**Algebra 1 Common Core Answers Student Edition Grade 8 – 9 Chapter 1 Foundations for Algebra Exercise 1.1**

Algebra 1 Common Core Answers Student Edition Grade 8 – 9

**Chapter 1 Foundations for Algebra Exercise 1.1 1LC**

**(a) **Given is to describe an expression as algebraic or numerical:-

To describe above expression as algebraic or numerical, we first define both algebraic and numerical expressions as below:

7÷2

**Algebraic expression-** It is a mathematical phrase which includes one or more variables.

For **example**:- 6n-8, 3h, \( \frac { x }{ 3 } \)

**Numerical expression-** It is a mathematical phrase which includes numbers and operations symbols, but no variables.

For **example**:- 7+2, 3×6-3, \( \frac { 14 }{ 7 } \)

Clearly, the above expression 7÷2 is a numerical expression as it includes no variable and just includes two numbers (7 and 2 )and one operational sign that is division (÷) and the value of this expression always remain constant.

Thus 7÷2 is a numerical expression.

**(b) **4m+6

To describe above expression as algebraic or numerical, we first define both algebraic and numerical expressions which is stated below:

**Algebraic expression-** It is a mathematical phrase which includes one or more variables.

For **example**:- 6n-8,3h,\( \frac { x }{ 3 } \)

**Numerical expression-** It is a mathematical phrase which includes numbers and operations symbols, but no variables

For **example**:- 7+2, 3×6-3, \( \frac { 14 }{ 7 } \)

Clearly, the above expression is an algebraic expression as it includes one variable (m) and the value of this expression keeps on changing with the change of value of variable.

Thus 4m+6 is a algebraic expression.

**(c) **2(5-4)

To describe above expression as algebraic or numerical, we first define both algebraic and numerical expressions which is stated below:

**Algebraic expression-** It is a mathematical phrase which includes one or more variables.

For **example**:- 6n-8,3h,\( \frac { x }{ 3 } \)

**Numerical expression-** It is a mathematical phrase which includes numbers and operations symbols, but no variables

For **example**:- 7+2, 3×6-3, \( \frac { 14 }{ 7 } \)

Clearly, the above expression is a numerical expression 2(5-4) as it includes no variable and just includes two numbers (2,5 and 4) and one operational sign that is division (× and -) and the value of this expression always remain constant.

Thus 2(5-4) is a algebraic expression.

**Chapter 1 Foundations for Algebra Exercise 1.1 2LC**

**a) **Given is to write an algebraic expression for the following phrase:-

The product of 9 and a number t =9t

It means to multiply 9 with t

Hence, the product of 9 and a number t = 9t

**b) **Given is to write an algebraic expression for the following phrase:-

The difference of a number x and \( \frac { 1 }{ 2 } \)

It means to subtract \( \frac { 1 }{ 2 } \) from x

Hence, the difference of a number x and \( \frac { 1 }{ 2 } \) \( =\quad x-\frac { 1 }{ 2 } \)

**c) **Given is to write an algebraic expression for the following phrase:-

The sum of a number m and 7.1

It means to add 7.1 to m

Hence, the sum of a number m and 7.1 = m+7.1

**d) **Given is to write an algebraic expression for the following phrase:-

The quotient of 207 and a number n

It means to divide 207 by n

Hence, the quotient of 207 and a number n = \( \frac { 207 }{ n } \)

**Chapter 1 Foundations for Algebra Exercise 1.1 3LC**

Given is to write a phrase for the following algebraic expression:-

**6c**

Clearly, 6c means the product of 6 and a number c

Thus, the phrase for 6c is the product of 6 and a number c

**Chapter 1 Foundations for Algebra Exercise 1.1 4LC**

Given is to write a phrase for the following algebraic expression:-

**x-1**

Clearly, x-1 is the difference of a number x and 1

Thus, the phrase for x-1 is the difference of a number x and 1

**Chapter 1 Foundations for Algebra Exercise 1.1 5LC**

Given is to write a phrase for the following algebraic expression:-

**\( \frac { t }{ 2 } \)**

Clearly, \( \frac { t }{ 2 } \) is the quotient of a number t and 2

Thus, the phrase for \( \frac { t }{ 2 } \) is the quotient of a number t and 2

**Chapter 1 Foundations for Algebra Exercise 1.1 6LC**

Given is to write a phrase for the following algebraic expression:-

**3t – 4**

Firstly, it means the product of 3 and a number t

Then, clearly it means the 4 less than a product of 3 and t

Thus, the phrase for 3t-4 is 4 less than a product of 3 and t

**Chapter 1 Foundations for Algebra Exercise 1.1 7LC**

Given to write a difference between numerical expression and algebraic expression:

**Chapter 1 Foundations for Algebra Exercise 1.1 8LC**

Given is to decide which expression out of given two expressions represent total cost to rent a truck by considering the given below table:

The given expressions are:-

**49n+0.75 or 49+0.75n**

After considering the value of the cost from the above table for different number of miles, it is understood that,

**$49 is constant for any number of miles.**

**$75 kepps on changing with number of miles.**

For n number of miles, the expression will become:

$49+($.75×n)

=$49+($.75×n)

=$49+$.75×n

The above expression for n number of miles can also be written as:

=49+0.75n

Thus, 49+0.75n is the expression that represents total cost to rent a truck that he drives n miles.

**Chapter 1 Foundations for Algebra Exercise 1.1 9E**

Given is to write an algebraic expression for the following phrase:-

4 more than p

It means to add 4 to p

Hence, 4 more than p = 4+p

**Chapter 1 Foundations for Algebra Exercise 1.1 10E**

Given is to write an algebraic expression for the following phrase:-

y minus 12

It means to subtract 12 from y

Hence, y minus 12 = y-12

**Chapter 1 Foundations for Algebra Exercise 1.1 11E**

Given is to write an algebraic expression for the following phrase:-

The quotient of n and 8

It means to divide n by 8

Hence, the quotient of n and 8 = \( \frac { t }{ 2 } \) or n÷8

**Chapter 1 Foundations for Algebra Exercise 1.1 12E**

Given is to write an algebraic expression for the following phrase:-

The product of 15 and c

It means to multiply 15 with c

Hence, The product of 15 and c = 15×c

**Chapter 1 Foundations for Algebra Exercise 1.1 13E**

Given is to write an algebraic expression for the following phrase:-

A number t divided by 82

It means to divide t by 82

Hence, a number t divided by 82 = \( \frac { t }{ 82 } \) or t÷82

**Chapter 1 Foundations for Algebra Exercise 1.1 14E**

Given is to write an algebraic expression for the following phrase:-

The sum of 13 and twice a number h

Firstly, it means multiply 2 with h which is equal to 2×h or 2h

Then, add 13 to this result

Hence, 13 more than twice a number h = 13+2h

**Chapter 1 Foundations for Algebra Exercise 1.1 15E**

Given is to write an algebraic expression for the following phrase:-

6.7 more than the product of 5 and n

Firstly, it means multiply 5 with n, which is equal to 5×n or 5n

Then, add 6.7 to this result

Hence, 6.7 more than the product of 5 and n = 6.7 + 5n

**Chapter 1 Foundations for Algebra Exercise 1.1 16E**

Given is to write an algebraic expression for the following phrase:-

9.85 less than the product of 37 and t

Firstly, it means multiply 37 with t which is equal to 37×t or 37t

Then, subtract 9.85 from this result

Hence, 9.85 less than the product of 37 and t = 37t – 9.85

**Chapter 1 Foundations for Algebra Exercise 1.1 17E**

Given is to write a phrase for the following algebraic expression:-

**q+5**

Clearly, it means the sum of a number q and 5

Thus, the phrase for **q+5** is the sum of a number q and 5

**Chapter 1 Foundations for Algebra Exercise 1.1 18E**

Given is to write a phrase for the following algebraic expression:-

**\( \frac { y }{ 5 } \)**

Clearly, it means the quotient of a number y and 5

Thus, the phrase for \( \frac { y }{ 5 } \) is the quotient of a number y and 5

**Chapter 1 Foundations for Algebra Exercise 1.1 19E**

Given is to write a phrase for the following algebraic expression:-

12x

Clearly, it means the product of 12 and a number x

Thus, the phrase for **12x** is the product of 12 and a number x

**Chapter 1 Foundations for Algebra Exercise 1.1 20E**

Given is to write a phrase for the following algebraic expression:-

**49+m**

Clearly, it means the sum of 49 and a number m

Thus, the phrase for **49+m** is the sum of 49 and a number m

**Chapter 1 Foundations for Algebra Exercise 1.1 21E**

Given is to write a phrase for the following algebraic expression:-

**9n+1**

Firstly, it means the product of 9 and a number n

Then, clearly it means the sum of 1 and 9 times a number n

Thus, the phrase for **9n+1** is the sum of 1 and 9 times a number n

**Chapter 1 Foundations for Algebra Exercise 1.1 22E**

Given is to write a phrase for the following algebraic expression:-

**\( \frac { z }{ 8 } -9 \)**

Firstly, it means the quotient of a number z and 8

Then, clearly it means 9 less than quotient of a number z and 8

Thus, the phrase for \( \frac { z }{ 8 } -9 \) is 9 less than quotient of a number z and 8

**Chapter 1 Foundations for Algebra Exercise 1.1 23E**

Given is to write a phrase for the following algebraic expression:-

**\( 15-\frac { 1\cdot 5 }{ d } \)**

Firstly, it means the quotient of 1.5 and a number d

Then, clearly it means the difference of 15 and the quotient of 1.5 and a number d

So, the phrase for \( 15-\frac { 1\cdot 5 }{ d } \) is the difference of 15 and the quotient of 1.5 and a number d

**Chapter 1 Foundations for Algebra Exercise 1.1 24E**

Given is to write a phrase for the following algebraic expression:-

**2(5-n)**

Firstly, it means the difference of 5 and a number n

Then, clearly it means the product of 2 and difference of 5 and a number n

Thus, the phrase for **2(5-n)** is the product of 2 and difference of 5 and a number n

**Chapter 1 Foundations for Algebra Exercise 1.1 25E**

Given is to write a rule in words and as an algebraic expression to model the relationship as per given below table:-

As per given question, he buys a bicycle and helmet on rent.

He pays **$9** for each hour he uses it and **$5** for helmet.

As per above given table,

**When he uses bicycle for 1 hour, the rental cost will become**

**(Rental cost of bicycle × number of hours) + rental cost helmet**

**where number of hour = 1**

Similarly, when he uses bicycle for 2 hours, the rental cost will become

**(Rental cost of bicycle × number of hours) + rental cost helmet**

**where number of hour = 2**

Next, when he uses bicycle for 3 hours, the rental cost will become

**(Rental cost of bicycle × number of hours) + rental cost helmet**

**where number of hour = 3**

Finally, when he uses bicycle for n hours, the rental cost will become

**(Rental cost of bicycle × number of hours) + rental cost helmet**

**where number of hour = n**

So, the expression for rental cost of bicycle for n number of hours **= ($9×n)+$5**

The expression **($9×n)+$5** in words can be stated as follows:-

Firstly, it means the product of 9 and a number n.

Then, add 5 to this result.

Thus, the phrase for **($9×n)+$5** is 5 more than the product of a number n and 9

**Chapter 1 Foundations for Algebra Exercise 1.1 26E**

Given is to write a rule in words and as an algebraic expression to model the relationship as per given below table:-

As per given question, a salesperson earned a weekly salary of **$150** and also paid **$2** for each pair of shoes he or she sold during the week.

As per above given table,

When he or she sold 5 pair of shoes, the total earned will become

**weekly salary + (extra amount he or she paid for selling number of pair of shoes × number of pair of shoes)**

**where number of pair of shoes sold = 5**

Similarly, when he or she sold 10 pair of shoes, the total earned will become

**weekly salary + (extra amount he or she paid for selling number of pair of shoes × number of pair of shoes)**

**where number of pair of shoes sold = 10**

Next, when he or she sold 15 pair of shoes, the total earned will become

**weekly salary + (extra amount he or she paid for selling number of pair of shoes × number of pair of shoes)**

**where number of pair of shoes sold = 15**

Finally, when he or she sold n pair of shoes, the total earned will become

**where number of pair of shoes sold = n**

So, the expression for total earned when he or she sold n number of pair of shoes

**=$150+($2×n)**

The expression **=$150+($2×n)** in words can be stated as follows:-

Firstly, it means the product of 2 and a number n.

Then, add 150 to this result.

Thus, the phrase for **$150+($2×n)** is 150 more than the product of 2 and a number n

**Chapter 1 Foundations for Algebra Exercise 1.1 27E**

Given is to write an algebraic expression for the following phrase:-

8 minus the product of 9 and r

Firstly, it means multiply 9 with r which is equal to 9×r or 9r

Then, subtract 8 from this result

Hence, 8 minus the product of 9 and r = 9r-8

**Chapter 1 Foundations for Algebra Exercise 1.1 28E**

Given is to write an algebraic expression for the following phrase:-

The sum of 15 and x, plus 7

Firstly, it means add 15 and a number x which is equal to 15+x

Then, add 7 to this result

Hence, the sum of 15 and x, plus 7 = (15+x)+7

**Chapter 1 Foundations for Algebra Exercise 1.1 29E**

Given is to write an algebraic expression for the following phrase:-

4 less than three seventh of y

Firstly, it means multiply a number y with \( \frac { 3 }{ 7 } \) which is equal to \( \frac { 3 }{ 7 } \times y \) or \( \frac { 3 }{ 7 } y \)

Then, subtract 4 from this result

Hence, 4 less than three seventh of \( y=\frac { 3 }{ 7 } y-4 \)

**Chapter 1 Foundations for Algebra Exercise 1.1 30E**

Given is to write an algebraic expression for the following phrase:-

The quotient of 12 and the product of 5 and t

Firstly, it means multiply 5 and a number t which is equal to 5×t or 5t

Then, divide 12 by this result

Hence, the quotient of12 and the product of 5 and \( t=\frac { 12 }{ 5t } \)

**Chapter 1 Foundations for Algebra Exercise 1.1 31E**

Given expression is:-

\( \frac { 5 }{ n} \)

For this expression, word phrase written by a student is:-

The quotient of n and 5

Clearly, the phrase written by him was incorrect as his phrase would represent the division of n and 5 which is given by:-

\( \frac { n }{ 5} \)

So, it is incorrect as per given expression.

Now, the given expression, that is, \( \frac { 5 }{ n} \) represent the division of 5 and n

So, **the correct phrase for the given expression is the quotient of 5 and n**

**Chapter 1 Foundations for Algebra Exercise 1.1 32E**

Given is to write an algebraic expression to give the rule for following statement:-

The number of bagels in any number b of baker’s dozen

Below is the given table showing number of bagels a shop gave him per baker’s dozen:

As per above given table,

When baker’s dozen is 1, the number of bagels will become

13×number of bangels,where number of bangels = 1

Similarly, when baker’s dozen is 2, the number of bagels will become

13×number of bangels,where number of bangels = 2

Next, when baker’s dozen is 3, the number of bagels will become

13×number of bangels,where number of bangels = 3

Finally, when baker’s dozen is n, the number of bagels will become

13×number of bangels,where number of bangels = b

So, the algebraic expression for number of bagels in b number of baker’s dozen = 13b

The pattern of increase in the number of bagels is described as:-

As per given table, the number of bagels increased with the multiple of 13 as

For 1 baker’s dozen, the numbers of bagels are 13 which is 13×1

For 2 baker’s dozen, the numbers of bagels are 26 which is 13×2

For 3 baker’s dozen, the numbers of bagels are 13 which is 13×3

And so on up to for b baker’s dozen, the numbers of bagels are 13b which is 13×b

Thus, **pattern of increase in the number of bagels is multiple of 13**

Operation on b to find the number of bagels is given by:-

Consider the expression to determine the operation on b to calculate the number of bagels 13b

Clearly, the above expression means multiply 13 with the number b to find number of bagels

Thus, **multiplication operation on b is performed to find number of bagels**

**Chapter 1 Foundations for Algebra Exercise 1.1 33E**

Each ticket cost is **$4.50**.

**a.** If n is the number of tickets purchased, Write an expression that gives the total cost of buying **n** tickets.

The expression is **4.50m**.

**b.** Suppose the total cost for n tickets is **$36**.Find the total cost if one more ticket is purchased.

The total cost is **$36 + $4.50 = $40.5**.

**Chapter 1 Foundations for Algebra Exercise 1.1 34E**

Given is to write an algebraic expression for the following statement:-

Number of boxes he will have wrapped when she has wrapped x number of boxes.

She and he wrapped gift boxes at the same pace.

In the first figure, only she wrapped 2 gift boxes.

In the second figure, she has wrapped 3 gift boxes and he has wrapped 1 gift box.

In the third figure, she has wrapped 4 gift boxes and he has wrapped 2 gift boxes.

As per above given statements, the expression will become

**number of boxes that he was wrapped = number of boxes that she has wrapped – 2**

Now, when she has wrapped x number of boxes, thus it means he has wrapped x – 2

Hence, the expression for number of boxes he wrapped when she wrapped x boxes = x – 2

**Chapter 1 Foundations for Algebra Exercise 1.1 35E**

Given is to write an expression that gives the value in dollars of d dimes:-

That is, we want to find the following conversion

**d dimes = ? dollars**

Now, as \(1dime=\frac { 1 }{ 10 } dollars\)

So, for d dimes multiply d on both sides, then expression will become,

\( d\quad dimes=d\times \frac { 1 }{ 10 } dollars \)

d dimes = d × 0.10 dollars

On multiplying, it becomes

d dimes = 0.10d dollars

Thus, the expression that gives the value in dollars of d dimes = 0.10d

Hence, **option A is correct**.

**Chapter 1 Foundations for Algebra Exercise 1.1 36E**

Given is to describe a real world situation and representation of variable for the following expression:

**5t**

The above expression 5t means the product of 5 and t.

If we take the variable t as the cost of 1kg of apples, then by above definition, 5t represents the cost of 5kg of apples.

Thus, in a real world situation 5t represents cost of 5kg of apples and t represents the cost of 1kg of apples

**Chapter 1 Foundations for Algebra Exercise 1.1 37E**

Given is to describe a real world situation and representation of variable for the following expression:

b+3

The above expression b+3 means the 3 more than b.

If the variable b represents as the age of a person X, then by above definition, b+3 represents the age of person Y who is 3 years elder to X.

Thus, in a real world situation b+3 represents age of person Y who is 3 years elder to person X and b represents the age of person X

**Chapter 1 Foundations for Algebra Exercise 1.1 38E**

Given is to describe a real world situation and representation of variable for the following expression:

\( \frac { 40 }{ h} \)

The above expression \( \frac { 40 }{ h} \) means the quotient of 40 and a number h.

If the variable h represents the number of persons, then by above definition, \( \frac { 40 }{ h} \) represents the amount of money received by 1 person when a total money of $40 is distributed among h number of persons.

Thus, in a real world situation \( \frac { 40 }{ h} \) represents the amount of money received by 1 person when a total money of is distributed $40 among h number of persons and h represents number of persons

**Chapter 1 Foundations for Algebra Exercise 1.1 39E**

Consider the following two expressions:

(5-2) ÷ n and (5÷n) – 2 ……(1)

(5÷n) – 2 ……(2)

Consider the following phrase:

2 less than 5 divided by a number n

Firstly, it means to divide 5 by a number n that is 5÷n and then, subtract 2 from the result.

Clearly, expression for the given phrase is as follows:

(5÷n) – 2

So, the expression (2) is a suitable expression for the given phrase.

The given expression (5-2) ÷ n describes as follows:

It means to subtract 2 from 5 and then, divide the result by n.

The expression (1) is not a suitable expression for the given phrase.

Therefore, it can be concluded that,

**The phrase is correctly represented by the expression (5÷n) – 2 and not by (5-2) ÷ n**

Next, is to decide whether verbal description lack precision.

**Yes, verbal description lack precision as they do not provide stress on the order of operations which is needed to be performed in a specific way to get the correct answer.**

**In the above phrase 2 less than 5 divided by a number n, it is not clear whether one has to apply subtraction first and then division or vice-versa.**

**Chapter 1 Foundations for Algebra Exercise 1.1 40E**

Given is to expression that could represent the given diagram:-

The above diagram is the tabular representation of a relationship between number x and 1

The diagram can be represented in two ways by counting the total number of 1’s either column wise (in a vertical direction) or row wise (in a horizontal direction)

Number of columns in the given figure =4

Number of rows in the given figure =3

First, we count the number of 1’s column wise.

Clearly in each column there are three 1’s.

So, counting 1’s in first column =3

As, each column has same number of 1’s and there are 4 columns

Thus, total number of 1’s in the given diagram =3+3+3+3

Adding we get,

Total number of 1’s in the given diagram =12

Secondly, we count the number of 1’s row wise.

Clearly in each row there are four 1’s.

So, counting 1’s in first row =4

As, each row has same number of 1’s and there are 3 rows.

Thus, total number of 1’s in the given diagram =4+4+4

Adding we get,

Total number of 1’s in the given diagram =12

Thus, the two different expressions to represent the given diagram are as follows:

Firstly, 3+3+3+3 and secondly 4+4+4

**Chapter 1 Foundations for Algebra Exercise 1.1 41E**

Given is to expression that could represent the given diagram:-

The above diagram is the tabular representation of a relationship between number x and 1

The diagram can be represented in two ways by counting the total number of 1’s either column wise (in a vertical direction) or row wise (in a horizontal direction)

Number of columns in the first row =4

Number of columns in the second row =2

First, we count the number of 1’s column wise.

In first column, there are two 1’s

In second column, there are two 1’s

In third column, there is only one 1

In fourth column, there is only one 1

So, counting 1’s in first column =2

Counting 1’s in second column =2

Counting 1’s in third column =1

Counting 1’s in fourth column =1

Thus, total number of 1’s in the given diagram =2+2+1+1

Adding we get,

Total number of 1’s in the given diagram =6

Secondly, we count the number of 1’s row wise.

In first row, there are 4 number of 1’s.

In second row, there are 2 number of 1’s.

Thus, total number of 1’s in the given diagram =4+2

Adding we get,

Total number of 1’s in the given diagram =6

Thus, the two different expressions to represent the given diagram are as follows:

Firstly, 2+2+1+1 and secondly 4+2

**Chapter 1 Foundations for Algebra Exercise 1.1 42E**

Given is to write an algebraic expression for the following phrase:-

2 less than the product of 3 and a number x

Firstly, it means to multiply 3 with x which is equal to 3×x or 3x

Then, subtract 2 from this result

Thus, 2 less than the product of 3 and a number x = 3x-2

Hence, **option A is correct**

**Chapter 1 Foundations for Algebra Exercise 1.1 43E**

Given is to write a phrase for the following algebraic expression:-

n÷8

We can also write above expression as \( \frac { n }{ 8} \)

Clearly, it means the quotient of a number n and 8

Thus, phrase for n÷8 is the quotient of a number n and 8

Hence, **option G is correct**.

**Chapter 1 Foundations for Algebra Exercise 1.1 44E**

Given is to write an algebraic expression for the following phrase:-

A state park charges are entrance fees + $18 for each night of camping

where an entrance fee is $20

Below is the given table showing the relationship for an entrance fees and charges for each night:-

By considering the above table, the expression for total cost of n nights of camping is:-

$18×n+$20

=$18n+$20

Or we can write an expression as 18n+20

Hence, **option B is correct**

**Chapter 1 Foundations for Algebra Exercise 1.1 45E**

Given expression is:

\( \frac { 1 }{ 4 } +\frac { 1 }{ 2 } \) …… (1)

First, we find LCD of the denominators of all above fractions

Listing all denominators: 4 and 2

Prime factors of 4

=2×2

Prime factors of 2 = 2×1

So, LCD of 4 and 2 =2×2

Thus, LCD of \( \frac { 1 }{ 4 } \) , \( \frac { 1 }{ 2 } \) is 4

Next, we convert each of these fraction to an equivalent fraction with the denominator same as their LCD, that is, 4.

\( \frac { 1 }{ 4 } \) is already in the equivalent form.

Next, to convert \( \frac { 1 }{ 2 } \)

put \( \frac { 1 }{ 2 } =\frac { ? }{ 4 } \)

To find the missing numerator, we observe the relation between denominators that is 2 and 4.

Clearly, \( \frac { 1 }{ 2 } =\frac { 1\times 2 }{ 2\times 2 } \)

\( =\frac { 2 }{ 4 } \)

Now writing the equivalent form of both the fractions in above expression (1), it becomes

\( \frac { 1 }{ 4 } +\frac { 1 }{ 2 } =\frac { 1 }{ 4 } +\frac { 2 }{ 4 } \)

Now, both the above fractions have same denominator that is 4. So they all are like fractions and to combine them, we just combine their numerators, keeping denominator same

\( \frac { 1 }{ 4 } +\frac { 1 }{ 2 } =\frac { 1+2 }{ 4 } \)

\( \frac { 1 }{ 4 } +\frac { 1 }{ 2 } =\frac { 3 }{ 4 } \)

Making prime factors of numerator and denominator, it becomes

\( \frac { 3\times 1 }{ 2\times 2 } \)

Cancelling the same factors in numerator and denominator, as it has no same factors then it becomes

\( =\frac { 3 }{ 4 } \)

Hence,

\( \frac { 1 }{ 4 } +\frac { 1 }{ 2 } =\frac { 3 }{ 4 } \)

**Chapter 1 Foundations for Algebra Exercise 1.1 46E**

Given expression is:

\( \frac { 9 }{ 14 } -\frac { 2 }{ 7 } \) …… (1)

First, we find LCD of the denominators of both the above fractions

Prime factors of 14 = 2×7

Prime factors of 7 = 7×1

So, LCD of \( \frac { 9 }{ 14 } \) and \( \frac { 2 }{ 7 } \) = 2×7 =14

Next, we convert each of these fraction to an equivalent fraction with the denominator same as their LCD, that is 14.

\( \frac { 9 }{ 14 } \) is already in the equivalent form.

Next, to convert \( \frac { 2 }{ 7 } \), put \( \frac { 2 }{ 7 } =\frac { ? }{ 14 } \)

Clearly, \( \frac { 2 }{ 7 } =\frac { 2\times 2 }{ 7\times 2 } \)

\( =\frac { 4 }{ 14 } \)

Now writing the equivalent form of both the fractions in above expression (1), it becomes

\( \frac { 9 }{ 14 } -\frac { 2 }{ 7 } =\frac { 9 }{ 14 } -\frac { 4 }{ 14 } \)

Now, both the above fractions have same denominator that is 14. So they both are like fractions and to combine them, we just combine their numerators, keeping the denominator same

\( \frac { 9 }{ 14 } -\frac { 2 }{ 7 } =\frac { 9-4 }{ 14 } \)

On subtraction the numerator, it becomes

\( =\frac { 5 }{ 14 } \)

Hence,

\( \frac { 9 }{ 14 } -\frac { 2 }{ 7 } =\frac { 5 }{ 14 } \)

**Chapter 1 Foundations for Algebra Exercise 1.1 47E**

Given expression is:

\( \frac { 2 }{ 5 } +\frac { 3 }{ 10 } \) …… (1)

First, we find LCD of the denominators of all above fractions

Listing all denominators: 5 and 10

Prime factors of 5 = 5×1

Prime factors of 10 = 2×5

So, LCD of 5 and 10 = 2×5 = 10

Thus, LCD of \( \frac { 2 }{ 5 } \), \( \frac { 3 }{ 10 } \) is 10

Next, we convert each of these fraction to an equivalent fraction with the denominator same as their LCD, that is, 10.

To convert \( \frac { 2 }{ 5 } \), put \( \frac { 2 }{ 5 } =\frac { ? }{ 10 } \)

To find the missing numerator, we observe the relation between denominators that is 5 and 10.

Clearly, \( \frac { 2 }{ 5 } =\frac { 2\times 2 }{ 5\times 2 } \)

Next, \( \frac { 3 }{ 10 } \) is already in the equivalent form.

Now writing the equivalent form of two fractions in above expression (1), it becomes

\( \frac { 2 }{ 5 } +\frac { 3 }{ 10 } =\frac { 4 }{ 10 } +\frac { 3 }{ 10 } \)

Now, both the above fractions have same denominator that is 10. So they all are like fractions and to combine them, we just combine their numerators, keeping denominator same

\( \frac { 2 }{ 5 } +\frac { 3 }{ 10 } =\frac { 4+3 }{ 10 } \)

On adding the terms of numerator, it becomes

\( =\frac { 7 }{ 10 } \)

Making prime factors of numerator and denominator, it becomes

\( =\frac { 7\times 1 }{ 2\times 5 } \)

Cancelling the same factors in numerator and denominator, as there is no same factor then, it becomes

\( =\frac { 7 }{ 10 } \)

Hence,

\( \frac { 2 }{ 5 } +\frac { 3 }{ 10 } =\frac { 7 }{ 10 } \)

**Chapter 1 Foundations for Algebra Exercise 1.1 48E**

Given expression is:

\( \frac { 5 }{ 6 } -\frac { 2 }{ 3 } \) …… (1)

First, we find LCD of the denominators of both the above fractions

Prime factors of 6 = 3×2

Prime factors of 3 = 3×1

So, LCD of \( \frac { 5 }{ 6 } \) and \( \frac { 2 }{ 3 } \) = 2×3×1 = 6

Next, we convert each of these fraction to an equivalent fraction with the denominator same as their LCD, that is 6.

\( \frac { 5 }{ 6 } \) is already in equivalent form.

Next, to convert \( \frac { 2 }{ 3 } \), put \( \frac { 2 }{ 3 } =\frac { ? }{ 6 } \)

Clearly, \( \frac { 2 }{ 3 } =\frac { 2\times 2 }{ 3\times 2 } \)

\( =\frac { 4 }{ 6 } \)

Now writing the equivalent form of both the fractions in above expression (1), it becomes

\( \frac { 5 }{ 6 } -\frac { 2 }{ 3 } =\frac { 5 }{ 6 } -\frac { 4 }{ 6 } \)

Now, both the above fractions have same denominator that is 6. So they both are like fractions and to combine them, we just combine their numerators, keeping denominator same

\( =\frac { 5-4 }{ 6 } \)

\( =\frac { 1}{ 6 } \)

Hence,

\( \frac { 5 }{ 6 } -\frac { 2 }{ 3 } =\frac { 1 }{ 6 } \)

**Chapter 1 Foundations for Algebra Exercise 1.1 49E**

Given numbers are 3 and 6

Prime factors of 3 = 1×3

Prime factors of 6 = 2×3

By definition of greatest common factor of two whole numbers, it contains each prime factor occurring that number of times equals to the minimum number of times it appears in the each of the whole numbers. Clearly, 3 just occurs once in both the whole number but 2 does not occurs once in 3

Then, greatest common factor of 3 and 6 =3

Hence, **greatest common factor of 3 and 6 is 3**

**Chapter 1 Foundations for Algebra Exercise 1.1 50E**

Given numbers are 12 and 15

Prime factors of 12 = 2×2×3

Prime factors of 15 = 3×5

By definition of greatest common factor of two whole numbers, it contains each prime factor occurring that number of times equals to the minimum number of times it appears in the each of the whole numbers. Clearly, 3 just occur once in both the whole number but 2 and 5 do not occur once in them.

Then, greatest common factor of 12 and 15 = 3

Hence, **greatest common factor of 12 and 15 is 3**

**Chapter 1 Foundations for Algebra Exercise 1.1 51E**

Given numbers are 7 and 11

Prime factors of 7 = 1×7

Prime factors of 11 = 1×11

By definition of greatest common factor of two whole numbers, it contains each prime factor occurring that number of times equals to the minimum number of times it appears in the each of the whole numbers. Clearly, 1 just occurs once in both the whole number but 7 and 11 does not occur once in them.

Then, greatest common factor of 7 and 11 = 1

Hence, **greatest common factor of 7 and 11 is 1**

**Chapter 1 Foundations for Algebra Exercise 1.1 52E**

Given numbers are 12 and 8

Prime factors of 12 = 2×2×3

Prime factors of 8 = 2×2×2

By definition of greatest common factor of two whole numbers, it contains each prime factor occurring that number of times equals to the minimum number of times it appears in the each of the whole numbers. Clearly, 2 occur twice in both the whole numbers but 3 do not occur once in them.

Then, greatest common factor of 12 and 8 = 2×2 = 4

Hence, **greatest common factor of 12 and 8 is 4**