enVisionmath 2.0: Grade 6, Volume 1 Chapter 4 Represent And Solve Equations And Inequalities
Section 4.9: Use Patterns To Write And Solve Equations
Page 233 Exercise 1 Answer
From the solve & Discuss It! problem, we know the number of candles is equal to 4 times the number of boxes.
If b is the number of boxes and c is the number of candles, then this relationship gives us the equation:
of candles = 4 x of boxes
c = 4b
Result
c = 4b
Page 233 Exercise 1 Answer
We are given the following table:
From the table, we can see that the number of candles is equal to 4 times the number of boxes since 2 x 4 = 8, 3 x 4 = 12 and 4 x 4 = 16.
To find the number of candles for 10 boxes, we must then multiply this number by 4:
4 x 10 = 40
Therefore, 10 boxes has 40 candles.
Result
The number of candles is equal to 4 times the number of boxes.
10 boxes has 40 candles.
Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions
Page 234 Exercise 1 Answer
24.4 ÷ 2 = 12.2
36.6 ÷ 3 = 12.2
61 ÷ 5 = 12.2
73.2 ÷ 6 = 12.2
Notice, when y is divided by s the result is 12.2, for all given values of s and y. In other words, y is equal to 12.2 times the value of s.
Thus, the rule can be written as:
y = 12.2s.
If the cyclist maintains this speed and rides for 8 seconds, that is s = 8,
y = 12.2s
= 12.2(8)
= 97.6,
he will cross 97.6 yards.
The equation y = 12.2s describes the pattern in the table because it is true for all the values of s and y that are given.
Result
97.6 yards.
Page 235 Exercise 2 Answer
From Example 2, we know the equation a = $75 − 5w represents the amount still owed, a, after Ethan has made $5 payments each week to his mother for w weeks.
Also from the Example, we also know that after 12 weeks, Ethan still owes $15. Since he makes payments of $5 each week and 15 ÷ 5 = 3, then Ethan will still need to make payments to his mother for 3 more weeks.
Result
After 12 weeks he still owes $15 so if he pays her $5 each week, he will still need to make payments to his mother for 3 more weeks.
Page 236 Exercise 1 Answer
We can use a pattern to write and solve an equation by comparing the values of the independent and dependent variables. Using our comparison, we can write the relationship between the variables in words. Using the relationship in words, we can then write the equation by replacing each quantity in the verbal description with its variable.
For example, consider the following relationship given in words:
The cost is 5 times the number of tickets.
Replacing the cost” with the variable $c$ and replacing the number of tickets” with the variable t then gives the equation c = 5t.
Result
We can use a pattern to write and solve an equation by comparing the values of the independent and dependent variables. Using our comparison, we can write the relationship between the variables in words. Using the relationship in words, we can then write the equation by replacing each quantity in the verbal description with its variable.
Page 236 Exercise 2 Answer
To find a pattern that relates the values in a table, we need to compare the values of the independent variable and the dependent variable. Using our comparison, we can determine what operation(s) and what number (s) must be used with the values of the independent variable to get the values of the dependent variable.
For example, consider the following table:
Comparing the values of the variables, we can see that the cost is always 5 times the number of tickets since 1 x 5 = 5, 2 x 5 = 10, 3 x 5 = 15, and 4 x5 = 20.
Result
To find a pattern that relates the values in a table, we need to compare the values of the independent variable and the dependent variable. Using our comparison, we can determine what operation(s) and what number(s) must be used with the values of the independent variable to get the values of the dependent variable.
Page 236 Exercise 3 Answer
In Example 2, the following table was made:
From this table, we can see that each time the number of weeks w increases by 1, the amount owed decreases by $5, which makes sense since he pays his mom $5 each week so the amount he still owes should also decrease by $5.
Result
As the number of weeks w increases by 1, the amount owed a decreases by $5.
Page 236 Exercise 4 Answer
We age given the following table:
We need to find a pattern for this table so we can write an equation. The y-values are bigger than the x-values so the operation might be addition or multiplication.
We know that 1 + 6 = 7 but 2 + 6 = 8 ≠ 12 so addition is not the pattern.
We know that 1 x 7 = 7 but 2 x 7 = 14 ≠ 12 so multiplication is also not the pattern.
The pattern will then need two operations in it.
To find one of those operations, we can determine how the value of y changes when the value of x increases by 1. From the table, the values of y increase by 5 each time x increases by 1 since 7 + 5 = 12, 12 + 5 = 17, and 17 + 5 = 22. This means that the pattern will have 5 times x in it.
Knowing that we need 5 times the value of x in the pattern, we can rewrite the values of y as:
7 = 5 + 2 = 5(1) + 2
12 = 10 + 2 = 5(2) + 2
17 = 15 + 2 = 5(3) + 2
22 = 20 + 2 = 5(4) + 2
y = 5(x) + 2
The pattern as an equation is then y = 5x + 2
Result
y = 5x + 2
Page 236 Exercise 5 Answer
We are given the following table:
From the table, we can see that Brenda’s age, b, is equal to Talia’s age, t, minus 5 since 7 – 5 = 2, 9 – 5 = 4, and 10 – 5 = 5. The pattern as an equation is then b = t – 5.
If Talia is 12, then t = 12. Using the equation gives:
b = t – 5 = 12 – 5 = 7
Therefore, Brenda is 7 when Talia is 12.
Result
Brenda’s age, b, is equal to Talia’s age, t, minus 5.
b = t − 5
Brenda is 7 when Talia is 12.
Page 236 Exercise 6 Answer
Use substitution – substitute x with its value, that is with 7 and 8, and calculate the value of y which depends on the value of x.
y = 2x − 7
= 2(7) − 7
= 14 − 7
= 7
If x = 7, then y = 7.
= 2x − 7
= 2(8) − 7
= 16 − 7
= 9
If x = 8, then y = 9.
Result
If x = 7, then y = 7. If x = 8, then y = 9.
Page 236 Exercise 7 Answer
The rule for the pattern as an equation is y = 2x − 7 so the rule for the pattern in words is:
The variable y is equal to the value of 2 times x minus 7.
Result
The variable y is equal to the value of 2 times x minus 7.
Page 237 Exercise 8 Answer
We are given the following table:
From the table, we can see that value of y is always 32 more than the value of x since:
33 = 1 + 32
34 = 2 + 32
35 = 3 + 32
36 = 4 + 32
37 = 5 + 32
The rule is then: The value of y is equal to the value of x plus 32.
As an equation, this rule is: y = x + 32
Result
Rule: The value of y is equal to the value of x plus 32.
Equation: y = x + 32
Page 237 Exercise 9 Answer
We are given the following table:
From the table, we can see that value of n is always 3 times the value of m since:
0 = 3 x 0
3 = 3 x 1
6 = 3 x 2
9 = 3 x 3
12 = 3 x 4
The rule is then: The value of n is equal to 3 times the value of m.
As an equation, this rule is: n = 3m.
Result
Rule: The value of n is equal to 3 times the value of m.
Equation: n = 3m
Page 237 Exercise 10 Answer
We are given the following table:
From the table, we can see that the value of k is always 15 less than the value of g:
17 = 32 – 15
22 = 37 – 15
27 = 42 – 15
The rule is then: The value of k is equal to the value of g minus 15.
The rule as an equation is: k = g – 15
To complete the table, we can substitute g = 47 and g = 52 into the equation and then simplify:
g = 47 : k = g – 15 = 47 – 15 = 32
g = 52 : k = g – 15 = 52 – 15 = 37
The completed table is then:
Result
Rule: The value of k is equal to the value of g minus 15.
Equation: k = g – 15
Page 237 Exercise 11 Answer
We are give the following table:
From the table, we can see that the value of y is always the value of x divided by 9:
0 = 0 ÷ 9
1 = 9 ÷ 9
2 = 18 ÷ 9
The rule is then: The value of y is equal to the value of x divided by 9.
The rule as an equation is: y = x ÷ 9
To complete the table, we can substitute x = 27 and x = 36 into the equation and then simplify:
x = 27 : y = x ÷ 9 = 27 ÷ 9 = 3
x = 36 : y = x ÷ 9 = 36 ÷ 9 = 4
The completed table is then:
Result
Rule: The value of y is equal to the value of x divided by 9.
Equation: y = x ÷ 9
Page 237 Exercise 12 Answer
We are given the following table:
To write the equation, it would help to find the cost for each teddy bear. To find the cost for 1 teddy bear, we can divide one of the costs from the table by its corresponding number of teddy bears:
$188,000 ÷ 4 = $47,000
Each teddy bear then costs $47,000 which means the cost is equal to 47,000 times the number of teddy bears:
188,000 = 47,000(4)
329,000 = 47,000(7)
517,000 = 47,000(11)
c = 47,000(n)
The equation is then c = 47,000n
Result
c = 47,000n
Page 237 Exercise 13a Answer
We are given the following table:
We are also given to the right of the table the following information:
General Admission: $8.00
Rides: $2.50
Since the cost of each ride is $2.50, then the cost for 5 rides is $2.50 more than the cost for 4 rides. The cost for 5 rides is then $18.00 + $2.50 = $20.50.
Since the cost of 1 ride is $2.50, then the cost of 2 rides is 2 x $2.50 = $5.00. The cost for 8 rides is then $5.00 more than the cost for 6 tides so the cost is $23.00 + $5.00 = $28.00.
The completed table is then:
Since the total cost is the cost for the rides plus the general admission of $8.00 and the cost for the rides is equal to 2.50 times the number of rides, we can rewrite the costs in the table as follows:
$15.50 = 7.50 + 8.00 = 2.50(3) + 8.00
$18.00 = 10.00 + 8.00 = 2.50(4) + 8.00
$20.50 = 12.50 + 8.00 = 2.50(5) + 8.00
$23.00 = 15.00 + 8.00 = 2.50(6) + 8.00
$28.00 = 20.00 + 8.00 = 2.50(8) + 8.00
c = 2.50(r) + 8.00
The pattern as an equation is then: c = 2.5r + 8
Result
c = 2.5r + 8
Page 237 Exercise 13b Answer
From part (a), we know the total cost, c, is represented by the equation c = 2.5r + 8 where r is the number of rides.
To find the cost for 12 rides, we can then substitute r = 12 into the equation and simplify:
c = 2.5r + 8 = 2.5(12) + 8 = 30 + 8 = 38
The cost for 12 rides is then $38.
Result
$38
Page 238 Exercise 14 Answer
Notice, for each w and z, the value of z is two less than the value of w. That is:
2 − 2 = 0
4 − 2 = 2
6 − 2 = 4
8 − 2 = 6
10 − 2 = 8.
Thus, the pattern is:
z = w − 2.
Result
z = w − 2
Page 238 Exercise 15 Answer
Comparing the values of x and y, we see that:
4 x 0 = 0
\(4 \times \frac{1}{2}=\frac{4}{1} \times \frac{1}{2}=\frac{4}{2}=2\)4 x 1 = 4
\(4 \times 1 \frac{1}{2}=\frac{4}{1} \times \frac{3}{2}=\frac{12}{2}=6\)4 x 2 = 8
\(4 \times 2 \frac{1}{2}=\frac{4}{1} \times \frac{5}{2}=\frac{20}{2}=10\)Thus, the value of y is 4 times the value of x so the pattern is y = 4x
Result
y = 4x
Page 238 Exercise 16 Answer
Use substitution – substitute d with its value, that is with 3 and 4, and calculate the value of t.
t = 5d + 5
= 5(3) + 5
= 15 + 5
= 20
If d = 3, then t = 20.
t = 5d + 5
= 5(4) + 5
= 20 + 5
= 25
If d = 4, then t = 25.
Result
If d = 3, then t = 20. If d = 4, then t = 25.
Page 238 Exercise 17 Answer
Use substitution – substitute x with its value, that is with 8 and 10, and calculate the value of y.
y = \(\frac{1}{2}\)x − 1
= \(\frac{1}{2}\)(8)−1
= 4 − 1
= 3
If x = 8, then y = 3.
y = \(\frac{1}{2}\)x − 1
= \(\frac{1}{2}\)(10)−1
= 5 − 1
= 4
If x = 10, then y = 4.
Result
If x = 8, then y = 3. If x = 10, then y = 4.
Page 238 Exercise 18 Answer
Use substitution – substitute x with its value, that is with 2 and 3, and calculate the value of y.
y = 2x + 1
= 2(2) + 1
= 4 + 1
= 5
If x = 2, then y = 5.
y = 2x + 1
= 2(3) + 1
= 6 + 1
= 7
If x = 3, then y = 7.
Result
If x = 2, then y = 5. If x = 3, then y = 7.
Page 238 Exercise 19 Answer
Use substitution – substitute a with its value and calculate the value of b.
b = \(\frac{a}{2}\) – 2
= \(\frac{17}{2}\) – 2
= \(\frac{17}{2}\) – \(\frac{4}{2}\)
= \(\frac{17-4}{2}\)
= \(\frac{13}{2}\)
= \(\frac{12}{2}\) + \(\frac{1}{2}\)
= \(6 \frac{1}{2}\)
If a = 17, then b = \(6 \frac{1}{2}\)
b = \(\frac{a}{2}\) – 2
= \(\frac{14}{2}\) – 2
= 7 -2
= 5
If a = 14, then b = 5
b = \(\frac{a}{2}\) – 2
= \(\frac{11}{2}\) – 2
= \(\frac{11}{2}\) – \(\frac{4}{2}\)
= \(\frac{11-4}{2}\)
= \(\frac{7}{2}\)
= \(\frac{6}{2}\) + \(\frac{1}{2}\)
= \(3 \frac{1}{2}\)
If a = 11, then b = \(3 \frac{1}{2}\)
b = \(\frac{a}{2}\) – 2
= \(\frac{8}{2}\) – 2
= 4 – 2
=2
If a = 8, then b = 2
b = \(b=\frac{a}{2}-2\)
= \(\frac{5}{2}\) – 2
= \(\frac{5}{2}\) – \(\frac{4}{2}\)
= \(\frac{5-4}{2}\)
= \(\frac{1}{2}\)
If a = 5, then b = \(\frac{1}{2}\)
Result
If a = 17, then b = \(6 \frac{1}{2}\). If a = 14, then b = 5. If a = 11, then b = \(3 \frac{1}{2}\). If a = 8, then b = 2. If a = 5, then b = \(\frac{1}{2}\).
Page 238 Exercise 20 Answer
Substitute h and d with corresponding values from the table to check if Maya is right.
h = d + 22
3 ≠ 33 + 22 = 55
5 ≠ 55 + 22 = 77
7 ≠ 77 + 22 = 99
9 ≠ 99 + 22 = 121
Maya is not correct. h = d + 22 is not true for any of the corresponding values of h and d.
d = 11h
33 = 11(3)
55 = 11(5)
77 = 11(7)
99 = 11(9)
The correct equation is d = 11h.
Result
Maya is not correct.
Page 238 Exercise 21 Answer
Find an equation which is true for the given values of t and c, than use it to answer the question.
26.25 ÷ 3 = 8.75
61.25 ÷ 7 = 8.75
78.75 ÷ 9 = 8.75
That is, for each value of t and c it is true c ÷ t = $8.75, or written differently c = $8.75t.
Use c = $8.75t to find the cost of 5 tickets, that is find the value of c if t = 5.
c = $8.75(5)
= $43.75
Result
The cost of 5 tickets is $43.75.