Chapter 4 Represent And Solve Equations And Inequalities
Section 4.5: Write And Solve Equations With Rational Numbers
Page 201 Exercise 1 Answer
When solving for unknowns involving money, we use the same process of using inverse operations to isolate the variable that we use when solving for unknowns involving whole numbers.
Page 201 Exercise 1a Answer
We know that:
cost of Lorna’s shirt + cost of Ben’s shirt = total cost
Since Lorna is on Team A, then the cost of Lorna’s shirt is $12.50. Since we don’t know what team Ben is on, let b represent the cost of Ben’s shirt. The total cost is $21.25 so the equation is:
12.50 + b = 21.25
Result
12.50 + b = 21.25 where b is the cost of Ben’s shirt.
Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions
Page 201 Exercise 1b Answer
If Dario is correct, then the price of Ben’s T-shirt must be $11.95 if Ben is on Team B.
From Part A, the equation we need to solve to find the cost of Ben’s shirt is 12.50 + b = 21.25. To solve for b, we need to subtract 12.50 on both sides:
12.50 + b − 12.50 = 21.25 − 12.50
b = 8.75
Since Ben’s T-shirt costs $8.75, then Ben is on Team C, not Team B, so Dario is incorrect.
Result
Dario is incorrect. Ben is on Team C since his T-shirt costs $8.75.
Page 202 Exercise 1 Answer
\(1 \frac{3}{8}+x=2 \frac{1}{4}\)\(1 \frac{3}{8}+x-1 \frac{3}{8}=2 \frac{1}{4}-1 \frac{3}{8}\)
\(x=1 \frac{5}{4}-1 \frac{3}{8}\)
x = \(1 \frac{10}{8}-1 \frac{3}{8}\)
x = \(\frac{10}{8}-\frac{3}{8}\)
x = \(\frac{7}{8}\)
The length of the shorter piece is \(\frac{7}{8}\) foot.
If the situation is different, that is we know the lenght of the shorter piece is \(\frac{7}{8}\) foot. Than the equation is:
\(\frac{7}{8}+x=2 \frac{1}{4}\)where x is now the length of the longer of the two pieces.
Result
x = \(\frac{7}{8}\)
Page 203 Exercise 2 Answer
When solving a multiplication equation with a rational coefficient, we isolate the variable by multiplying both sides of the equation by the reciprocal of the coefficient.
For the equation \(\frac{5}{9}\)y = 25, the coefficient is \(\frac{5}{9}\), which has a reciprocal of \(\frac{9}{5}\) so solving the equation gives:
\(\frac{5}{9}\)y = 25 Given equation.
\(\frac{9}{5} \cdot \frac{5}{9} y=\frac{9}{5} \cdot 25\) Multiply both sides by \(\frac{9}{5}\).
\(y=\frac{225}{5}=45\) Simplify.
Result
y = 45
Page 203 Exercise 3 Answer
We know that:
(number of apples) . (price per apple) = total cost
Molly buys 8 apples for a total cost of $3.60. Let m be the cost of each apple. The equation is then:
8m = 3.60
To solve this equation we need to divide both sides by 8:
8m ÷ 8 = 3.60 ÷ 8
m = 0.45
Molly then paid $.45 for each apple.
Result
$0.45
Page 204 Exercise 4 Answer
Since Carmen spent $12.50 in total, the sum $6.35 + c equals $12.50.
The equation is:
$6.35 + c = $12.50.
Solve the equation to find the value of c, that is the cost of the compass.
$6.35 + c = $12.50
$6.35 + c – $6.35 = $12.50 – $6.35 (Subtract $6.35 from both sides.)
c = $6.15
To cost of the compass is $6.15.
Result
To cost of the compass is $6.15.
Page 205 Exercise 1 Answer
To write equations involving rational numbers, we use the same process as when writing equations involving whole numbers, which is using the given relationship to determine what operation is needed for the equation. We also use the same process when solving, which is using properties of equality and inverse relationships.
Page 205 Exercise 2 Answer
When solving equations, we need to isolate the variable by performing an operation that will undo” the operation we already have. Since inverse operations undo” each other, then inverse relationships are important for solving equations.
Result
When solving equations, we need to isolate the variable by performing an operation that will undo” the operation we already have. Since inverse operations undo” each other, then inverse relationships are important for solving equations.
Page 205 Exercise 3 Answer
When solving equations, you must always perform an operation on both sides of the equation to keep it balanced.
To solve the equation, 3.5 must be added to both sides of the equation. Johnny is then incorrect because he said he only added 3.5 to the left side, not both sides.
Result
Johhny is incorrect since 3.5 must be added to both sides of the equation, not just the left side.
Page 205 Exercise 4 Answer
When adding or subtracting fractions and mixed numbers, we always need to get a common denominator. This means that when solving addition and subtraction equations with mixed numbers, we need to rewrite the mixed number and fraction to have a common denominator before we can add or subtract.
For example, to solve y + \(\frac{3}{4}\) = \(4 \frac{1}{2}\), we must rewrite the mixed number \(4 \frac{1}{2}\) as \(4 \frac{2}{4}\) so it will have a common denominator with \(\frac{3}{4}\).
When we multiply fractions, we must always convert mixed numbers to improper fractions. This means that when solving multiplication division equations with mixed numbers, we need to rewrite the mixed number as an improper fraction before we can multiply or divide.
For example, to solve \(\frac{3}{4}\)y = \(4 \frac{1}{2}\), we must rewrite the mixed number \(4 \frac{1}{2}\) as \(\frac{9}{2}\).
Result
When solving addition and subtraction equations, you need to write the mixed number and fraction with a common denominator. When solving multiplication and division equations, you need to write the mixed number as an improper fraction.
Page 205 Exercise 5 Answer
To solve equations, we isolate the variable using inverse relationships and the properties of equality. To solve addition equations, we use the Subtraction Property of Equality, and to solve subtraction equations, we use the Addition Property of Equality since addition and subtraction are inverse operations. To solve multiplication equations, we use the Division Property of Equality, and to solve division equations, we use the Multiplication Property of Equality since multiplication and division are inverse operations.
Page 205 Exercise 6 Answer
\(t-\frac{2}{3}=25 \frac{3}{4}\) Given equation.
\(t-\frac{2}{3}+\frac{2}{3}=25 \frac{3}{4}+\frac{2}{3}\) Add \(\frac{2}{3}\) on both sides.
\(t=25 \frac{9}{12}+\frac{8}{12}\) Get a common denominator.
t = \(25 \frac{17}{12}\) Add.
t = \(26 \frac{5}{12}\) Regroup.
Result
t = \(26 \frac{5}{12}\)
Page 205 Exercise 7 Answer
\(\frac{f}{2} \times 2=\frac{5}{8} \times 2\) (Multiply both side by 2.)
\(f=\frac{5}{8} \times \frac{2}{1}\) \(f=\frac{5 \times 2}{8 \times 1}\)\(\frac{10}{8}\) (Multiply the fractions.)
\(f=\frac{8}{8}+\frac{2}{8}\)f = \(1 \frac{1}{4}\) (Simplify.)
Result
f = \(1 \frac{1}{4}\)
Page 205 Exercise 8 Answer
13.27 + 24.45 = t − 24.45 + 24.45 (Add 24.45 to both sides.)
37.72 = t (Evaluate.)
Result
t = 37.72
Page 205 Exercise 9 Answer
r ÷ 5.5 × 5.5 = 18.2 × 5.5 (Multiply both sides by 5.5.)
r = 100.1 (Evaluate.)
Result
r = 100.1
Page 205 Exercise 10 Answer
\(\frac{7}{10}+\frac{3}{5}=x-\frac{3}{5}+\frac{3}{5}\) (Add \(\frac{3}{5}\) to both sides.)
\(\frac{7}{10}+\frac{3 \times 2}{5 \times 2}=x\)\(\frac{7}{10}\) + \(\frac{6}{10}\) = x (Add the fractions.)
\(\frac{13}{10}\)\(\frac{10}{10}\) + \(\frac{3}{10}\) = x
\(1 \frac{3}{10}\) = x
Result
x = \(1 \frac{3}{10}\)
Page 205 Exercise 11 Answer
1.8x ÷ 1.8 = 40.14 ÷ 1.8 (Divide both sides by 1.8.)
x = 22.3 (Evaluate.)
Result
x = 22.3
Page 205 Exercise 12 Answer
17.3 + v − 17.3 = 22.32 − 17.3 (Subtract 17.3 from both sides.)
v = 5.02 (Evaluate.)
Result
v = 5.02
Page 205 Exercise 13 Answer
\(9 \times \frac{8}{3}=\frac{3}{8} y \times \frac{8}{3}\) (Multiply both sides by \(\frac{8}{3}\).)
\(\frac{9}{1} \times \frac{8}{3}=y\) (Multiply the fractions.)
\(\frac{9 \times 8}{1 \times 3}=y\)\(\frac{72}{3}\) = y
24 = y (Evaluate.)
Result
y = 24
Page 205 Exercise 14 Answer
\(\frac{7}{4}+z=\frac{8}{3}\) (Rewrite the mixed numbers as fractions.)
\(\frac{7}{4}+z-\frac{7}{4}=\frac{8}{3}-\frac{7}{4}\) (Subtract \(\frac{7}{4}\) from both sides.)
\(z=\frac{8 \times 4}{3 \times 4}-\frac{7 \times 3}{4 \times 3}\)
z = \(\frac{32}{12}\) – \(\frac{21}{12}\)
z = \(\frac{11}{12}\)
Result
z = \(\frac{11}{12}\)
Page 206 Exercise 15 Answer
w − 3.2 + 3.2 = 5.6 + 3.2
w = 8.8 (Add 3.2 to both sides.)
Result
w = 8.8
Page 206 Exercise 16 Answer
9.6 ÷ 1.6 = 1.6y ÷ 1.6 (Divide both sides by 1.6.)
6 = y (Evaluate.)
Result
y = 6
Page 206 Exercise 17 Answer
48.55 + k − 48.55 = 61.77 − 48.55 (Subtract 48.55 to both sides.)
k = 13.22 (Evaluate.)
Result
k = 13.22
Page 206 Exercise 18 Answer
m ÷ 3.54 × 3.54 = 1.5 × 3.54 (Multiply both sides by 3.54.)
m = 5.31 (Evaluate.)
Result
m = 5.31
Page 206 Exercise 19 Answer
\(7 \frac{1}{9}=2 \frac{4}{5}+m\) Given equation.
\(7 \frac{1}{9}-2 \frac{4}{5}=2 \frac{4}{5}+m-2 \frac{4}{5}\) Subtract \(2 \frac{4}{5}\) on both sides.
\(7 \frac{5}{45}-2 \frac{36}{45}=m\) Get a common denominator.
\(6 \frac{50}{45}-2 \frac{36}{45}=m\) Regroup 7 \(7 \frac{5}{45}\).
\(4 \frac{14}{45}\) = m Subtract.
Result
m = \(4 \frac{14}{45}\)
Page 206 Exercise 20 Answer
\(a+3 \frac{1}{4}=5 \frac{2}{9}\) Given equation.
\(a+3 \frac{1}{4}-3 \frac{1}{4}=5 \frac{2}{9}-3 \frac{1}{4}\) Subtract \(3 \frac{1}{4}\) on both sides.
\(a=5 \frac{8}{36}-3 \frac{9}{36}\) Get a common denominator.
\(a=4 \frac{44}{36}-3 \frac{9}{36}\) Regroup \(5 \frac{8}{36}\)
a = \(1 \frac{35}{36}\) Subtract.
Result
a = \(1 \frac{35}{36}\)
Page 206 Exercise 21 Answer
\(\frac{1}{8}\) x y x 8 = 4 x 8 (Multiply both sides by 8.)
y = 32 (Evaluate.)
Result
y = 32
Page 206 Exercise 22 Answer
\(k-6 \frac{3}{8}=4 \frac{6}{7}\) Given equation.
\(k-6 \frac{3}{8}+6 \frac{3}{8}=4 \frac{6}{7}+6 \frac{3}{8}\) Add \(6 \frac{3}{8}\) on both sides.
\(k=4 \frac{48}{56}+6 \frac{21}{56}\) Get a common denominator.
k = \(10 \frac{69}{56}\) Add.
k = \(11 \frac{13}{56}\) Regroup.
Result
k = \(11 \frac{13}{56}\)
Page 206 Exercise 23 Answer
Mr. Marlon bought four tickets. Let’s mark the cost of one ticket x. Since the total cost is $210. Thus, the equation which represents this situation is:
4x = 210.
To answer the question find the solution of the equation.
4x = 210
x = 210 ÷ 4 (Divide both sides by 4.)
x = \(\frac{105}{2}\)
The cost of one ticket is $52.50.
Result
$52.50
Page 206 Exercise 24 Answer
The right sides of the given equations are equal.
However, the solution of the equation \(\frac{5}{9}\)m = \(2 \frac{3}{4}\) must be greater than the solution of the equation \(\frac{5}{8}\)m = \(2 \frac{3}{4}\).
To solve the equations, we multiply both sides by the reciprocals of the coefficients, which are \(\frac{8}{5}\) and \(\frac{9}{5}\).
\(\frac{9}{5}\) is greater than \(\frac{8}{5}\) which means that \(2 \frac{3}{4}\) x \(\frac{9}{5}\), which is the solution of the second equation, is greater than the \(2 \frac{3}{4}\) x \(\frac{8}{5}\) which is the solution of the first equation.
Result
The solution of the equation \(\frac{5}{9}\)m = \(2 \frac{3}{4}\) must be greater than the solution of the equation \(\frac{5}{8}\) m = \(2 \frac{3}{4}\) since the reciprocal of \(\frac{5}{9}\) is greater than the reciprocal of \(\frac{5}{8}\).
Page 206 Exercise 25 Answer
It is given that the record is 21 ft \(2 \frac{1}{4}\) in and that Tim’s best long jump is 20 ft \(9 \frac{1}{2}\) in.
Since the long jumps are measured to the nearest quarter inch, Tim must jump one-quarter inch farther than the current record in order to beat it. Tim must then jump a new record distance of:
\(21 \mathrm{ft} 2 \frac{1}{4} \mathrm{in}+\frac{1}{4} \mathrm{in}=21 \mathrm{ft} 2 \frac{2}{4} \text { in }=21 \mathrm{ft} 2 \frac{1}{2} \text { in }\)To find how much farther Tim must jump to break the record, we then need to subtract the new record and Tim’s current long jump distance.
Before we subtract, we should convert the two lengths to inches. Since there are 12 inches in 1 foot, we can multiply the number of feet by 12 to convert it to inches:
new record:
\(21 \mathrm{ft} 2 \frac{1}{2} \text { in }=\left(21 \times 12+2 \frac{1}{2}\right) \text { in }=\left(252+2 \frac{1}{2}\right) \text { in }=254 \frac{1}{2} \text { in }\)current distance:
\(20 \mathrm{ft} 9 \frac{1}{2} \text { in }=\left(20 \times 12+9 \frac{1}{2}\right) \text { in }=\left(240+9 \frac{1}{2}\right) \text { in }=249 \frac{1}{2} \text { in }\)
The difference of the new record and his current long jump distance is then:
\(254 \frac{1}{2} \text { in }-249 \frac{1}{2} \text { in }=5 \text { in }\)Tim must then jump 5 inches farther to break the record.
Result
5 inches farther
Page 207 Exercise 26 Answer
It takes one gallon of fuel to move the space shuttle 40 feet.
To calculate how many gallons it takes to move it 3 miles, first rewrite 3 miles in feet. Since 1 mile is 5,280 feet, 3 miles is 3 times that.
3 gal = 3 x 5,280 ft
= 15,840 ft
Since it take 1 gallon per 40 feet, divide 15,840 by 4.
15,840 ÷ 40 = 396
It takes 396 gallons of fuel to move the space shuttle 3 miles.
Result
396 gallons.
Page 207 Exercise 27 Answer
The solution of b × \(\frac{5}{6}\) = 25 must be greater than 25 since the solution is b multiplied by a number less than one – that is, we can take less than the whole number, specifically \(\frac{5}{6}\) of the number b, and it will be equal to 25 so it must be greater.
Solve and compare to check the answer.
b x \(\frac{5}{6}\) = 25
\(b \times \frac{5}{6} \times \frac{6}{5}=25 \times \frac{6}{5}\) (Multiply by a reciprocal of \(\frac{5}{6}\).)
\(b=\frac{25}{1} \times \frac{6}{5}\)
\(b=\frac{25 \times 6}{1 \times 5}\)
b = \(\frac{150}{5}\)
b = 30 > 25
The solution of the given equation is 30 which is greater than 25.
Result
The solution is 30 which is greater than 25.
Page 207 Exercise 28 Answer
Area of a rectangle is given as a product of its width and length, that is if w is its width and l is its length, the area is A = w × l.
Since the length and the area are given, substitute and solve for w to answer the question.
w x l = A
\(w \times \frac{3}{7} \mathrm{ft}=2 \mathrm{ft}^2\)
\(w \times \frac{3}{7} \times \frac{7}{3}=2 \times \frac{7}{3}\) (Multiply by a reciprocal of \(\frac{3}{7}\).)
w = \(\frac{2}{1}\) x \(\frac{7}{3}\) (Rewrite the whole number as a fraction.)
\(w=\frac{2 \times 7}{1 \times 3}\)w = \(\frac{14}{3}\)(Multiply the fractions.)
w = \(\frac{12}{3}+\frac{2}{3}\)
w = \(4 \frac{2}{3}\)ft (Rewrite as a mixed number.)
The width of the rectangle is \(4 \frac{2}{3}\)ft.
Result
\(4 \frac{2}{3}\) ft.
Page 207 Exercise 29 Answer
Since Helen needs to fill the pool with \(10 \frac{1}{2}\) gallons of water that is one side of the equation. The other side is the product of how much she can carry in one trip – that is \(1 \frac{7}{8}\) of a gallon, and x – the number of trips she must make.
To answer the question solve the following equation:
\(1 \frac{7}{8}\)x = \(10 \frac{1}{2}\).
\(1 \frac{7}{8}\)x = \(10 \frac{1}{2}\)
\(\frac{15}{8}\)x = \(\frac{21}{2}\) (Rewrite mixed numbers as fractions)
\(\frac{15}{8} x \times \frac{8}{15}=\frac{21}{2} \times \frac{8}{15}\) \(x=\frac{21 \times 8}{2 \times 15}\)x = \(\frac{168}{30}\)
x = \(\frac{28}{5}\)
\(x=\frac{25}{5}+\frac{3}{5}\)x = \(5 \frac{3}{5}\)
Helen needs to make \(5 \frac{3}{5}\) trips, since that is not a whole number, she must make at least 6 trips to fill the whole pool.
Result
6 trips
Page 207 Exercise 30 Answer
When the pool was full there was \(10 \frac{1}{2}\) gallons of water in it, that is one side of the equation. The other side is the sum of the water which was spashed, g gallons, and the water still left in the pool, that is \(7 \frac{7}{8}\)
The equation which describes the situation is:
\(g+7 \frac{7}{8}=10 \frac{1}{2}\)
To answer the question solve the equation.
\(g+7 \frac{7}{8}=10 \frac{1}{2}\)
\(g+\frac{63}{8}=\frac{21}{2}\) (Rewrite mixed numbers as fractions.)
\(g+\frac{63}{8}-\frac{63}{8}=\frac{21}{2}-\frac{63}{8}\) (Subtract \(\frac{63}{8}\) from both sides.)
g = \(\frac{84}{8}\) – \(\frac{63}{8}\)
g = \(\frac{84}{8}\) – \(\frac{63}{8}\)
g = \(\frac{84-63}{8}\)
g = \(\frac{21}{8}\)
g = \(\frac{16}{8}\) + \(\frac{5}{8}\)
g = \(2 \frac{5}{8}\)
\(2 \frac{5}{8}\) gallons was splashed out of the pool.
Result
\(2 \frac{5}{8}\) gallons.
Page 207 Exercise 31 Answer
\(2 \frac{1}{2}\)y = \(\frac{5}{8}\) (The first step is obviously the equation itself.)
\(\frac{5}{2}\)y = \(\frac{5}{8}\) (The second step is rewriting mixed numbers as fractions.)
y = \(\frac{5}{8}\) . \(\frac{2}{5}\) (The third step is multiplying by a reciprocal of \(\frac{5}{2}\).)
y = \(\frac{10}{40}\) or \(\frac{1}{4}\) (The last step is the value of y.)
Result
\(2 \frac{1}{2}\)y = \(\frac{5}{8}\)
\(\frac{5}{2}\)y = \(\frac{5}{8}\)
y = \(\frac{5}{8}\) . \(\frac{2}{5}\)
y = \(\frac{10}{40}\) or \(\frac{1}{4}\)
Page 208 Exercise 32 Answer
Solve the given equation, start by dividing both sides by 4.5.
4.5n ÷ 4.5 = 8,640 ÷ 4.5
n = 1920
Nontasters have 1,920 taste buds.
Result
n = 1,920
Page 208 Exercise 33 Answer
Since there are 2.25 times more women supertasters than men, and m represents the number of men classified as supertasters, the product 2.25m is the number of women supertasters, that is 72.
2.25m = 72
2.25m ÷ 2.25 = 72 ÷ 2.25(Divide both sides by 2.25.)
m = 32
There are 32 men classified as supertasters.
Result
32
Page 208 Exercise 34 Answer
Five times the fraction f equals \(\frac{1}{8}\). That is 5f = \(\frac{1}{8}\).
5f = \(\frac{1}{8}\)
5f ÷ 5 = \(\frac{1}{8}\) ÷ 5 (Divide both sides by 5.)
f = \(\frac{1}{8}\) x \(\frac{1}{5}\) (Instead of dividing by 5, multiply by its reciprocal.)
f = \(\frac{1}{40}\)
To solve the equation divide both sides by five. Since dividing by a number is the same as multiplying by the reciprocal of that number, we can multiply \(\frac{1}{8}\) by \(\frac{1}{5}\), which is a reciprocal of 5, and get the same result.
Result
\(\frac{1}{40}\)Page 208 Exercise 35 Answer
To answer the question solve the equation, start by rewriting the mixed numbers as fractions.
\(\frac{43}{8}\) + m = 8
m = 8 – \(\frac{43}{8}\) (Subtract \(\frac{43}{8}\) from both sides.)
m = \(\frac{64}{8}\) – \(\frac{43}{8}\) (Rewrite 8 as a fraction with 8 as a denominator.)
m = \(\frac{64-43}{8}\) (Subtract the fractions.)
m = \(\frac{21}{8}\)
m = \(\frac{16}{8}+\frac{5}{8}\)
m = \(2 \frac{5}{8}\)
Yelena needs to swim \(2 \frac{5}{8}\) more miles.
Result
\(2 \frac{5}{8}\) miles.
Page 208 Exercise 36 Answer
Any equation written using addition can be rewritten as an equivalent equation using subtraction. One way is to subtract the same number from both sides, for example:
m + 4.56 = 55.7
m + 4.56 − 55.7 = 55.7 − 55.7
m − 51.14 = 0.
Another way is to subtract the opposite number of the one you are adding. Since −4.56 is the opposite of 4.56, the equation above would look like:
m + 4.56 = 55.7
m − (−4.56) = 55.7
Result
Yes because you can subtract the same number from both sides to get a subtraction equation or you can rewrite the addition as subtracting the opposite number.
Page 208 Exercise 37 Answer
Oscar’s expression is true now, when he is 12 years old and his sister is 6, however it will not be true in a year.
If Oscar’s age is a and his sister’s age is s, using Oscar’s expression the equation showing the relationship between their age would be:
a ÷ 2 = s.
This is true when a = 12 and s = 6, that is, it is true now.
In a year Oscar will be 13 and his sister 7, that is a = 13 and s = 7, than the eqaution
13 ÷ 2 = 7
not be true.
Result
No.
Page 208 Exercise 38 Answer
Solve the equation to answer the question, start by dividing both sides by 0.26.
0.26y ÷ 0.26 = 0.676 ÷ 0.26
y = 2.6
The correct answer is marked C.
Result
C 2.6
Page 208 Exercise 39 Answer
Solve the equation to answer the question – start by subtracting 0.435 from both sides.
0.435 + x − 0.435 = 0.92 − 0.435
x = 0.485
The correct answer is marked D.
Result
D 0.485