Content

• Steps
• Part 1 of 3: Writing the Equation
• Part 2 of 3: Solving the Equation
• Part 3 of 3: Verifying the Solution
• Tips

An equation with modulus (absolute value) is any equation in which a variable or expression is enclosed in modular brackets. The absolute value of the variable ${ displaystyle x}$ denoted as $x$and the modulus is always positive (except for zero, which is neither positive nor negative). An absolute value equation can be solved like any other mathematical equation, but a modulus equation can have two endpoints because you have to solve the positive and negative equations.

Steps

Part 1 of 3: Writing the Equation

1. 1 Understand the mathematical definition of a module. It is defined like this: ${ displaystyle | p | = { begin {cases} p & { text {if}} p geq 0 - p & { text {if}} p0 end {cases}}}$... This means that if the number ${ displaystyle p}$ positively, the modulus is ${ displaystyle p}$... If the number ${ displaystyle p}$ negative, the modulus is ${ displaystyle -p}$... Since minus by minus gives plus, the modulus ${ displaystyle -p}$ positive.
   • For example, | 9 | = 9; | -9 | = - (- 9) = 9.
2. 2 Understand the concept of absolute value from a geometric point of view. The absolute value of a number is equal to the distance between the origin and this number. A module is denoted by modular quotes that enclose a number, variable, or expression ($displaystyle$). The absolute value of a number is always positive.
   • For example, $=3$ and $3$... Both numbers -3 and 3 are at a distance of three units from 0.
3. 3 Isolate the module in the equation. The absolute value must be on one side of the equation. Any numbers or terms outside the modular brackets must be carried over to the other side of the equation. Please note that the modulus cannot be equal to a negative number, therefore, if after isolating the modulus it is equal to a negative number, such an equation has no solution.
   • For example, given the equation $6x-2$; to isolate the module, subtract 3 from both sides of the equation:
     $+3=7$
     $+3-3=7-3$
     $displaystyle$

Part 2 of 3: Solving the Equation

1. 1 Write down the equation for a positive value. Equations with modulus have two solutions. To write a positive equation, get rid of the modular brackets and then solve the resulting equation (as usual).
   • For example, a positive equation for $displaystyle$ is an ${ displaystyle 6x-2 = 4}$.
2. 2 Solve a positive equation. To do this, calculate the value of the variable using mathematical operations. This is how you find the first possible solution to the equation.
   • For example:
     ${ displaystyle 6x-2 = 4}$
     ${ displaystyle 6x-2 + 2 = 4 + 2}$
     ${ displaystyle 6x = 6}$
     ${ displaystyle { frac {6x} {6}} = { frac {6} {6}}}$
     ${ displaystyle x = 1}$
3. 3 Write down the equation for the negative value. To write a negative equation, get rid of the modular brackets, and on the other side of the equation, precede the number or expression with a minus sign.
   • For example, a negative equation for $=4$ is an ${ displaystyle 6x-2 = -4}$.
4. 4 Solve the negative equation. To do this, calculate the value of the variable using mathematical operations. This is how you find the second possible solution to the equation.
   • For example:
     ${ displaystyle 6x-2 = -4}$
     ${ displaystyle 6x-2 + 2 = -4 + 2}$
     ${ displaystyle 6x = -2}$
     ${ displaystyle { frac {6x} {6}} = { frac {-2} {6}}}$
     ${ displaystyle x = { frac {-1} {3}}}$

Part 3 of 3: Verifying the Solution

1. 1 Check the result of solving the positive equation. To do this, substitute the resulting value into the original equation, that is, substitute the value ${ displaystyle x}$found as a result of solving the positive equation into the original equation with modulus. If equality is true, the decision is correct.
   • For example, if, as a result of solving a positive equation, you find that ${ displaystyle x = 1}$, substitute ${ displaystyle 1}$ to the original equation:
     $6x-2$
     $displaystyle$
     $displaystyle$
     $=4$
2. 2 Check the result of solving the negative equation. If one of the solutions is correct, this does not mean that the second solution will also be correct. So substitute the value ${ displaystyle x}$, found as a result of solving the negative equation, into the original equation with modulus.
   • For example, if, as a result of solving a negative equation, you find that ${ displaystyle x = { frac {-1} {3}}}$, substitute ${ displaystyle { frac {-1} {3}}}$ to the original equation:
     $6x-2$
     ${ displaystyle | 6 ({ frac {-1} {3}}) - 2 | = 4}$
     $-2-2$
     $=4$
3. 3 Pay attention to valid solutions. The solution to an equation is valid (correct) if equality is satisfied when substituted into the original equation. Note that an equation can have two, one, or no valid solutions.
   • In our example $=4$ and $-4$, that is, equality is observed and both decisions are valid. Thus, the equation $6x-2$ has two possible solutions: ${ displaystyle x = 1}$, ${ displaystyle x = { frac {-1} {3}}}$.

Tips

• Remember that modular brackets differ from other types of brackets in appearance and functionality.