#### Question

5 Equations to Show Solution Sets

35x+7=0

|x|-2=0

x+2=x+\sqrt{4}

x^2-5x+6 =0

x^3-x=0

Question

5 Equations to Show Solution Sets

35x+7=0

|x|-2=0

x+2=x+\sqrt{4}

x^2-5x+6 =0

x^3-x=0

Solving an equation is to find a value for variables that makes the equation hold true in its domain. Depending on the given conditions, there may be more than solutions to the equation. A solution set is values for all set of all variables that will satisfy the equation.

The factors that have impact on the solution set are not only types of equations, but also given domain and other conditions. For example, linear equation normally has one solution. When it comes to linear equations with variable involving absolute value, there may be two values in solution set. If the equation has no solution, the solution set is empty.

One of the best practice to make sure everything is correct is to plug in each value to verify if the equation is true.

true and thus the value is a solution.

35x+7=0

For a general linear equation, there's only one value in solution set.

To find the solution for linear equation, move the constant term to the right hand side. And leave the term with variable in the left hand side. Then,

35x = -7

Divide both sides with the coefficient of the term with variable, here 35

\dfrac{35x}{35} = \dfrac{-7}{35}

Then the solution set is determined

x = -\dfrac{1}{5}

The solution set for the linear equation is denoted as \{-\dfrac{1}{5} \}

|x|-2=0

For a linear equation with absolute value, there may be more than one values in solution set.

To solve a linear equation with absolute value, isolate the absolute value on one side of the equation, that is

|x|=2

Then set the variable equal to both the positive and negative value of the constant term.

x=\pm2

In this way, the absolute value sign is removed and the equation turns to two general linear equations.

The solution set for the linear equation with absolute value is denoted as \{2,-2\}.

x+2=x+\sqrt{4}

Solving the above equation shows that both sides are actually equivalent. The equation is true for all real numbers.

The solution set is denoted as ℝ (the set of all real numbers)

x^2-5x+6 =0

For a quadratic equation, there are normally two distinct roots. Sometimes, the two roots are duplicate.

Solve the quadratic equation by using factorization method.

(x-2)(x-3)=0

Solve for x,

x_1=2 \\ x_2=3

The solution set is denoted as \{2,3\}.

x^3-x=0

For cubic equation, there are three roots. Some of the roots may be complex numbers or duplicate real numbers.

Solving the cubic equation by factorization.

x(x-1)(x+1)=0

Solving for x gives three distinct real roots.

x_1=0 \\ x_2=1\\x_3=-1

The solution set is denoted as \{0,1,-1\}.