4 ways to use the distributive property to solve an equation

2023 Author: Sybil Carroll | [email protected]. Last modified: 2023-05-21 06:08

In math, the distributive property is a rule that helps you simplify an equation in parentheses. At an early stage, you learned that you must first solve the operations inside the parentheses, but in the case of algebraic expressions, that rule does not always hold. The distributive property allows you to multiply the term outside the parentheses by the terms inside. Make sure to perform the operation properly so that you do not lose any data and you can correctly solve the equation. You can also use the distributive property to simplify equations that include fractions.

Steps

Method 1 of 4: Use the Basic Distributive Property

Step 1. Multiply the term outside the parentheses by each of the terms inside

To do this, you will basically have to distribute the outer term in those that are inside. Multiply that outer term by the first one inside the parentheses and then do the same with the second. If there are more than two terms, keep distributing the outer term until there are none left. Keep the sign (either plus or minus) inside the parentheses.

• 2 (x − 3) = 10 { displaystyle 2 (x-3) = 10}

• 2(x)−(2)(3)=10{displaystyle 2(x)-(2)(3)=10}
• 2x−6=10{displaystyle 2x-6=10}

Step 2. Combine like terms

Before you can solve the equation, you must combine the like terms. Combine all the number terms with each other. Besides, it combines all the variable terms. To simplify the equation, order the terms so that the variables are on one side of the equals sign and the constants (numbers alone) are on the other.

• 2x − 6 = 10 { displaystyle 2x-6 = 10}

  …..(problema original)

• 2x−6(+6)=10(+6){displaystyle 2x-6(+6)=10(+6)}

….. (suma 6 en ambos lados)

• 2x=16{displaystyle 2x=16}

….. (variable a la izquierda; constante a la derecha)

Step 3. Solve the equation

Find x { displaystyle x}

al dividir ambos lados de la ecuación entre el coeficiente en frente de la variable.

• 2x=16{displaystyle 2x=16}

…..(problema original)

• 2x/2=16/2{displaystyle 2x/2=16/2}

…..(divide ambos lados entre 2)

• x=8{displaystyle x=8}

…..(solución)

Método 2 de 4: Distribuir coeficientes negativos

Step 1. Distribute a negative number along with its sign

If you have a negative number that is multiplied by one or more terms inside the parentheses, be sure to also distribute its negative sign between each of the inner terms.

• Remember the basic rules for multiplying negative numbers:

  • Neg. x Neg. = Pos.
  • Neg. x Pos. = Neg.

• Look at the following example:

  • −4 (9−3x) = 48 { displaystyle -4 (9-3x) = 48}

    ….. (problema original)

  • −4(9)−(−4)(3x)=48{displaystyle -4(9)-(-4)(3x)=48}

  …..(distribuye (-4) en cada término)

  • −36−(−12x)=48{displaystyle -36-(-12x)=48}

  …..(simplifica la multiplicación)

  • −36+12x=48{displaystyle -36+12x=48}

  …..(observa que ese “menos -12” se convierte en +12)

Step 2. Combine like terms

When you finish the distribution, you should simplify the equation by moving all the variable terms to one side of the equal sign, as well as all those numbers without a variable to the other side. Do this by a combination of addition or subtraction.

• −36 + 12x = 48 { displaystyle -36 + 12x = 48}

  …..(problema original)

• −36(+36)+12x=48+36{displaystyle -36(+36)+12x=48+36}

…..(suma 36 a cada lado)

• 12x=84{displaystyle 12x=84}

…..(simplifica la suma para aislar la variable)

Step 3. Divide the equation to find the final solution

Solve the equation by dividing both sides of the equation by the same coefficient as the variable. You should get a single variable on one side of the equation with the result on the other.

• 12x = 84 { displaystyle 12x = 84}

  …..(problema original)

• 12x/12=84/12{displaystyle 12x/12=84/12}

…..(divide ambos lados entre 12)

• x=7{displaystyle x=7}

…..(solución)

Step 4. Consider the subtraction as if it were an addition (-1)

In an algebraic problem, whenever you see a minus sign, especially if it comes before a parenthesis, you must imagine that it says + (-1). This will help you to correctly distribute the negative to all the terms within the parentheses. Then solve the problem as you did before.

• For example, consider the following problem: 4x− (x + 2) = 4 { displaystyle 4x- (x + 2) = 4}

  . Para asegurarte de distribuir adecuadamente el negativo, reescribe el problema de la siguiente manera:

  • 4x+(−1)(x+2)=4{displaystyle 4x+(-1)(x+2)=4}

• Luego distribuye el (-1) hacia los términos dentro de los paréntesis de la siguiente manera:

  • 4x+(−1)(x+2)=4{displaystyle 4x+(-1)(x+2)=4}

  …..(problema revisado)

  • 4x−x−2=4{displaystyle 4x-x-2=4}

  …..(multiplica (-1) por x y por 2)

  • 3x−2=4{displaystyle 3x-2=4}

  …..(combina los términos)

  • 3x−2+2=4+2{displaystyle 3x-2+2=4+2}

  …..(suma 2 en ambos lados)

  • 3x=6{displaystyle 3x=6}

  …..(simplifica los términos)

  • 3x/3=6/3{displaystyle 3x/3=6/3}

  …..(divide ambos lados entre 3)

  • x=2{displaystyle x=2}

  …..(solución)

Método 3 de 4: Utilizar la propiedad distributiva para simplificar las fracciones

Step 1. Identify the fractional coefficients or the constants

Sometimes you may have a problem that contains fractions as coefficients and constants. You can leave them as is and apply the basic algebraic rules to solve the problem. However, if you use the distributive property, you can simplify the solution by converting the fractions to integrals.

• Consider the following example: x − 3 = x3 + 16 { displaystyle x-3 = { frac {x} {3}} + { frac {1} {6}}}

  . Las fracciones en este problema son x3{displaystyle {frac {x}{3}}}

  y 16{displaystyle {frac {1}{6}}}

Step 2. Find the least common multiple (LCM) of all the denominators

In this step, you can ignore all integrals. Just look at the fractions and find the LCM of all the denominators. To do this, you will need the lowest number that is evenly divisible by the denominators of the fractions in the equation. In this example, the denominators are 3 and 6; therefore, the LCM is 6.

Step 3. Multiply all the terms in the equation by the LCM

Remember that you can perform any operation you want on an algebraic equation as long as you do it on both sides. Multiply all the terms in the equation by the LCM to cancel the fractions and “convert” them to integrals. Enclose the left and right sides of the equation in parentheses and make the distribution:

• x − 3 = x3 + 16 { displaystyle x-3 = { frac {x} {3}} + { frac {1} {6}}}

  …..(ecuación original)

• (x−3)=(x3+16){displaystyle (x-3)=({frac {x}{3}}+{frac {1}{6}})}

…..(introduce los paréntesis)

• 6(x−3)=6(x3+16){displaystyle 6(x-3)=6({frac {x}{3}}+{frac {1}{6}})}

…..(multiplica ambos lados por el MCM)

• 6x−6(3)=6(x3)+6(16){displaystyle 6x-6(3)=6({frac {x}{3}})+6({frac {1}{6}})}

…..(distribuye la multiplicación)

• 6x−18=2x+1{displaystyle 6x-18=2x+1}

…..(simplifica la multiplicación)

Step 4. Combine like terms

Combine all the terms so that all variables appear on one side of the equation and all constants appear on the other. Use the basic operations of addition and subtraction to shift terms from one side to the other.

• 6x − 18 = 2x + 1 { displaystyle 6x-18 = 2x + 1}

  …..(problema simplificado)

• 6x−2x−18=2x−2x+1{displaystyle 6x-2x-18=2x-2x+1}

…..(resta 2x en ambos lados)

• 4x−18=1{displaystyle 4x-18=1}

…..(simplifica la resta)

• 4x−18+18=1+18{displaystyle 4x-18+18=1+18}

…..(suma 18 en ambos lados)

• 4x=19{displaystyle 4x=19}

…..(simplifica la suma)

Step 5. Solve the equation

Find the final solution by dividing both sides of the equation by the coefficient of the variable. In this way, there should be a single "x" term on one side of the equation and the numerical solution on the other.

• 4x = 19 { displaystyle 4x = 19}

  …..(problema revisado)

• 4x/4=19/4{displaystyle 4x/4=19/4}

…..(divide ambos lados entre 4)

• x=194 or 434{displaystyle x={frac {19}{4}}{text{ or }}4{frac {3}{4}}}

…..(solución final)

Método 4 de 4: Distribuir una fracción extensa

Step 1. Interpret a large fraction as if a distributive division

Sometimes you may have a problem that includes multiple terms in the numerator of a fraction over a single denominator. You will have to consider this as a distributive problem and apply the denominator in each term of the numerator. You can then rewrite the fraction to show the distribution as follows:

• 4x + 82 = 4 { displaystyle { frac {4x + 8} {2}} = 4}

  …..(problema original)

• 4x2+82=4{displaystyle {frac {4x}{2}}+{frac {8}{2}}=4}

…..(distribuye el denominador en cada término del numerador)

Step 2. Simplify each numerator as a separate fraction

After distributing the denominator in each term, simplify each of these individually.

• 4x2 + 82 = 4 { displaystyle { frac {4x} {2}} + { frac {8} {2}} = 4}

  …..(problema revisado)

• 2x+4=4{displaystyle 2x+4=4}

…..(simplifica las fracciones)

Step 3. Isolate the variable

Solve the problem by isolating the variable on one side of the equation and moving the constant terms to the other. Do this using a combination of addition and subtraction as needed.

• 2x + 4 = 4 { displaystyle 2x + 4 = 4}

  …..(problema revisado)

• 2x+4−4=4−4{displaystyle 2x+4-4=4-4}

…..(resta 4 en ambos lados)

• 2x=0{displaystyle 2x=0}

…..(aísla “x” en un lado)

Step 4. Divide by the coefficient to solve the problem

In the last step, divide by the coefficient of the variable. This should give you the final solution, with the only variable on one side and the number solution on the other.

• 2x = 0 { displaystyle 2x = 0}

  …..(problema revisado)

• 2x/2=0/2{displaystyle 2x/2=0/2}

…..(divide ambos lados entre 2)

• x=0{displaystyle x=0}

…..(solución)

Step 5. Avoid the common pitfall of dividing only one term

It can be tempting (but wrong) to divide the first numerator by the denominator and cancel the fraction. In the case of the problem above, an error like this would look like this:

• 4x + 82 = 4 { displaystyle { frac {4x + 8} {2}} = 4}

  …..(problema original)

• 2x+8=4{displaystyle 2x+8=4}

…..(divides solo 4x entre 2 en lugar de entre el numerador total)

• 2x+8−8=4−8{displaystyle 2x+8-8=4-8}
• 2x=−4{displaystyle 2x=-4}
• x=−2{displaystyle x=-2}

….. (solución incorrecta)

Step 6. Check the solution

You can verify the operation by entering the solution in the original problem. When simplifying, you should get a true statement. If you simplify and get an incorrect statement, it means that the solution was incorrect. In this example, test the two solutions of x = 0 and x = -2 to determine which of them is correct.

• Start with the solution x = 0:

  • 4x + 82 = 4 { displaystyle { frac {4x + 8} {2}} = 4}

    …..(problema original)

  • 4(0)+82=4{displaystyle {frac {4(0)+8}{2}}=4}

  …..(reemplaza x con 0)

  • 0+82=4{displaystyle {frac {0+8}{2}}=4}
  • 82=4{displaystyle {frac {8}{2}}=4}
  • 4=4{displaystyle 4=4}

  …..(declaración verdadera y, por ende, solución correcta)

• prueba la solución “falsa” de x = -2:

  • 4x+82=4{displaystyle {frac {4x+8}{2}}=4}

  …..(problema original)

  • 4(−2)+82=4{displaystyle {frac {4(-2)+8}{2}}=4}

  …..(reemplaza x con -2)

  • −8+82=4{displaystyle {frac {-8+8}{2}}=4}
  • 02=4{displaystyle {frac {0}{2}}=4}
  • 0=4{displaystyle 0=4}

  …..(declaración falsa, por ende, x = -2 es falso)