Chapter 4 Represent And Solve Equations And Inequalities
Section 4.0: Review What You Know
Page 175 Exercise 1 Answer
In the expression 6x, x is a variable.
Variable is a symbol, usually a letter, which represents a number, called the value of the variable. Variable can be either arbitrary, not fully specified, or unknown.
Result
Variable
Page 175 Exercise 2 Answer
x + 5 is an algebraic expression.
An algebraic expression as a type of math expression that has at least one variable and at least one operation.
Result
Algebraic expression.
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Page 175 Exercise 3 Answer
Evaluate an expression to find its value.
To evaluate an expression one can use substitution to replace a variable with its numerical value. Then use the order of operations to simplify.
Result
Evaluate.
Page 175 Exercise 4 Answer
The expression on each side of the equal sign in an equation are equal.
For example,
4 + 5 = 3 + 6,
both sides are equal to 9.
Result
Equation.
Page 175 Exercise 5 Answer
An equation is given:
6 + 2 = 2 + 6.
Both 2 + 6 and 6 + 2 is equal to 8, thus the equation is true.
Result
True.
Page 175 Exercise 6 Answer
An equation is given:
2.5 − 1 = 1 − 2.5.
The left side, 2.5 − 1 equals 1.5, and the right side, 1 − 2.5 equals − 1.5, which means that the left and the right side are not equal. The equation is false.
Result
False.
Page 175 Exercise 7 Answer
An equation is given:
\(\frac{1}{2} \times 3=3 \times \frac{1}{2}\)Both sides are equal to \(\frac{3}{2}\). Thus, the equation is true.
Result
True.
Page 175 Exercise 8 Answer
An equation is given:
\(\frac{3}{4} \div 5=\frac{3}{4} \times \frac{1}{5}\)Since dividing by 5 is equal to multiplying by its reciprocal, which is \(\frac{1}{5}\), both sides are equal to \(\frac{3}{20}\).
Thus, the equation is true.
Result
True.
Page 175 Exercise 9 Answer
An equation is given:
5 ÷ \(\frac{1}{3}\) = \(\frac{5}{3}\).
Since dividing by a number is equal to multiplying by its reciprocal, the left side is equal to 5 × 3, that is 15. The left and the right side are not equal. Thus, the equation is false.
Result
False.
Page 175 Exercise 10 Answer
An equation is given:
\(\frac{2}{3}\) x 5 = \(\frac{10}{15}\)
When multiplying a fraction by a whole number the result is a fraction where the denominator is the same and the numerator is the product of the whole number and the numerator. So, the left side is equal to \(\frac{10}{3}\). The left and the right side are not equal. Thus, the equation is false.
Result
False.
Page 175 Exercise 11 Answer
To evaluate the expression x − 2 use substitution, that is substitute x with its value x = 8.
x − 2 = 8 − 2 (Substitute the variable with its value.)
= 6 (Find the difference.)
Result
6
Page 175 Exercise 12 Answer
To evaluate the expression 2b use substitution, that is substitute b with its value b = 9.
2b = 2(9) (Substitute the variable with its value.)
= 18 (Find the product.)
Result
18
Page 175 Exercise 13 Answer
To evaluate the expression \(3 \frac{3}{4}\) + y use substitution, that is substitute y with its value y = \(\frac{5}{6}\).
\(3 \frac{3}{4}\) + y = \(3 \frac{3}{4}\) + \(\frac{5}{6}\) (Substitute the variable with its value.)
= \(\frac{15}{4}\) + \(\frac{5}{6}\) (Rewrite the mixed number as a fraction.)
= \(\frac{45}{12}\) + \(\frac{10}{12}\) (Rewrite friactions with 12 as denominators.)
= \(\frac{55}{12}\) (Find the sum.)
= \(4 \frac{7}{12}\) (Rewrite the result as a mixed number.)
Result
Page 175 Exercise 14 Answer
To evaluate the expression \(\frac{15}{x}\) use substitution, that is substitute x with its value x = 3.
\(\frac{15}{x}\) = \(\frac{15}{3}\) (Substitute the variable with its value.)
= 5 (Find the quotient.)
Result
5
Page 175 Exercise 15 Answer
To evaluate the expression 5.6t use substitution, that is substitute t with its value t = 0.7.
5.6t = 5.6(0.7) (Substitute the variable with its value.)
= 3.92 (Find the product.)
Result
3.92
Page 175 Exercise 16 Answer
To evaluate the expression 4x use substitution, that is substitute x with its value x = \(\frac{1}{2}\).
4x = 4(\(\frac{1}{2}\)) (Substitute the variable with its value.)
= 2 (Find the product.)
Result
2
Page 175 Exercise 17 Answer
Follow the order of operations, first evaluate parentheses and brackets from inside out, then evaluate powers, finally calculate any products and quotients, sums and differences.
[(33 ÷ 3) + 1] − 22 = [11 + 1] − 22 (Find the quotient.)
= 12 – 22 (Find the sum.)
= 12 – 4 (Evaluate the power.)
= 8 (Find the difference.)
Result
8
Page 175 Exercise 18 Answer
To plot the point A(−6,2) find −6 on the x-axis, since it is its x-coordinate, and find 2 on the y-axis, since it is its y-coordinate. Follow the gridlines from these points to where they meet. Mark that spot A.
Result
To plot the point A(−6,2) find −6 on the x-axis and find 2 on the y-axis. Follow the gridlines from these points to where they meet. Mark that spot A.