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Evaluate, Simplify, and Translate Expressions

Evaluate algebraic expressions

Evaluate Algebraic Expressions

In the last section, we simplified expressions using the order of operations. In this section, we’ll evaluate expressions—again following the order of operations.

To evaluate an algebraic expression means to find the value of the expression when the variable is replaced by a given number. To evaluate an expression, we substitute the given number for the variable in the expression and then simplify the expression using the order of operations.

Example

Try it.

Evaluate \(x+7\) when

  1. ⓐ \(\ x=3\)
  2. ⓑ \(\ x=12\)

Solution

ⓐ To evaluate, substitute \(3\) for \(x\) in the expression, and then simplify.

Substitute.
Add.

When \(x=3,\) the expression \(x+7\) has a value of \(10.\)

ⓑ To evaluate, substitute \(12\) for \(x\) in the expression, and then simplify.

Substitute.
Add.

When \(x=12,\) the expression \(x+7\) has a value of \(19.\)

Notice that we got different results for parts ⓐ and ⓑ even though we started with the same expression. This is because the values used for \(x\) were different. When we evaluate an expression, the value varies depending on the value used for the variable.

Example

Try it.

Evaluate \(9x-2,\text{when}\\)

  1. ⓐ \(\ x=5\ \\)
  2. ⓑ \(\ x=1\)

Solution

Remember \(ab\) means \(a\) times \(b,\) so \(9x\) means \(9\) times \(x.\)

ⓐ To evaluate the expression when \(x=5,\) we substitute \(5\) for \(x,\) and then simplify.

Multiply.
Subtract.

ⓑ To evaluate the expression when \(x=1,\) we substitute \(1\) for \(x,\) and then simplify.

Multiply.
Subtract.

Notice that in part ⓐ that we wrote \(9⋅5\) and in part ⓑ we wrote \(9(1).\) Both the dot and the parentheses tell us to multiply.

Example

Try it.

Evaluate \({x}^{2}\) when \(x=10.\)

Solution

We substitute \(10\) for \(x,\) and then simplify the expression.

Use the definition of exponent.
Multiply.

When \(x=10,\) the expression \({x}^{2}\) has a value of \(100.\)

Example

Try it.

\(\text{Evaluate}\ {2}^{x}\ \text{when}\ x=5.\)

Solution

In this expression, the variable is an exponent.

Use the definition of exponent.
Multiply.

When \(x=5,\) the expression \({2}^{x}\) has a value of \(32.\)

Condensed — the full section is in OpenStax Prealgebra 2e.

Identify Terms, Coefficients, and Like Terms

Algebraic expressions are made up of terms. A term is a constant or the product of a constant and one or more variables. Some examples of terms are \(7,y,5{x}^{2},9a,\text{and}\ 13xy.\)

The constant that multiplies the variable(s) in a term is called the coefficient. We can think of the coefficient as the number in front of the variable. The coefficient of the term \(3x\) is \(3.\) When we write \(x,\) the coefficient is \(1,\) since \(x=1⋅x.\) gives the coefficients for each of the terms in the left column.

TermCoefficient
\(9a\)\(9\)
\(y\)\(1\)
\(5{x}^{2}\)\(5\)

An algebraic expression may consist of one or more terms added or subtracted. In this chapter, we will only work with terms that are added together. gives some examples of algebraic expressions with various numbers of terms. Notice that we include the operation before a term with it.

ExpressionTerms
\(7\)\(7\)
\(y\)\(y\)
\(x+7\)\(x,7\)
\(2x+7y+4\)\(2x,7y,4\)
\(3{x}^{2}+4{x}^{2}+5y+3\)\(3{x}^{2},4{x}^{2},5y,3\)
Example

Try it.

Identify each term in the expression \(9b+15{x}^{2}+a+6.\) Then identify the coefficient of each term.

Solution

The expression has four terms. They are \(9b,15{x}^{2},a,\) and \(6.\)

The coefficient of \(9b\) is \(9.\)

The coefficient of \(15{x}^{2}\) is \(15.\)

Remember that if no number is written before a variable, the coefficient is \(1.\) So the coefficient of \(a\) is \(1.\)

The coefficient of a constant is the constant, so the coefficient of \(6\) is \(6.\)

Some terms share common traits. Look at the following terms. Which ones seem to have traits in common?

\[5x,7,{n}^{2},4,3x,9{n}^{2}\]

Which of these terms are like terms?

  • The terms \(7\) and \(4\) are both constant terms.
  • The terms \(5x\) and \(3x\) are both terms with \(x.\)
  • The terms \({n}^{2}\) and \(9{n}^{2}\) both have \({n}^{2}.\)

Terms are called like terms if they have the same variables and exponents. All constant terms are also like terms. So among the terms \(5x,7,{n}^{2},4,3x,9{n}^{2},\)

\[7\ \text{and}\ 4\ \text{are like terms.}\]\[5x\ \text{and}\ 3x\ \text{are like terms.}\]\[{n}^{2}\ \text{and}\ 9{n}^{2}\ \text{are like terms.}\]

Condensed — the full section is in OpenStax Prealgebra 2e.

Simplify Expressions by Combining Like Terms

We can simplify an expression by combining the like terms. What do you think \(3x+6x\) would simplify to? If you thought \(9x,\) you would be right!

We can see why this works by writing both terms as addition problems.

Add the coefficients and keep the same variable. It doesn’t matter what \(x\) is. If you have \(3\) of something and add \(6\) more of the same thing, the result is \(9\) of them. For example, \(3\) oranges plus \(6\) oranges is \(9\) oranges. We will discuss the mathematical properties behind this later.

The expression \(3x+6x\) has only two terms. When an expression contains more terms, it may be helpful to rearrange the terms so that like terms are together. The Commutative Property of Addition says that we can change the order of addends without changing the sum. So we could rearrange the following expression before combining like terms.

Now it is easier to see the like terms to be combined.

Example

Try it.

Simplify the expression: \(3x+7+4x+5.\)

Solution
Identify the like terms.
Rearrange the expression, so the like terms are together.
Add the coefficients of the like terms.
The original expression is simplified to...

When any of the terms have negative coefficients, the procedure is the same, except that you have to subtract instead of adding to combine like terms.

Example

Try it.

Simplify the expression: \(7{x}^{2}+8x-{x}^{2}-4x.\)

Solution
Identify the like terms.
Rearrange the expression so like terms are together.
Add the coefficients of the like terms.

These are not like terms and cannot be combined. So \(6{x}^{2}+4x\) is in simplest form.

Translate Words to Algebraic Expressions

In the previous section, we listed many operation symbols that are used in algebra, and then we translated expressions and equations into word phrases and sentences. Now we’ll reverse the process and translate word phrases into algebraic expressions. The symbols and variables we’ve talked about will help us do that. They are summarized in .

OperationPhraseExpression
Addition\(a\) plus \(b\)
the sum of \(a\) and \(b\)
\(a\) increased by \(b\)
\(b\) more than \(a\)
the total of \(a\) and \(b\)
\(b\) added to \(a\)
\(a+b\)
Subtraction\(a\) minus \(b\)
the difference of \(a\) and \(b\)
\(b\) subtracted from \(a\)
\(a\) decreased by \(b\)
\(b\) less than \(a\)
\(a-b\)
Multiplication\(a\) times \(b\)
the product of \(a\) and \(b\)
\(a⋅b\), \(ab\), \(a(b)\), \((a)(b)\)
Division\(a\) divided by \(b\)
the quotient of \(a\) and \(b\)
the ratio of \(a\) and \(b\)
\(b\) divided into \(a\)
\(a\div b\), \(a/b\), \(\frac{a}{b}\), \(ba\)

Look closely at these phrases using the four operations:

  • the sum of \(a\) and \(b\)
  • the difference of \(a\) and \(b\)
  • the product of \(a\) and \(b\)
  • the quotient of \(a\) and \(b\)

Each phrase tells you to operate on two numbers. Look for the words of and and to find the numbers.

Example

Try it.

Translate each word phrase into an algebraic expression:

  1. ⓐ the difference of \(20\) and \(4\)
  2. ⓑ the quotient of \(10x\) and \(3\)

Solution

ⓐ The key word is difference, which tells us the operation is subtraction. Look for the words of and and to find the numbers to subtract.

\(\begin{array}{l} \\ \text{the difference}\ \text{of}\ 20\ and\ 4 \\ 20\ \text{minus}\ 4 \\ 20-4\end{array}\)

ⓑ The key word is quotient, which tells us the operation is division.

\(\begin{array}{l} \\ \text{the quotient of}\ 10x\ \text{and}\ 3 \\ \text{divide}\ 10x\ \text{by}\ 3 \\ 10x\div 3\end{array}\)

This can also be written as \(\begin{array}{l}10x/3\ \text{or}\ \frac{10x}{3}\end{array}\)

How old will you be in eight years? What age is eight more years than your age now? Did you add \(8\) to your present age? Eight more than means eight added to your present age.

How old were you seven years ago? This is seven years less than your age now. You subtract \(7\) from your present age. Seven less than means seven subtracted from your present age.

Condensed — the full section is in OpenStax Prealgebra 2e.

Key Concepts

  • Combine like terms.
    1. Identify like terms.
    2. Rearrange the expression so like terms are together.
    3. Add the coefficients of the like terms

Evaluate, Simplify, and Translate Expressions

Evaluate Algebraic Expressions

In the following exercises, evaluate the expression for the given value.

Try it.

\(7x+8\ \text{when}\ x=2\)

Solution

22

Try it.

\(9x+7\ \text{when}\ x=3\)

Try it.

\(5x-4\ \text{when}\ x=6\)

Solution

26

Try it.

\(8x-6\ \text{when}\ x=7\)

Try it.

\({x}^{2}\ \text{when}\ x=12\)

Solution

144

Try it.

\({x}^{3}\ \text{when}\ x=5\)

Try it.

\({x}^{5}\ \text{when}\ x=2\)

Solution

32

Try it.

\({x}^{4}\ \text{when}\ x=3\)

Try it.

\({3}^{x}\ \text{when}\ x=3\)

Solution

27

Try it.

\({4}^{x}\ \text{when}\ x=2\)

Try it.

\({x}^{2}+3x-7\ \text{when}\ x=4\)

Solution

21

Try it.

\({x}^{2}+5x-8\ \text{when}\ x=6\)

Try it.

\(2x+4y-5\ \text{when}\ x=7,y=8\)

Solution

41

Try it.

\(6x+3y-9\ \text{when}\ x=6,y=9\)

Try it.

\({(x-y)}^{2}\ \text{when}\ x=10,y=7\)

Solution

9

Try it.

\({(x+y)}^{2}\ \text{when}\ x=6,y=9\)

Solution

225

Try it.

\({a}^{2}+{b}^{2}\ \text{when}\ a=3,b=8\)

Solution

73

Try it.

\({r}^{2}-{s}^{2}\ \text{when}\ r=12,s=5\)

Try it.

\(2l+2w\ \text{when}\ l=15,w=12\)

Solution

54

Try it.

\(2l+2w\ \text{when}\ l=18,w=14\)

Identify Terms, Coefficients, and Like Terms

In the following exercises, list the terms in the given expression.

Try it.

\(15{x}^{2}+6x+2\)

Solution

15x2, 6x, 2

Try it.

\(11{x}^{2}+8x+5\)


Try it.

\(10{y}^{3}+y+2\)

Solution

10y3, y, 2

Try it.

\(9{y}^{3}+y+5\)

In the following exercises, identify the coefficient of the given term.

Try it.

\(8a\)

Solution

8

Try it.

\(13m\)

Try it.

\(5{r}^{2}\)

Solution

5

Try it.

\(6{x}^{3}\)

In the following exercises, identify all sets of like terms.

Try it.

\({x}^{3},8x,14,8y,5,8{x}^{3}\)

Solution

x3 and 8x3; 14 and 5

Try it.

\(6z,3{w}^{2},1,6{z}^{2},4z,{w}^{2}\)

Try it.

\(9a,{a}^{2},16ab,16{b}^{2},4ab,9{b}^{2}\)

Solution

16ab and 4ab; 16b2 and 9b2

Try it.

\(3,25{r}^{2},10s,10r,4{r}^{2},3s\)

Simplify Expressions by Combining Like Terms

In the following exercises, simplify the given expression by combining like terms.

Try it.

\(10x+3x\)

Solution

13x

Try it.

\(15x+4x\)

Try it.

\(17a+9a\)

Solution

26a

Try it.

\(18z+9z\)

Try it.

\(4c+2c+c\)

Solution

7c

Try it.

\(6y+4y+y\)

Try it.

\(9x+3x+8\)

Solution

12x + 8

Try it.

\(8a+5a+9\)

Try it.

\(7u+2+3u+1\)

Solution

10u + 3

Try it.

\(8d+6+2d+5\)

Try it.

\(7p+6+5p+4\)

Solution

12p + 10

Try it.

\(8x+7+4x-5\)

Try it.

\(10a+7+5a-2+7a-4\)


Solution

22a + 1

Try it.

\(7c+4+6c-3+9c-1\)

Try it.

\(3{x}^{2}+12x+11+14{x}^{2}+8x+5\)

Solution

17x2 + 20x + 16

Try it.

\(5{b}^{2}+9b+10+2{b}^{2}+3b-4\)

Translate English Phrases into Algebraic Expressions

In the following exercises, translate the given word phrase into an algebraic expression.

Try it.

The sum of 8 and 12

Solution

8 + 12

Try it.

The sum of 9 and 1

Try it.

The difference of 14 and 9

Solution

14 − 9

Try it.

8 less than 19

Try it.

The product of 9 and 7

Solution

9 ⋅ 7

Try it.

The product of 8 and 7

Try it.

The quotient of 36 and 9

Solution

36 ÷ 9

Try it.

The quotient of 42 and 7

Try it.

The difference of \(x\) and \(4\)

Solution

x − 4

Try it.

\(3\) less than \(x\)

Try it.

The product of \(6\) and \(y\)

Solution

6y

Try it.

The product of \(9\) and \(y\)

Try it.

The sum of \(8x\) and \(3x\)

Solution

8x + 3x

Try it.

The sum of \(13x\) and \(3x\)

Try it.

The quotient of \(y\) and \(3\)

Solution

y ÷ 3

Try it.

The quotient of \(y\) and \(8\)

Try it.

Eight times the difference of \(y\) and nine

Solution

8 (y − 9)

Try it.

Seven times the difference of \(y\) and one

Try it.

Five times the sum of \(x\) and \(y\)

Solution

5 (x + y)

Try it.

Nine times five less than twice \(x\)

In the following exercises, write an algebraic expression.

Try it.

Adele bought a skirt and a blouse. The skirt cost \(\text{\$15}\) more than the blouse. Let \(b\) represent the cost of the blouse. Write an expression for the cost of the skirt.

Solution

b + 15

Try it.

Eric has rock and classical CDs in his car. The number of rock CDs is \(3\) more than the number of classical CDs. Let \(c\) represent the number of classical CDs. Write an expression for the number of rock CDs.

Try it.

The number of girls in a second-grade class is \(4\) less than the number of boys. Let \(b\) represent the number of boys. Write an expression for the number of girls.

Solution

b − 4

Try it.

Marcella has \(6\) fewer male cousins than female cousins. Let \(f\) represent the number of female cousins. Write an expression for the number of boy cousins.

Try it.

Greg has nickels and pennies in his pocket. The number of pennies is seven less than twice the number of nickels. Let \(n\) represent the number of nickels. Write an expression for the number of pennies.

Solution

2n − 7

Try it.

Jeannette has \(\text{\$5}\) and \(\text{\$10}\) bills in her wallet. The number of fives is three more than six times the number of tens. Let \(t\) represent the number of tens. Write an expression for the number of fives.

Condensed — the full section is in OpenStax Prealgebra 2e.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

  1. Is \(n\div 5\) an expression or an equation?
    If you missed this problem, review .

    答えを明らかにしろ

    expression

  2. Simplify \({4}^{5}.\)
    If you missed this problem, review .

    答えを明らかにしろ

    \(1,024\)

  3. Simplify \(1+8⋅9.\)
    If you missed this problem, review .

    答えを明らかにしろ

    \(73\)

  4. Evaluate \(x+7\) when

    1. ⓐ \(\ x=3\)
    2. ⓑ \(\ x=12\)

    答えを明らかにしろ

    ⓐ To evaluate, substitute \(3\) for \(x\) in the expression, and then simplify.

    Substitute.
    Add.

    When \(x=3,\) the expression \(x+7\) has a value of \(10.\)

    ⓑ To evaluate, substitute \(12\) for \(x\) in the expression, and then simplify.

    Substitute.
    Add.

    When \(x=12,\) the expression \(x+7\) has a value of \(19.\)

    Notice that we got different results for parts ⓐ and ⓑ even though we started with the same expression. This is because the values used for \(x\) were different. When we evaluate an expression, the value varies depending on the value used for the variable.

  5. Evaluate:

    \(y+4\ \text{when}\\)

    1. ⓐ \(\ y=6\ \\)
    2. ⓑ \(\ y=15\)

    答えを明らかにしろ

    1. ⓐ 10
    2. ⓑ 19

  6. Evaluate:

    \(a-5\ \text{when}\\)

    1. ⓐ \(\ a=9\ \\)
    2. ⓑ \(\ a=17\)

    答えを明らかにしろ
    1. ⓐ 4
    2. ⓑ 12
  7. Evaluate \(9x-2,\text{when}\\)

    1. ⓐ \(\ x=5\ \\)
    2. ⓑ \(\ x=1\)

    答えを明らかにしろ

    Remember \(ab\) means \(a\) times \(b,\) so \(9x\) means \(9\) times \(x.\)

    ⓐ To evaluate the expression when \(x=5,\) we substitute \(5\) for \(x,\) and then simplify.

    Multiply.
    Subtract.

    ⓑ To evaluate the expression when \(x=1,\) we substitute \(1\) for \(x,\) and then simplify.

    Multiply.
    Subtract.

    Notice that in part ⓐ that we wrote \(9⋅5\) and in part ⓑ we wrote \(9(1).\) Both the dot and the parentheses tell us to multiply.

  8. Evaluate:

    \(8x-3,\text{when}\\)

    1. ⓐ \(\ x=2\ \\)
    2. ⓑ \(\ x=1\)

    答えを明らかにしろ

    1. ⓐ 13
    2. ⓑ 5

  9. Evaluate:

    \(4y-4,\text{when}\\)

    1. ⓐ \(\ y=3\ \\)
    2. ⓑ \(\ y=5\)

    答えを明らかにしろ

    1. ⓐ 8
    2. ⓑ 16

  10. Evaluate \({x}^{2}\) when \(x=10.\)

    答えを明らかにしろ

    We substitute \(10\) for \(x,\) and then simplify the expression.

    Use the definition of exponent.
    Multiply.

    When \(x=10,\) the expression \({x}^{2}\) has a value of \(100.\)

  11. Evaluate:

    \({x}^{2}\ \text{when}\ x=8.\)

    答えを明らかにしろ

    64

  12. Evaluate:

    \({x}^{3}\ \text{when}\ x=6.\)

    答えを明らかにしろ

    216

  13. \(\text{Evaluate}\ {2}^{x}\ \text{when}\ x=5.\)

    答えを明らかにしろ

    In this expression, the variable is an exponent.

    Use the definition of exponent.
    Multiply.

    When \(x=5,\) the expression \({2}^{x}\) has a value of \(32.\)

  14. Evaluate:

    \({2}^{x}\ \text{when}\ x=6.\)

    答えを明らかにしろ

    64

  15. Evaluate:

    \({3}^{x}\ \text{when}\ x=4.\)

    答えを明らかにしろ

    81

  16. \(\text{Evaluate}\ 3x+4y-6\ \text{when}\ x=10\ \text{and}\ y=2.\)

    答えを明らかにしろ

    This expression contains two variables, so we must make two substitutions.

    Multiply.
    Add and subtract left to right.

    When \(x=10\) and \(y=2,\) the expression \(3x+4y-6\) has a value of \(32.\)

  17. Evaluate:

    \(2x+5y-4\ \text{when}\ x=11\ \text{and}\ y=3\)

    答えを明らかにしろ

    33

  18. Evaluate:

    \(5x-2y-9\ \text{when}\ x=7\ \text{and}\ y=8\)

    答えを明らかにしろ

    10

  19. \(\text{Evaluate}\ 2{x}^{2}+3x+8\ \text{when}\ x=4.\)

    答えを明らかにしろ

    We need to be careful when an expression has a variable with an exponent. In this expression, \(2{x}^{2}\) means \(2⋅x⋅x\) and is different from the expression \({(2x)}^{2},\) which means \(2x⋅2x.\)

    Simplify \({4}^{2}\).
    Multiply.
    Add.

  20. Evaluate:

    \(3{x}^{2}+4x+1\ \text{when}\ x=3.\)

    答えを明らかにしろ

    40

  21. Evaluate:

    \(6{x}^{2}-4x-7\ \text{when}\ x=2.\)

    答えを明らかにしろ

    9

  22. Identify each term in the expression \(9b+15{x}^{2}+a+6.\) Then identify the coefficient of each term.

    答えを明らかにしろ

    The expression has four terms. They are \(9b,15{x}^{2},a,\) and \(6.\)

    The coefficient of \(9b\) is \(9.\)

    The coefficient of \(15{x}^{2}\) is \(15.\)

    Remember that if no number is written before a variable, the coefficient is \(1.\) So the coefficient of \(a\) is \(1.\)

    The coefficient of a constant is the constant, so the coefficient of \(6\) is \(6.\)

  23. Identify all terms in the given expression, and their coefficients:

    \(4x+3b+2\)

    答えを明らかにしろ

    The terms are 4x, 3b, and 2. The coefficients are 4, 3, and 2.

  24. Identify all terms in the given expression, and their coefficients:

    \(9a+13{a}^{2}+{a}^{3}\)

    答えを明らかにしろ

    The terms are 9a, 13a2, and a3, The coefficients are 9, 13, and 1.

  25. Identify the like terms:

    1. ⓐ \(\ {y}^{3},7{x}^{2},14,23,4{y}^{3},9x,5{x}^{2}\)
    2. ⓑ \(\ 4{x}^{2}+2x+5{x}^{2}+6x+40x+8xy\)

    答えを明らかにしろ

    ⓐ \(\ {y}^{3},7{x}^{2},14,23,4{y}^{3},9x,5{x}^{2}\)

    Look at the variables and exponents. The expression contains \({y}^{3},{x}^{2},x,\) and constants.

    The terms \({y}^{3}\) and \(4{y}^{3}\) are like terms because they both have \({y}^{3}.\)

    The terms \(7{x}^{2}\) and \(5{x}^{2}\) are like terms because they both have \({x}^{2}.\)

    The terms \(14\) and \(23\) are like terms because they are both constants.

    The term \(9x\) does not have any like terms in this list since no other terms have the variable \(x\) raised to the power of \(1.\)

    ⓑ \(\ 4{x}^{2}+2x+5{x}^{2}+6x+40x+8xy\)

    Look at the variables and exponents. The expression contains the terms \(4{x}^{2},2x,5{x}^{2},6x,40x,\text{and}\ 8xy\)

    The terms \(4{x}^{2}\) and \(5{x}^{2}\) are like terms because they both have \({x}^{2}.\)

    The terms \(2x,6x,\text{and}\ 40x\) are like terms because they all have \(x.\)

    The term \(8xy\) has no like terms in the given expression because no other terms contain the two variables \(xy.\)

  26. Identify the like terms in the list or the expression:

    \(9,2{x}^{3},{y}^{2},8{x}^{3},15,9y,11{y}^{2}\)

    答えを明らかにしろ

    9 and 15; 2x3 and 8x3; y2 and 11y2

  27. Identify the like terms in the list or the expression:

    \(4{x}^{3}+8{x}^{2}+19+3{x}^{2}+24+6{x}^{3}\)

    答えを明らかにしろ

    4x3 and 6x3; 8x2 and 3x2; 19 and 24

  28. Simplify the expression: \(3x+7+4x+5.\)

    答えを明らかにしろ
    Identify the like terms.
    Rearrange the expression, so the like terms are together.
    Add the coefficients of the like terms.
    The original expression is simplified to...
  29. Simplify:

    \(7x+9+9x+8\)

    答えを明らかにしろ

    16x + 17

  30. Simplify:

    \(5y+2+8y+4y+5\)

    答えを明らかにしろ

    17y + 7

  31. Simplify the expression: \(7{x}^{2}+8x-{x}^{2}-4x.\)

    答えを明らかにしろ
    Identify the like terms.
    Rearrange the expression so like terms are together.
    Add the coefficients of the like terms.

    These are not like terms and cannot be combined. So \(6{x}^{2}+4x\) is in simplest form.

  32. Simplify:

    \(3{x}^{2}+9x+{x}^{2}+5x\)

    答えを明らかにしろ

    4x2 + 14x

  33. Simplify:

    \(11{y}^{2}+8y+{y}^{2}+7y\)

    答えを明らかにしろ

    12y2 + 15y

  34. Translate each word phrase into an algebraic expression:

    1. ⓐ the difference of \(20\) and \(4\)
    2. ⓑ the quotient of \(10x\) and \(3\)

    答えを明らかにしろ

    ⓐ The key word is difference, which tells us the operation is subtraction. Look for the words of and and to find the numbers to subtract.

    \(\begin{array}{l} \\ \text{the difference}\ \text{of}\ 20\ and\ 4 \\ 20\ \text{minus}\ 4 \\ 20-4\end{array}\)

    ⓑ The key word is quotient, which tells us the operation is division.

    \(\begin{array}{l} \\ \text{the quotient of}\ 10x\ \text{and}\ 3 \\ \text{divide}\ 10x\ \text{by}\ 3 \\ 10x\div 3\end{array}\)

    This can also be written as \(\begin{array}{l}10x/3\ \text{or}\ \frac{10x}{3}\end{array}\)

  35. Translate the given word phrase into an algebraic expression:

    1. ⓐ the difference of \(47\) and \(41\)
    2. ⓑ the quotient of \(5x\) and \(2\)

    答えを明らかにしろ

    1. ⓐ 47 − 41
    2. ⓑ 5x ÷ 2

  36. Translate the given word phrase into an algebraic expression:

    1. ⓐ the sum of \(17\) and \(19\)
    2. ⓑ the product of \(7\) and \(x\)

    答えを明らかにしろ

    1. ⓐ 17 + 19
    2. ⓑ 7x

  37. Translate each word phrase into an algebraic expression:

    1. ⓐ Eight more than \(y\)
    2. ⓑ Seven less than \(9z\)

    答えを明らかにしろ

    ⓐ The key words are more than. They tell us the operation is addition. More than means “added to”.

    \(\begin{array}{l}\text{Eight more than}\ y \\ \text{Eight added to}\ y \\ y+8\end{array}\)

    ⓑ The key words are less than. They tell us the operation is subtraction. Less than means “subtracted from”.

    \(\begin{array}{l}\text{Seven less than}\ 9z \\ \text{Seven subtracted from}\ 9z \\ 9z-7\end{array}\)

  38. Translate each word phrase into an algebraic expression:

    1. ⓐ Eleven more than \(x\)
    2. ⓑ Fourteen less than \(11a\)

    答えを明らかにしろ

    1. x + 11
    2. ⓑ 11a − 14

  39. Translate each word phrase into an algebraic expression:

    1. ⓐ \(19\) more than \(j\)
    2. ⓑ \(21\) less than \(2x\)

    答えを明らかにしろ

    1. j + 19
    2. ⓑ 2x − 21

  40. Translate each word phrase into an algebraic expression:

    1. ⓐ five times the sum of \(m\) and \(n\)
    2. ⓑ the sum of five times \(m\) and \(n\)

    答えを明らかにしろ

    ⓐ There are two operation words: times tells us to multiply and sum tells us to add. Because we are multiplying \(5\) times the sum, we need parentheses around the sum of \(m\) and \(n.\)

    five times the sum of \(m\) and \(n\)
    \(\begin{array}{l} \\ \\ 5(m+n)\end{array}\)

    ⓑ To take a sum, we look for the words of and and to see what is being added. Here we are taking the sum of five times \(m\) and \(n.\)

    the sum of five times \(m\) and \(n\)
    \(\begin{array}{l} \\ \\ 5m+n\end{array}\)

    Notice how the use of parentheses changes the result. In part ⓐ , we add first and in part ⓑ , we multiply first.

Symbols used here

\pm
plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.
\neq
not equal
The two sides are different.
\leq,\ \geq
less/greater than or equal
Inequalities that allow equality; < and > exclude it.
\approx
approximately equal
Equal to the precision shown, not exactly.
\sqrt{x},\ \sqrt[n]{x}
square root, n-th root
The non-negative number whose square (n-th power) is x.
a \bmod n
remainder
What is left after dividing a by n.
\%
per cent
Per hundred: 15% = 15/100.
a : b,\ \frac{a}{b}
ratio, fraction
a for every b; a divided by b.

How to: Evaluate, Simplify, and Translate Expressions

  1. Evaluate algebraic expressions
  2. Identify terms, coefficients, and like terms
  3. Simplify expressions by combining like terms
  4. Translate word phrases to algebraic expressions

Questions people ask

Why does the order of operations matter?

Because 2 + 3 × 4 would otherwise be two different numbers. The convention (brackets, exponents, multiplication and division, addition and subtraction) exists so every reader gets the same value from the same expression.

How do I check an arithmetic answer?

Estimate first (round every number and compute roughly), then compare. If the estimate and the exact answer disagree by more than a little, one of them is wrong. The solver shows every operation, so you can find which line went astray.

Why are fractions harder than decimals?

They are not harder, they are more exact: 1/3 is a precise number, 0.333 is an approximation. Fractions need a common denominator to add, which is the one extra step people trip on.

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Parts of this page are adapted from OpenStax Prealgebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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