 1.\end{cases}$

Step 1: Write the coefficient matrix $A$ and constants vector $\vec{b}$ :

$A = \begin{bmatrix}1 & 1 & 1 \\2 & 2 & 2 \\1 & -1 & 1\end{bmatrix}, \quad\vec{b} = \begin{bmatrix}2 \ 5 \ 1\end{bmatrix}.$

Step 2: Compute $\det(A)$ :

$\begin{split}\det(A) &= 1(2 \times 1 - 2 \times -1) \\&- 1(2 \times 1 - 2 \times 1) \\&+ 1(2 \times -1 - 2 \times 1).\end{split}$

$\det(A) = 1(2 + 2) - 1(2 - 2) + 1(-2 - 2) = 4 - 0 - 4 = 0.$

Since $\det(A) = 0$ , the system is inconsistent and has no solution.

Below, you can see a plot of three planes defined by this system of equations. You can change the viewpoint by dragging the plot for a better perspective. As you observe, these three planes do not intersect at any point, indicating that there is no solution for this system of equations.

Example 4: Special Case – Infinitely Many Solutions

Solve the system:
$\begin{cases}x + y + z &= 3, \\2x + 2y + 2z &= 6, \\x - y + z &= 1.\end{cases}$

Step 1: Write the coefficient matrix $A$ and constants vector $\vec{b}$ :

$A = \begin{bmatrix}1 & 1 & 1 \\2 & 2 & 2 \\1 & -1 & 1\end{bmatrix}, \quad\vec{b} = \begin{bmatrix}3 \ 6 \ 1\end{bmatrix}.$

Step 2: Compute $\det(A)$ :

$\begin{split}\det(A) =& 1(2 \times 1 - 2 \times -1) \\&- 1(2 \times 1 - 2 \times 1) \\&+ 1(2 \times -1 - 2 \times 1).\end{split}$

$\begin{split}\det(A) &= 1(2 + 2) - 1(2 - 2) + 1(-2 - 2) \\&= 4 - 0 - 4 \\&= 0.\end{split}$

Since $\det(A) = 0$ , we check consistency.

The second equation is a multiple of the first, indicating a dependency between them. As a result, the system has infinitely many solutions. Essentially, instead of three distinct equations, we only have two, which means there are two planes that intersect along a line, resulting in infinitely many solutions.Below, you can see a plot of three planes defined by this system of equations. You can change the viewpoint by dragging the plot for a better perspective. To observe the relationship between the planes, you can toggle their visibility by clicking on their names in the legend. You’ll notice that the first and second planes overlap completely. The infinitely many solutions for this system of equations can be expressed as the line $x = 2t, \, y = 1, \, z = 2 - 2t$

Python Code

Here is the Python code to define and solve the examples above . It includes calculations for determinants and solutions for each case.
```
import numpy as np

# Function to compute determinant of a matrix
def determinant(matrix):
    return round(np.linalg.det(matrix), 2)

# Function to solve a system using Cramer's Rule
def cramers_rule(A, b):
    det_A = determinant(A)
    if det_A == 0:
        print("Det(A) = 0, system may have no solution or infinitely many solutions.")
        return None

    solutions = []
    n = len(b)
    for i in range(n):
        Ai = np.copy(A)
        Ai[:, i] = b
        det_Ai = determinant(Ai)
        solutions.append(det_Ai / det_A)
    return solutions

# Example 1: Solving a 2x2 System
A1 = np.array([[1, 2],
               [3, -1]])
b1 = np.array([5, 4])

print("Example 1: Solving a 2x2 System")
solutions_1 = cramers_rule(A1, b1)
if solutions_1:
    print(f"Solutions: x1 = {solutions_1[0]}, x2 = {solutions_1[1]}")

# Example 2: Solving a 3x3 System
A2 = np.array([[2, 1, -1],
               [3, -2, 4],
               [1, 3, -2]])
b2 = np.array([1, 7, -3])

print("\nExample 2: Solving a 3x3 System")
solutions_2 = cramers_rule(A2, b2)
if solutions_2:
    print(f"Solutions: x1 = {solutions_2[0]}, x2 = {solutions_2[1]}, x3 = {solutions_2[2]}")

# Example 3: No Solution
A3 = np.array([[1, 1, 1],
               [2, 2, 2],
               [1, -1, 1]])
b3 = np.array([2, 5, 1])

print("\nExample 3: No Solution")
det_A3 = determinant(A3)
if det_A3 == 0:
    print("Det(A) = 0. The system has no solution (inconsistent).")

# Example 4: Infinitely Many Solutions
A4 = np.array([[1, 1, 1],
               [2, 2, 2],
               [1, -1, 1]])
b4 = np.array([3, 6, 1])

print("\nExample 4: Infinitely Many Solutions")
det_A4 = determinant(A4)
if det_A4 == 0:
    print("Det(A) = 0. The system has infinitely many solutions (dependent equations).")
```
December 27, 2024
Solving Systems of Equations Using the Elimination Method
When solving systems of linear equations, the elimination method is one of the most popular and systematic approaches. This method involves adding or subtracting equations to eliminate one of the variables, allowing us to solve for the other. Once one variable is found, we substitute it back into one of the original equations to find the second variable.

Steps for Solving Using the Elimination Method

Here are the general steps for solving a system of equations using the elimination method:
1. Align the equations so that like terms (variables and constants) are in columns.
2. Eliminate one variable by adding or subtracting the equations. If necessary, multiply one or both equations by a constant to align coefficients.
3. Solve for the remaining variable.
4. Substitute the solution back into one of the original equations to find the other variable.
5. Verify the solution by substituting the values into both original equations.
Example 1: A Simple Elimination Scenario

Solve the following system of equations:

$\begin{aligned}[t] 2x + 3y &= 8 \\ 4x - 3y &= 10 \end{aligned}$

Step 1: Align the equations

The equations are already aligned:

$\begin{aligned}2x + 3y &= 8 \\ 4x - 3y &= 10 \end{aligned}$

Step 2: Eliminate one variable

Notice that the coefficients of $y$ are opposites: $+3y$ in the first equation and $−3y$ in the second equation. To eliminate $y$ , add the two equations:

$(2x+3y)+(4x -3y)=8+10$

Simplify:

$6x=18$

Step 3: Solve for $x$

Divide both sides by $6$ :

$x=3$

Step 4: Solve for $y$

Substitute $x=3$ into one of the original equations. Let’s use the first equation:

$2(3)+3y=8$

Simplify:

$6+3y=8$

Subtract $6$ from both sides:

$3y=2$

Divide by $3$ :

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

Step 5: Verify the solution

Substitute $x=3$ and $y=\frac{2}{3}$ into both original equations to confirm they work:
1. For $2x+3y=8$ : $2(3) + 3\left(\frac{2}{3}\right) = 8 \quad \text{✓}$
2. For $4x−3y=10$ : $4(3) - 3\left(\frac{2}{3}\right) = 10 \quad \text{✓}$
Thus, the solution is: $\boxed{(x, y) = (3, \frac{2}{3})}$

Visualization

In the plot below, you can see the two lines represented by the equations. The point of intersection, marked on the graph, represents the solution to the system of equations.

Example 2: Multiplying to Align Coefficients

Solve the following system of equations:

$\begin{aligned}[t] 2y + 3x&= 12 \\ 5x - 4y &= -2 \end{aligned}$

Step 1: Align the equations

The equations are aligned below:

$\begin{aligned}[t] 3x + 2y &= 12 \\ 5x - 4y &= -2 \end{aligned}$

Step 2: Eliminate one variable

We need to align the coefficients of either $x$ or $y$ . Let’s eliminate $y$ . To do this, multiply the first equation by $2$ and the second equation by $1$ (so that the coefficients of $y$ are opposites):

$\begin{aligned}[t] 2(3x + 2y) &= 2(12) \\ 1(5x - 4y) &= 1(-2) \end{aligned}$

This gives us:

$\begin{aligned}[t] 6x + 4y &= 24 \\ 5x - 4y &= -2 \end{aligned}$

Now, add the equations to eliminate $y$ :

$(6x + 4y) + (5x - 4y) = 24 + (-2)$

Simplify: $11x = 22$

Step 3: Solve for $x$

Divide both sides by $11$ : $x = 2$

Step 4: Solve for $y$

Substitute $x=2$ into one of the original equations. Let’s use the first equation:

$3(2) + 2y = 12$

Simplify:

$6 + 2y = 12$

Subtract $6$ from both sides:

$2y = 6$

Divide by $2$ :

$y = 3$

Step 5: Verify the solution

Substitute $x = 2$ and $y=3$ into both original equations to confirm they work:
1. For $3x+2y=12$ : $3(2) + 2(3) = 12 \quad \text{✓}$
2. For $5x−4y=−2$ : $5(2) - 4(3) = -2 \quad \text{✓}$
Thus, the solution is: $\boxed{(x, y) = (2, 3)}$

Visualization

The plot below illustrates the two lines corresponding to the equations. The intersection point indicates the solution to the system of equations.

Example 3: Infinite Solutions

Sometimes, a system of equations can have infinite solutions (the equations represent the same line). Let’s look at an example.

Solve the following system:

$\begin{aligned}[t] 2x + 4y &= 8 \\ x + 2y &= 4 \end{aligned}$

Step 1: Align the equations

The equations are already aligned:

$\begin{aligned} 2x + 4y &= 8 \\ x + 2y &= 4 \end{aligned}$

Step 2: Eliminate one variable

Notice that the second equation is just half of the first. Multiply the second equation by $2$ :

$\begin{aligned} 2x + 4y &= 8 \\ 2x + 4y &= 8 \end{aligned}$

Subtract the second equation from the first:

$(2x + 4y) - (2x + 4y) = 8 - 8$

Simplify:

$0 = 0$

This is a true statement, meaning the two equations represent the same line. Therefore, there are infinite solutions.

Visualization

The plot below shows the two equations as overlapping lines, indicating that every point on the line is a solution to the system.

Example 4: No Solution

Also, a system of equations can have no solution (the lines are parallel). Here is an example.

Solve the following system of equations:

$\begin{aligned}[t] 2x + 3y &= 6 \\ 6y + 4x &= 15 \end{aligned}$

Step 1: Align the equations

The equations are aligned below:

$\begin{aligned}[t] 2x + 3y &= 6 \ 4x + 6y &= 15 \end{aligned}$

Step 2: Eliminate one variable

To eliminate one variable, we need to make the coefficients of either $x$ or $y$ the same. Let’s eliminate $x$ . Multiply the first equation by $2$ so that the coefficients of $x$ : match:

$\begin{aligned}[t] 2(2x + 3y) &= 2(6) \\ 4x + 6y &= 15 \end{aligned}$

This gives us:

$\begin{aligned}[t] 4x + 6y &= 12 \\ 4x + 6y &= 15 \end{aligned}$

Now subtract the first equation from the second:

$(4x + 6y) - (4x + 6y) = 15 - 12$

Simplify:

$0 = 3$

This is a false statement, meaning the system of equations has no solution. The two lines are parallel and never intersect.

Visualization

The plot below illustrates the two parallel lines, confirming that there is no point of intersection. This graphical representation helps to visualize why the system has no solution.

Example 5: Solving a 3×3 System of Equations

Solve the following system of equations:
$\begin{cases}x + 2y + z &= 8 \\2x + y - z &= 3 \\3x - y + 2z &= 7\end{cases}$

Step 1: Align the equations
The equations are already aligned:
$\begin{cases}x + 2y + z &= 8 \\2x + y - z &= 3 \\3x - y + 2z &= 7\end{cases}$

Step 2: Eliminate one variable
We will eliminate ( z ) first. To do this, we can use the first two equations. We can add the first equation to the second equation after adjusting the coefficients of ( z ).

First, we can rewrite the second equation to align ( z ):
$2x + y - z = 3 \quad \text{(no change needed)}$

Now, we can add the first equation to the second equation:
$\begin{aligned}(x + 2y + z) + (2x + y - z) &= 8 + 3 \\3x + 3y &= 11\end{aligned}$

Simplify:

(Eq 4) $x + y = \frac{11}{3} \quad$

Next, we will eliminate ( z ) using the first and third equations. We can rewrite the first equation to align ( z ):
$2x + 4y + 2z = 16 \quad \text{(multiplied by two)}$

Now, we can add the first equation to the third equation after adjusting the coefficients of ( z ):
$(3x - y + 2z) - (2x + 4y + 2z) &= 7 - 16$

Simplify:

(Eq 5) $x - 5y = -9 \quad$

Step 3: Solve for one variable
Now we have two new equations (Equation 4 and Equation 5):
1. $x + y = \frac{11}{3}$ (Equation 4)
2. $x - 5y &= -9$ (Equation 5)
We can solve Equation 4 for $x$ :
$x = \frac{11}{3} - y$

Substituting $x$ into Equation 5:

$\left(\frac{11}{3} - y\right) - 5y = -9$

Simplify:

$\frac{11}{3} - y - 5y = -9$

Combine like terms:

$\frac{11}{3} - 6y = -9$

Multiply through by 3 to eliminate the fraction:

$11 - 18y = -27$

Subtract 11 from both sides:

$-18y = -38$

Divide by -18:

$y = \frac{19}{9}$

Step 4: Solve for $x$ and $z$
Substituting $y = \frac{19}{9}$ back into Equation 4:
$x + \frac{19}{9} = \frac{11}{3}$

Subtract $\frac{19}{9}$ from both sides:

$x = \frac{11}{3} - \frac{19}{9} = \frac{33-19}{9} = \frac{14}{9}$

Now, substitute $x = \frac{14}{9}$ and $y = \frac{19}{9}$ back into the first original equation to find $z$ :
$\frac{14}{9} + 2\left( \frac{19}{9}\right) + z = 8$

Simplify:

$\frac{14}{9} + \frac{38}{9} + z = 8$

Combine the fractions:

$\frac{52}{9} + z = 8$

Subtract $\frac{52}{9}$ from both sides:

$z = 8 - \frac{52}{9} = \frac{72}{9} - \frac{52}{9} = \frac{20}{9}$

Step 5: Verify the solution
Substitute $x = \frac{14}{9}$ , $y = \frac{19}{9}$ , and $z = \frac{20}{9}$ into all original equations to confirm they work:
1. For $x + 2y + z = 8$ :
  $\begin{aligned}&\frac{14}{9} + 2\left(\frac{19}{9} \right) + \frac{20}{9} \\&= \frac{14}{9} + \frac{38}{9} + \frac{20}{9} \\&= \frac{14+38+20}{9} \\&= 8 \quad \text{\checkmark}\end{aligned}$
2. For $2x + y - z = 3$ :
  $\begin{aligned}&2(\frac{14}{9}) + \frac{19}{9} - \frac{20}{9} \\&= \frac{28}{9} + \frac{19}{9} - \frac{20}{9} \\&= - \frac{28+19-20}{9} \\&= 3 \quad \text{\checkmark}\end{aligned}$
3. For $3x - y + 2z = 7$ :
  $\begin{aligned}&3(\frac{14}{9}) - \frac{19}{9} + 2\left(\frac{20}{9} \right) \\&= \frac{42}{9} - \frac{19}{9} + \frac{40}{9} \\&= \frac{42-19+40}{3} \\&= 7 \quad \text{\checkmark}\end{aligned}$
Thus, the solution is: $\boxed{(x, y, z) = \left(\frac{14}{9}, \frac{19}{9}, \frac{20}{9}\right)}$

Visualization

The plot below illustrates the three planes represented by the equations. The point of intersection, marked on the graph, represents the unique solution to the system of equations.

Key Takeaways
- The elimination method is a powerful way to solve systems of equations by eliminating one variable at a time.
- Always align your equations properly and, if necessary, multiply through by constants to align coefficients.
- Verify your solution by substituting the values into both original equations.
- Keep an eye out for special cases like infinite solutions or no solution.
December 23, 2024
Solving Quadratic Equations II: Taking Square Roots

A quadratic equation can be solved by taking the square root of both sides of the equation. This method uses the square root property,
$y^2=z \to y=\pm \sqrt{z}$
Before taking the square root, the equation must be arranged with the x2 term isolated on the left- hand side of the equation and its coefficient reduced to 1. There are four steps in solving quadratic equations by this method:
(more…)

July 31, 2010
Solving Quadratic Equations I: Factoring (Grouping)

Sometimes, quadratic equations can be solved by factorizing, which is also called grouping. The solve by factoring process is usually consists of three major steps:

All terms should be moved to one side of the equation using addition or subtraction. Rewrite the equation so that the left side of the equation is set equal to 0. For example, if the original equation is $x^2 = 5x - 3$ , it should be rewritten as $x^2 - 5x + 3 = 0$ .

The equation should be factored completely. It will have two factors.

Each factor should be set equal to zero. Since factors are of first power, they are easy to be solved.

All of the solutions should be combined to obtain the full solution set for the original equation.

The catch of this method is the finding of factors. Except some trivial cases, factoring quadratic equations is not easier than using other methods to solve the quadratic equations. Here are a few tips to help you factorize a quadratic equations:

(more…)

July 15, 2010
Solving Equations II: Radical Equations

A radical equation is an equation in which a variable appears under a radical sign. It may also have more than one radical. Let’s see some examples of radical equations:

$\sqrt{x-1}=2$

$\sqrt{2x+4}=\sqrt{5}$

$\sqrt{4x+1}+2x=-1$

$\sqrt{x+1}=\sqrt{2x+1}+10$

$\sqrt{10-x}+\sqrt{x-5}=3$

$\sqrt{x^2+3}=3$

$\sqrt[3]{2x-4}=-4$

$\sqrt{\sqrt{x+2}+7}=4$

$\sqrt[3]{x^2-1}=1$

$\sqrt[3]{6x^2+2}-1=x$

(more…)

July 10, 2010
Solving Equations I: Linear Equations

In a linear equation, each term is either a constant or the product of a constant and a single variable of degree 1. It can have one or more variables. Here are some linear equations:

$x+1=0$
$2x+4=6$
$-3x+1=x+4$
$1+5x=0$
$2-3x=3+2x$
$\frac{1}{2}x+\frac{3}{4}=\frac{1}{3}$
$\frac{x-1}{3}+\frac{1}{4}=\frac{5}{2}$
$\frac{x}{c}+b=a$ where $c\neq 0$ is a constant
$\sqrt{3}x-\sqrt{2}=\sqrt{5}$

However, the following equations are not linear:
(more…)

July 10, 2010