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Lesson Explainer: Evaluating Algebraic Expressions Mathematics • 6th Grade

In this explainer, we will learn how to evaluate simple algebraic expressions with one or multiple variables and how to apply this in real-world problems.

We first recall that we can represent unknown values as variables and that we can represent relationships involving one or more variables as algebraic expressions. Since these algebraic expressions represent relationships involving numbers, we can evaluate these expressions for specific values of the variables. One method that we can use to perform this evaluation is known as substitution.

For example, imagine we are told that a temperature, 𝑇, measured in degree Celsius can be approximated in degreesΒ kelvin by the expression 𝑇+273. We can determine the value of, say, 35∘C in kelvins by setting the value of 𝑇 to 35∘C in the expression. We call this substituting this value into the expression.

We get 𝑇+273=35+273=308.K

There are many other uses for this type of evaluation. For example, we know that the area of a square is given by its length squared and that the area of a right triangle is one-half the product of the lengths of its legs. We can use this to determine an expression for the area of the following shape.

The area of the square is given by π‘ŽοŠ¨ and the right triangle is half the square, so its area is 12π‘ŽοŠ¨. Hence, the area of the entire shape is π‘Ž+12π‘ŽοŠ¨οŠ¨. We can then use this to determine the area of this shape for a given value of π‘Ž. For example, if we are given that the value of π‘Ž is 4, then we will substitute π‘Ž=4 into this expression and evaluate to get π‘Ž+12π‘Ž=4+12ο€Ή4=16+12Γ—16=16+8=24.squareunits

We use square units since this represents an area.

Let’s now see an example where we will evaluate an algebraic expression with the given value of a single variable.

Example 1: Evaluating an Algebraic Expression with One Variable

Evaluate 14+π‘¦οŠ¨ if 𝑦=6.

Answer

We substitute the value of 𝑦=6 into the expression to get 14+6. We know that 6=6Γ—6=36, so we have 14+6=14+36=50.

In our next example, we will use a given expression and value to convert a temperature from degrees Celsius to degrees Fahrenheit.

Example 2: Relating an Algebraic Expression with One Variable to a Real-World Problem

The expression 9𝐢+1605 can be used to convert the temperature in degrees Celsius, 𝐢, into degrees Fahrenheit. If a thermometer shows a temperature of 60∘C, determine the temperature in degrees Fahrenheit.

Answer

Since the given expression converts a temperature, 𝐢, given in degrees Celsius into degrees Fahrenheit, we need to substitute 𝐢=60 into the given expression. This gives us 9(60)+1605.

To evaluate this, we first note that 9Γ—60=540, giving =540+1605.

Next, we calculate that 540+160=700 to get =7005.

Finally, evaluate the division to get =140.∘F

We note it is important to include the units of degrees Fahrenheit since this is a temperature.

In our next example, we will evaluate an algebraic expression that has two variables, each with a given value.

Example 3: Evaluating an Algebraic Expression with Two Variables

Evaluate π‘Žβˆ’π‘ if π‘Ž=17 and 𝑏=βˆ’57.

Answer

We need to substitute the value of π‘Ž as 17 and that of 𝑏 as βˆ’57 into the expression. We can substitute π‘Ž=17 to get π‘Žβˆ’π‘=17βˆ’π‘.

Then, we substitute 𝑏=βˆ’57, giving us π‘Žβˆ’π‘=17βˆ’ο€Όβˆ’57.

We know that subtracting a negative number is the same as adding the positive of that number, so =17+57.

Finally, since the denominators are equal, we have =1+57=67.

In our next example, we will use a given algebraic expression and the values of two variables to determine the average gross profit of a family per month.

Example 4: Relating an Algebraic Expression with Two Variables to a Real-World Problem

Last year, a family’s total income was $46β€Žβ€‰β€Ž000, while their total expenses were $48β€Žβ€‰β€Ž100. Use the expression πΌβˆ’πΈ12, where 𝐼 represents total income and 𝐸 represents total expenses, to find the average difference between the family’s income and its expenses each month.

Answer

We need to substitute the values 𝐼=46000 and 𝐸=48100 into the expression and then evaluate. We have πΌβˆ’πΈ12=46000βˆ’4810012.

We can evaluate the numerator to get =βˆ’210012.

Finally, we evaluate the division by 12, giving us =βˆ’$175.

Since this represents an amount in dollar, we add the dollar sign to this value.

Hence, the gross average profit of the family per month is βˆ’$175.

In our next example, we will evaluate an expression involving substituting in the values of two variables.

Example 5: Evaluating an Algebraic Expression with Two Variables

Evaluate 7π‘žπ‘ž+2(𝑝+5) for 𝑝=3 and π‘ž=12.

Answer

We need to substitute the values 𝑝=3 and π‘ž=12 into the expression and then evaluate. Substituting these values into the expression gives us =7π‘žπ‘ž+2(𝑝+5)=7(12)12+2(3+5).

In the numerator, we have 7Γ—12=84. In the denominator, we have 12+2Γ—(3+5)=28. Hence, =8428.

This quotient can be simplified to reach our final answer, 3.

In our final example, we will evaluate an algebraic expression by substituting in the values of three variables.

Example 6: Evaluating an Algebraic Expression with Three Variables

Evaluate 𝑝𝑛𝑑+11 for 𝑛=4, 𝑝=5, and 𝑑=5.

Answer

We need to substitute the values 𝑛=4, 𝑝=5, and 𝑑=5 into the expression and then evaluate. Substituting these values into the expression gives us 𝑝𝑛𝑑+11=5ο€Ή45+11.

We can evaluate the numerator as 5Γ—4=5Γ—16=80 and the denominator as 5+11=16. Hence, =8016=8Γ—108Γ—2=102=5.

Let’s finish by recapping some of the important points from this explainer.

Key Points

  • We can evaluate algebraic expressions at specific values of the variables by substituting the values of the variables into the expression and then calculating.
  • We need to be careful of the order of operations when evaluating the expressions.
  • In real-world scenarios, we are often calculating values that are physical quantities. This means we need to be mindful that these values will need units.

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