NCERT Textbook: Linear Equations in One Variable

 Page 1


? ???????? ? ? ? ? ? ? ? ? ? ?????????????? ? ? ? ?
2.1  Introduction
In the earlier classes, you have come across several algebraic expressions and equations.
Some examples of expressions we have so far worked with are:
5x, 2x – 3, 3x + y, 2xy + 5, xyz + x + y + z, x
2
 + 1, y + y
2
Some examples of equations are: 5x = 25, 2x – 3 = 9, 
5 37
2 , 6 10 2
22
yz + = + =-
You would remember that equations use the equality (=) sign; it is missing in expressions.
Of these given expressions, many have more than one variable. For example, 2 xy + 5
has two variables. We however, restrict to expressions with only one variable when we
form equations. Moreover, the expressions we use to form equations are linear . This means
that the highest power of the variable appearing in the expression is 1.
These are linear expressions:
2x, 2x + 1, 3y – 7, 12 – 5z, 
5
( – 4) 10
4
x +
These are not linear expressions:
x
2
 + 1, y + y
2
, 1 + z + z
2
 + z
3
(since highest power of variable > 1)
Here we will deal with equations with linear expressions in one variable only. Such
equations are known as linear equations in one variable. The simple equations which
you studied in the earlier classes were all of this type.
Let us briefly revise what we know:
(a) An algebraic equation is an equality
involving variables. It  has an equality sign.
The expression on the left of the equality sign
is the Left Hand Side (LHS). The expression
on the right of the equality sign is the Right
Hand Side (RHS).
? ? ? ? ? ? ?
?
??????? ?????????? ??
? ? ? ? ? ? ? ? ? ? ? ?
2x – 3 = 7
2x – 3 = LHS
7 = RHS
Reprint 2024-25
Page 2


? ???????? ? ? ? ? ? ? ? ? ? ?????????????? ? ? ? ?
2.1  Introduction
In the earlier classes, you have come across several algebraic expressions and equations.
Some examples of expressions we have so far worked with are:
5x, 2x – 3, 3x + y, 2xy + 5, xyz + x + y + z, x
2
 + 1, y + y
2
Some examples of equations are: 5x = 25, 2x – 3 = 9, 
5 37
2 , 6 10 2
22
yz + = + =-
You would remember that equations use the equality (=) sign; it is missing in expressions.
Of these given expressions, many have more than one variable. For example, 2 xy + 5
has two variables. We however, restrict to expressions with only one variable when we
form equations. Moreover, the expressions we use to form equations are linear . This means
that the highest power of the variable appearing in the expression is 1.
These are linear expressions:
2x, 2x + 1, 3y – 7, 12 – 5z, 
5
( – 4) 10
4
x +
These are not linear expressions:
x
2
 + 1, y + y
2
, 1 + z + z
2
 + z
3
(since highest power of variable > 1)
Here we will deal with equations with linear expressions in one variable only. Such
equations are known as linear equations in one variable. The simple equations which
you studied in the earlier classes were all of this type.
Let us briefly revise what we know:
(a) An algebraic equation is an equality
involving variables. It  has an equality sign.
The expression on the left of the equality sign
is the Left Hand Side (LHS). The expression
on the right of the equality sign is the Right
Hand Side (RHS).
? ? ? ? ? ? ?
?
??????? ?????????? ??
? ? ? ? ? ? ? ? ? ? ? ?
2x – 3 = 7
2x – 3 = LHS
7 = RHS
Reprint 2024-25
??? ? ? ? ????? ????
(b) In an equation the values of
the expressions on the LHS
and RHS are equal. This
happens to be true only for
certain values of the variable.
These values are the
solutions of the equation.
(c) How to find the solution of an equation?
We assume that the two sides of the equation are balanced.
We perform the same mathematical operations on both
sides of the equation, so that the balance is not disturbed.
A few such steps give the solution.
2.2 Solving Equations having the Variable on
both Sides
An equation is the equality of the values of two expressions. In the equation 2x – 3 = 7,
the two expressions are  2x – 3 and 7.  In most examples that we have come across so
far, the RHS is just a number.  But this need not always be so; both sides could have
expressions with variables.  For example, the equation  2x – 3 = x + 2 has expressions
with a variable on both sides; the expression on the LHS is (2x – 3) and the expression
on the RHS is (x + 2).
• We now discuss how to solve such equations which have expressions with the variable
on both sides.
Example 1: Solve 2x – 3 = x + 2
Solution: We have
2x = x + 2 + 3
or 2x = x + 5
or 2x – x = x + 5 – x (subtracting x from both sides)
or x = 5 (solution)
Here we subtracted from both sides of the equation, not a number (constant), but a
term involving the variable. We can do this as variables are also numbers. Also, note that
subtracting x from both sides amounts to transposing x to LHS.
Example 2: Solve 5x + 
73
14
22
x =-
Solution: Multiply both sides of the equation by 2. We get
25
7
2
×+
?
?
?
?
?
?
x
 =
2
3
2
14 ×-
?
?
?
?
?
?
x
x = 5 is the solution of the equation
2x – 3 = 7. For  x = 5,
LHS = 2 × 5 – 3 = 7 = RHS
On the other hand x = 10 is not a solution of the
equation. For x = 10,  LHS = 2 × 10 –3 = 17.
This is not equal to the RHS
Reprint 2024-25
Page 3


? ???????? ? ? ? ? ? ? ? ? ? ?????????????? ? ? ? ?
2.1  Introduction
In the earlier classes, you have come across several algebraic expressions and equations.
Some examples of expressions we have so far worked with are:
5x, 2x – 3, 3x + y, 2xy + 5, xyz + x + y + z, x
2
 + 1, y + y
2
Some examples of equations are: 5x = 25, 2x – 3 = 9, 
5 37
2 , 6 10 2
22
yz + = + =-
You would remember that equations use the equality (=) sign; it is missing in expressions.
Of these given expressions, many have more than one variable. For example, 2 xy + 5
has two variables. We however, restrict to expressions with only one variable when we
form equations. Moreover, the expressions we use to form equations are linear . This means
that the highest power of the variable appearing in the expression is 1.
These are linear expressions:
2x, 2x + 1, 3y – 7, 12 – 5z, 
5
( – 4) 10
4
x +
These are not linear expressions:
x
2
 + 1, y + y
2
, 1 + z + z
2
 + z
3
(since highest power of variable > 1)
Here we will deal with equations with linear expressions in one variable only. Such
equations are known as linear equations in one variable. The simple equations which
you studied in the earlier classes were all of this type.
Let us briefly revise what we know:
(a) An algebraic equation is an equality
involving variables. It  has an equality sign.
The expression on the left of the equality sign
is the Left Hand Side (LHS). The expression
on the right of the equality sign is the Right
Hand Side (RHS).
? ? ? ? ? ? ?
?
??????? ?????????? ??
? ? ? ? ? ? ? ? ? ? ? ?
2x – 3 = 7
2x – 3 = LHS
7 = RHS
Reprint 2024-25
??? ? ? ? ????? ????
(b) In an equation the values of
the expressions on the LHS
and RHS are equal. This
happens to be true only for
certain values of the variable.
These values are the
solutions of the equation.
(c) How to find the solution of an equation?
We assume that the two sides of the equation are balanced.
We perform the same mathematical operations on both
sides of the equation, so that the balance is not disturbed.
A few such steps give the solution.
2.2 Solving Equations having the Variable on
both Sides
An equation is the equality of the values of two expressions. In the equation 2x – 3 = 7,
the two expressions are  2x – 3 and 7.  In most examples that we have come across so
far, the RHS is just a number.  But this need not always be so; both sides could have
expressions with variables.  For example, the equation  2x – 3 = x + 2 has expressions
with a variable on both sides; the expression on the LHS is (2x – 3) and the expression
on the RHS is (x + 2).
• We now discuss how to solve such equations which have expressions with the variable
on both sides.
Example 1: Solve 2x – 3 = x + 2
Solution: We have
2x = x + 2 + 3
or 2x = x + 5
or 2x – x = x + 5 – x (subtracting x from both sides)
or x = 5 (solution)
Here we subtracted from both sides of the equation, not a number (constant), but a
term involving the variable. We can do this as variables are also numbers. Also, note that
subtracting x from both sides amounts to transposing x to LHS.
Example 2: Solve 5x + 
73
14
22
x =-
Solution: Multiply both sides of the equation by 2. We get
25
7
2
×+
?
?
?
?
?
?
x
 =
2
3
2
14 ×-
?
?
?
?
?
?
x
x = 5 is the solution of the equation
2x – 3 = 7. For  x = 5,
LHS = 2 × 5 – 3 = 7 = RHS
On the other hand x = 10 is not a solution of the
equation. For x = 10,  LHS = 2 × 10 –3 = 17.
This is not equal to the RHS
Reprint 2024-25
? ???????? ? ? ? ? ? ? ? ? ? ?????????????? ? ? ? ?
Why 6? Because it is the
smallest multiple (or LCM)
of the given denominators.
(2 × 5x) + 
2
7
2
×
?
?
?
?
?
? = 2
3
2
2 14 ×
?
?
?
?
?
?
-× x ()
or 10x + 7 = 3x – 28
or 10x – 3x + 7 = – 28 (transposing 3x to LHS)
or 7x + 7 = – 28
or 7x = – 28 – 7
or 7x = – 35
or x =
35
7
-
or x = – 5 (solution)
EXERCISE 2.1
Solve the following equations and check your results.
1. 3x = 2x + 18 2. 5t – 3 = 3t – 5 3. 5x + 9 = 5 + 3x
4. 4z + 3 = 6 + 2z 5. 2x – 1 = 14 – x 6. 8x + 4 = 3 (x – 1) + 7
7. x = 
4
5
 (x + 10) 8.
2
3
x
 + 1 = 
7
3
15
x
+
9. 2y + 
5
3
 = 
26
3
y -
10. 3m = 5 m – 
8
5
2.3  Reducing Equations to Simpler Form
Example 16: Solve 
61 3
1
36
xx +-
+=
Solution: Multiplying both sides of the equation by 6,
6 (6 1)
61
3
x+
+×
 =
6( 3)
6
x-
or 2 (6x + 1) + 6 = x – 3
or 12x + 2 + 6 = x – 3 (opening the brackets )
or 12x + 8 = x – 3
or 12x – x + 8 = – 3
or 11x + 8 = – 3
or 11x = –3 – 8
or 11x = –11
or x = – 1 (required solution)
Reprint 2024-25
Page 4


? ???????? ? ? ? ? ? ? ? ? ? ?????????????? ? ? ? ?
2.1  Introduction
In the earlier classes, you have come across several algebraic expressions and equations.
Some examples of expressions we have so far worked with are:
5x, 2x – 3, 3x + y, 2xy + 5, xyz + x + y + z, x
2
 + 1, y + y
2
Some examples of equations are: 5x = 25, 2x – 3 = 9, 
5 37
2 , 6 10 2
22
yz + = + =-
You would remember that equations use the equality (=) sign; it is missing in expressions.
Of these given expressions, many have more than one variable. For example, 2 xy + 5
has two variables. We however, restrict to expressions with only one variable when we
form equations. Moreover, the expressions we use to form equations are linear . This means
that the highest power of the variable appearing in the expression is 1.
These are linear expressions:
2x, 2x + 1, 3y – 7, 12 – 5z, 
5
( – 4) 10
4
x +
These are not linear expressions:
x
2
 + 1, y + y
2
, 1 + z + z
2
 + z
3
(since highest power of variable > 1)
Here we will deal with equations with linear expressions in one variable only. Such
equations are known as linear equations in one variable. The simple equations which
you studied in the earlier classes were all of this type.
Let us briefly revise what we know:
(a) An algebraic equation is an equality
involving variables. It  has an equality sign.
The expression on the left of the equality sign
is the Left Hand Side (LHS). The expression
on the right of the equality sign is the Right
Hand Side (RHS).
? ? ? ? ? ? ?
?
??????? ?????????? ??
? ? ? ? ? ? ? ? ? ? ? ?
2x – 3 = 7
2x – 3 = LHS
7 = RHS
Reprint 2024-25
??? ? ? ? ????? ????
(b) In an equation the values of
the expressions on the LHS
and RHS are equal. This
happens to be true only for
certain values of the variable.
These values are the
solutions of the equation.
(c) How to find the solution of an equation?
We assume that the two sides of the equation are balanced.
We perform the same mathematical operations on both
sides of the equation, so that the balance is not disturbed.
A few such steps give the solution.
2.2 Solving Equations having the Variable on
both Sides
An equation is the equality of the values of two expressions. In the equation 2x – 3 = 7,
the two expressions are  2x – 3 and 7.  In most examples that we have come across so
far, the RHS is just a number.  But this need not always be so; both sides could have
expressions with variables.  For example, the equation  2x – 3 = x + 2 has expressions
with a variable on both sides; the expression on the LHS is (2x – 3) and the expression
on the RHS is (x + 2).
• We now discuss how to solve such equations which have expressions with the variable
on both sides.
Example 1: Solve 2x – 3 = x + 2
Solution: We have
2x = x + 2 + 3
or 2x = x + 5
or 2x – x = x + 5 – x (subtracting x from both sides)
or x = 5 (solution)
Here we subtracted from both sides of the equation, not a number (constant), but a
term involving the variable. We can do this as variables are also numbers. Also, note that
subtracting x from both sides amounts to transposing x to LHS.
Example 2: Solve 5x + 
73
14
22
x =-
Solution: Multiply both sides of the equation by 2. We get
25
7
2
×+
?
?
?
?
?
?
x
 =
2
3
2
14 ×-
?
?
?
?
?
?
x
x = 5 is the solution of the equation
2x – 3 = 7. For  x = 5,
LHS = 2 × 5 – 3 = 7 = RHS
On the other hand x = 10 is not a solution of the
equation. For x = 10,  LHS = 2 × 10 –3 = 17.
This is not equal to the RHS
Reprint 2024-25
? ???????? ? ? ? ? ? ? ? ? ? ?????????????? ? ? ? ?
Why 6? Because it is the
smallest multiple (or LCM)
of the given denominators.
(2 × 5x) + 
2
7
2
×
?
?
?
?
?
? = 2
3
2
2 14 ×
?
?
?
?
?
?
-× x ()
or 10x + 7 = 3x – 28
or 10x – 3x + 7 = – 28 (transposing 3x to LHS)
or 7x + 7 = – 28
or 7x = – 28 – 7
or 7x = – 35
or x =
35
7
-
or x = – 5 (solution)
EXERCISE 2.1
Solve the following equations and check your results.
1. 3x = 2x + 18 2. 5t – 3 = 3t – 5 3. 5x + 9 = 5 + 3x
4. 4z + 3 = 6 + 2z 5. 2x – 1 = 14 – x 6. 8x + 4 = 3 (x – 1) + 7
7. x = 
4
5
 (x + 10) 8.
2
3
x
 + 1 = 
7
3
15
x
+
9. 2y + 
5
3
 = 
26
3
y -
10. 3m = 5 m – 
8
5
2.3  Reducing Equations to Simpler Form
Example 16: Solve 
61 3
1
36
xx +-
+=
Solution: Multiplying both sides of the equation by 6,
6 (6 1)
61
3
x+
+×
 =
6( 3)
6
x-
or 2 (6x + 1) + 6 = x – 3
or 12x + 2 + 6 = x – 3 (opening the brackets )
or 12x + 8 = x – 3
or 12x – x + 8 = – 3
or 11x + 8 = – 3
or 11x = –3 – 8
or 11x = –11
or x = – 1 (required solution)
Reprint 2024-25
??? ? ? ? ????? ????
Check: LHS =
6( 1) 1 6 1
11
33
- + -+
+= +
 = 
5 35 32
33 3 3
- -+ -
+= =
RHS =
( 1) 3 4 2
6 63
-- - -
==
LHS = RHS. (as required)
Example 17: Solve  5x – 2 (2x – 7)  =  2 (3x – 1) + 
7
2
Solution:  Let us open the brackets,
LHS = 5x – 4x + 14  = x + 14
RHS = 6x – 2 + 
7
2
 =
47 3
66
22 2
xx -+ = +
The equation is x + 14 =6x + 
3
2
or 14 = 6x – x + 
3
2
or 14 = 5x + 
3
2
or 14 – 
3
2
 = 5x (transposing 
3
2
)
or
28 3
2
-
 = 5x
or
25
2
 = 5x
or x =
25 1 5 5 5
2 52 52
×
×= =
×
Therefore, required solution is x = 
5
2
.
Check: LHS = 
=
25 25 25
2(5 7) 2( 2) 4
2 22
- - = - -= +
 = 
25 8 33
22
+
=
RHS =
=
26 7 33
22
+
= = LHS.   (as required)
Did you observe how we
simplified the form of the given
equation? Here, we had to
multiply both sides of the
equation by the LCM of the
denominators of the terms in the
expressions of the equation.
Note, in this example we
brought the equation to a
simpler form by opening
brackets and combining like
terms on both sides of the
equation.
Reprint 2024-25
Page 5


? ???????? ? ? ? ? ? ? ? ? ? ?????????????? ? ? ? ?
2.1  Introduction
In the earlier classes, you have come across several algebraic expressions and equations.
Some examples of expressions we have so far worked with are:
5x, 2x – 3, 3x + y, 2xy + 5, xyz + x + y + z, x
2
 + 1, y + y
2
Some examples of equations are: 5x = 25, 2x – 3 = 9, 
5 37
2 , 6 10 2
22
yz + = + =-
You would remember that equations use the equality (=) sign; it is missing in expressions.
Of these given expressions, many have more than one variable. For example, 2 xy + 5
has two variables. We however, restrict to expressions with only one variable when we
form equations. Moreover, the expressions we use to form equations are linear . This means
that the highest power of the variable appearing in the expression is 1.
These are linear expressions:
2x, 2x + 1, 3y – 7, 12 – 5z, 
5
( – 4) 10
4
x +
These are not linear expressions:
x
2
 + 1, y + y
2
, 1 + z + z
2
 + z
3
(since highest power of variable > 1)
Here we will deal with equations with linear expressions in one variable only. Such
equations are known as linear equations in one variable. The simple equations which
you studied in the earlier classes were all of this type.
Let us briefly revise what we know:
(a) An algebraic equation is an equality
involving variables. It  has an equality sign.
The expression on the left of the equality sign
is the Left Hand Side (LHS). The expression
on the right of the equality sign is the Right
Hand Side (RHS).
? ? ? ? ? ? ?
?
??????? ?????????? ??
? ? ? ? ? ? ? ? ? ? ? ?
2x – 3 = 7
2x – 3 = LHS
7 = RHS
Reprint 2024-25
??? ? ? ? ????? ????
(b) In an equation the values of
the expressions on the LHS
and RHS are equal. This
happens to be true only for
certain values of the variable.
These values are the
solutions of the equation.
(c) How to find the solution of an equation?
We assume that the two sides of the equation are balanced.
We perform the same mathematical operations on both
sides of the equation, so that the balance is not disturbed.
A few such steps give the solution.
2.2 Solving Equations having the Variable on
both Sides
An equation is the equality of the values of two expressions. In the equation 2x – 3 = 7,
the two expressions are  2x – 3 and 7.  In most examples that we have come across so
far, the RHS is just a number.  But this need not always be so; both sides could have
expressions with variables.  For example, the equation  2x – 3 = x + 2 has expressions
with a variable on both sides; the expression on the LHS is (2x – 3) and the expression
on the RHS is (x + 2).
• We now discuss how to solve such equations which have expressions with the variable
on both sides.
Example 1: Solve 2x – 3 = x + 2
Solution: We have
2x = x + 2 + 3
or 2x = x + 5
or 2x – x = x + 5 – x (subtracting x from both sides)
or x = 5 (solution)
Here we subtracted from both sides of the equation, not a number (constant), but a
term involving the variable. We can do this as variables are also numbers. Also, note that
subtracting x from both sides amounts to transposing x to LHS.
Example 2: Solve 5x + 
73
14
22
x =-
Solution: Multiply both sides of the equation by 2. We get
25
7
2
×+
?
?
?
?
?
?
x
 =
2
3
2
14 ×-
?
?
?
?
?
?
x
x = 5 is the solution of the equation
2x – 3 = 7. For  x = 5,
LHS = 2 × 5 – 3 = 7 = RHS
On the other hand x = 10 is not a solution of the
equation. For x = 10,  LHS = 2 × 10 –3 = 17.
This is not equal to the RHS
Reprint 2024-25
? ???????? ? ? ? ? ? ? ? ? ? ?????????????? ? ? ? ?
Why 6? Because it is the
smallest multiple (or LCM)
of the given denominators.
(2 × 5x) + 
2
7
2
×
?
?
?
?
?
? = 2
3
2
2 14 ×
?
?
?
?
?
?
-× x ()
or 10x + 7 = 3x – 28
or 10x – 3x + 7 = – 28 (transposing 3x to LHS)
or 7x + 7 = – 28
or 7x = – 28 – 7
or 7x = – 35
or x =
35
7
-
or x = – 5 (solution)
EXERCISE 2.1
Solve the following equations and check your results.
1. 3x = 2x + 18 2. 5t – 3 = 3t – 5 3. 5x + 9 = 5 + 3x
4. 4z + 3 = 6 + 2z 5. 2x – 1 = 14 – x 6. 8x + 4 = 3 (x – 1) + 7
7. x = 
4
5
 (x + 10) 8.
2
3
x
 + 1 = 
7
3
15
x
+
9. 2y + 
5
3
 = 
26
3
y -
10. 3m = 5 m – 
8
5
2.3  Reducing Equations to Simpler Form
Example 16: Solve 
61 3
1
36
xx +-
+=
Solution: Multiplying both sides of the equation by 6,
6 (6 1)
61
3
x+
+×
 =
6( 3)
6
x-
or 2 (6x + 1) + 6 = x – 3
or 12x + 2 + 6 = x – 3 (opening the brackets )
or 12x + 8 = x – 3
or 12x – x + 8 = – 3
or 11x + 8 = – 3
or 11x = –3 – 8
or 11x = –11
or x = – 1 (required solution)
Reprint 2024-25
??? ? ? ? ????? ????
Check: LHS =
6( 1) 1 6 1
11
33
- + -+
+= +
 = 
5 35 32
33 3 3
- -+ -
+= =
RHS =
( 1) 3 4 2
6 63
-- - -
==
LHS = RHS. (as required)
Example 17: Solve  5x – 2 (2x – 7)  =  2 (3x – 1) + 
7
2
Solution:  Let us open the brackets,
LHS = 5x – 4x + 14  = x + 14
RHS = 6x – 2 + 
7
2
 =
47 3
66
22 2
xx -+ = +
The equation is x + 14 =6x + 
3
2
or 14 = 6x – x + 
3
2
or 14 = 5x + 
3
2
or 14 – 
3
2
 = 5x (transposing 
3
2
)
or
28 3
2
-
 = 5x
or
25
2
 = 5x
or x =
25 1 5 5 5
2 52 52
×
×= =
×
Therefore, required solution is x = 
5
2
.
Check: LHS = 
=
25 25 25
2(5 7) 2( 2) 4
2 22
- - = - -= +
 = 
25 8 33
22
+
=
RHS =
=
26 7 33
22
+
= = LHS.   (as required)
Did you observe how we
simplified the form of the given
equation? Here, we had to
multiply both sides of the
equation by the LCM of the
denominators of the terms in the
expressions of the equation.
Note, in this example we
brought the equation to a
simpler form by opening
brackets and combining like
terms on both sides of the
equation.
Reprint 2024-25
? ???????? ? ? ? ? ? ? ? ? ? ?????????????? ? ? ? ?
WHAT HAVE WE DISCUSSED?
1. An algebraic equation is an equality involving variables. It says that the value of the expression on
one side of the equality sign is equal to the value of the expression on the other side.
2. The equations we study in Classes VI, VII and VIII are linear equations in one variable. In such
equations, the expressions which form the equation contain only one variable. Further, the equations
are linear, i.e., the highest power of the variable appearing in the equation is 1.
3. An equation may have linear expressions on both sides. Equations that we studied in Classes VI
and VII had just a number on one side of the equation.
4. Just as numbers, variables can, also, be transposed from one side of the equation to the other .
5. Occasionally, the expressions forming equations have to be simplified before we can solve them
by usual methods. Some equations may not even be linear to begin with, but they can be brought
to a linear form by multiplying both sides of the equation by a suitable expression.
6. The utility of linear equations is in their diverse applications; different problems on numbers, ages,
perimeters, combination of currency notes, and so on can be solved using linear equations.
EXERCISE 2.2
Solve the following linear equations.
1.
11
25 3 4
xx
-= +
2.
35
21
24 6
nn n
-+=
3.
8 17 5
7
362
xx
x+- = -
4.
53
35
xx --
=
5.
32 23 2
4 33
tt
t
-+
- =-
6.
12
1
23
mm
m
--
- =-
Simplify and solve the following linear equations.
7. 3(t – 3) = 5(2t + 1) 8. 15(y – 4) –2(y – 9) + 5(y + 6) = 0
9. 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
10. 0.25(4f – 3) = 0.05(10f – 9)
Reprint 2024-25