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# Solving Equations Involving Addition and Multiplication

• Here, we will learn to solve equations involving two operations i.e. addition and multiplication.
• We will use inverse operations to solve equations.

For example:

Solve for $$a;$$

$$3a+6 = 9$$

Here, we have to remove $$6$$ and $$3$$ from the given equation.

We know that the inverse operation of addition is subtraction.

Thus, we will subtract $$6$$ from both the sides.

$$3a+ \not{6}- \not{6} = 9 -6$$

$$\Rightarrow 3a = 3$$

We know that the inverse operation of multiplication is division.

Thus, we will divide by $$3$$ on both the sides to get only the variable on one side.

$$\dfrac{ \not{3}a}{ \not{3}}=\dfrac{ \not{3}}{ \not{3}}$$

$$\Rightarrow a=1$$

#### Solve for $$y$$ :  $$4y+5 = 29$$

A $$29$$

B $$4$$

C $$5$$

D $$6$$

×

Given equation: $$4y + 5 = 29$$

Here, we will use inverse operations to remove $$4$$ and $$5$$ from the given equation.

First, we will subtract $$5$$ from both the sides.

[Inverse operation of addition is subtraction]

$$4y\, + \not{5}\,-\not{5}= 29-5$$

$$\Rightarrow 4y = 24$$

We will divide by $$4$$ on both the sides to get only the variable on one side.

[Inverse operation of multiplication is division]

$$\dfrac{ \not{4}y}{ \not{4}} = \dfrac{24}{4}$$

$$\Rightarrow y = 6$$

Hence, option (D) is correct.

### Solve for $$y$$ :  $$4y+5 = 29$$

A

$$29$$

.

B

$$4$$

C

$$5$$

D

$$6$$

Option D is Correct

# Solving Equations Involving Subtraction and Multiplication

• Here, we will learn to solve equations involving two operations i.e. subtraction and multiplication.
• We will use inverse operations to solve equations.

For example:

Solve for $$b;$$

$$2b-9=17$$

Here, we have to remove $$2$$ and $$9$$ from the above equation.

To remove these, we will use inverse operations. Thus, we will add $$9$$ to both the sides.

[Inverse operation of subtraction is addition]

$$2b- \not{9}+ \not{9}=17+9$$

$$\Rightarrow 2b=26$$

Now, we will divide by $$2$$ on both the sides to get only the variable on one side.

[Inverse operation of multiplication is division]

$$\dfrac{ \not{2}b}{ \not{2}} = \dfrac{26}{2}$$

$$\Rightarrow b=13$$

#### Solve for $$c$$ :  $$12c-4=44$$

A $$12$$

B $$4$$

C $$44$$

D $$13$$

×

Given equation: $$12c-4=44$$

Here, we will use inverse operations to remove $$12$$ and $$4$$ from the given equation.

We will add $$4$$ to both the sides.

[Inverse operation of subtraction is addition]

$$12c- \not{4}+ \not{4}=44+4$$

$$\Rightarrow 12c = 48$$

We will divide by $$12$$ on both the sides to get only the variable on one side.

[Inverse operation of multiplication is division]

$$\dfrac{12c}{12} = \dfrac{48}{12}$$

$$\Rightarrow c = 4$$

Hence, option (B) is correct.

### Solve for $$c$$ :  $$12c-4=44$$

A

$$12$$

.

B

$$4$$

C

$$44$$

D

$$13$$

Option B is Correct

# Solving Equations Involving Addition and Division

• Here, we will learn to solve equations involving two operations i.e. addition and division.
• We will use inverse operations to solve equations.

For example:

Solve for $$d;$$

$$\dfrac{d}{3}+ 4 = 8$$

Here, we have to remove $$3$$ and $$4$$ from the given equation.

To remove these, we will use inverse operations.

Thus, we will subtract $$4$$ from both the sides.

[Inverse operation of addition is subtraction]

$$\dfrac{d}{3}+ \not{4}- \not{4} = 8-4$$

$$\Rightarrow \dfrac{d}{3} = 4$$

• Now, we will multiply by $$3$$ on both the sides to get only the variable on one side.

[Inverse operation of division is multiplication]

$$\dfrac{ \not{3}d}{ \not{3}} = 4×3$$

$$\Rightarrow d=12$$

#### Solve for $$k$$ :   $$\dfrac{k}{19} + 6 =10$$

A $$19$$

B $$20$$

C $$76$$

D $$10$$

×

Given equation: $$\dfrac{k}{19} + 6 = 10$$

We will use inverse operations to remove $$6$$ and $$19$$ from the given equation.

First, we will subtract $$6$$ from both the sides.

[Inverse operation of addition is subtraction]

$$\dfrac{k}{19} + \not{6} - \not{6}=10-6$$

$$\Rightarrow \dfrac{k}{19} = 4$$

Now, we  will multiply by $$19$$ on both the sides to get only the variable on one side.

[Inverse operation of division is multiplication]

$$\dfrac{19k}{19} = 4×19$$

$$\Rightarrow k = 76$$

Hence, option (C) is correct.

### Solve for $$k$$ :   $$\dfrac{k}{19} + 6 =10$$

A

$$19$$

.

B

$$20$$

C

$$76$$

D

$$10$$

Option C is Correct

# Solving Equations Involving Addition and Subtraction

• Now, it's time to learn how to solve equations involving multiple operations.
• Here, we will deal with two operations i.e. addition and subtraction.
• As always, we will use inverse operations to solve equations.

For example:

Solve for $$x;$$

$$x+5-7=4$$

Simplify using order of operations,

$$\Rightarrow\;x-2=4$$

To solve for $$x,$$ add $$2$$ to both the sides to get only the variable on one side.

$$\Rightarrow\;x-\not2+\not2=4+2$$   [Inverse operation of subtraction is addition]

$$\Rightarrow\;x=6$$

#### Solve for $$z$$ : $$z-2+6=9$$

A $$5$$

B $$6$$

C $$9$$

D $$2$$

×

Given equation: $$z-2+6=9$$

Simplifying using order of operations,

$$z+4=9$$

To solve for $$z,$$ we will subtract $$4$$ from both the sides to get only the variable on one side.

$$z+\not{4}-\not{4}=9-4$$   [Inverse operation of addition is subtraction]

$$\Rightarrow\;z=5$$

Hence, option (A) is correct.

### Solve for $$z$$ : $$z-2+6=9$$

A

$$5$$

.

B

$$6$$

C

$$9$$

D

$$2$$

Option A is Correct

# Solving Equations Involving Subtraction and Division

• Here, we will learn to solve equations involving two operations i.e. subtraction and division.
• We will use inverse operations to solve equations.

For example:

Solve for $$y;$$

$$\dfrac{y}{7} - 2=13$$

Here, we have to remove $$7$$ and $$2$$ from the given equation.

To remove these, we will use inverse operations.

Thus, we will add $$2$$ to both the sides.

[Inverse operation of subtraction is addition]

$$\dfrac{y}{7} - \not{2}+ \not{2} = 13+2$$

$$\Rightarrow \dfrac{y}{7} = 15$$

Now, we will multiply by $$7$$ on both the sides to get only the variable on one side.

[Inverse operation of division is multiplication]

$$\dfrac{7y}{7} = 15×7$$

$$\Rightarrow y = 105$$

#### Solve for $$z$$ :   $$\dfrac{z}{-19}-2=5$$

A $$-133$$

B $$-19$$

C $$5$$

D $$-100$$

×

Given equation: $$\dfrac{z}{-19}-2=5$$

We will use inverse operations to remove $$-19$$ and $$-2$$ from the given equation.

First, we will add $$2$$ to both the sides.

[Inverse operation of subtraction is addition]

$$\dfrac{z}{-19} - \not{2}+ \not{2} = 5+2$$

$$\Rightarrow\dfrac{z}{-19} = 7$$

Now, we will multiply by  $$(-19)$$ on both the sides to get only the variable on one side.

[Inverse operation of division is multiplication]

$$\dfrac{-19z}{-19} = 7 × (-19)$$

$$\Rightarrow z = - 133$$

Hence, option (A) is correct.

### Solve for $$z$$ :   $$\dfrac{z}{-19}-2=5$$

A

$$-133$$

.

B

$$-19$$

C

$$5$$

D

$$-100$$

Option A is Correct

# Solving Equations Involving Multiplication and Division

• Here, we will learn to solve equations involving two operations i.e. multiplication and division.
• We will use inverse operations to solve equations.

For example:

Solve for $$a;$$

$$\dfrac{2a}{3} = 6$$

Here, we have to remove $$2$$ and $$3$$ from the given equation.

To remove these, we will use inverse operations.

Thus, we will multiply $$3$$ by $$6$$ which is on the right side of the equation.

[Inverse operation of division is multiplication]

$$2a=3×6$$

$$\Rightarrow 2a= 18$$

Now, we will divide by $$2$$ on both the sides to get only the variable on one side.

[Inverse operation of multiplication is division]

$$\dfrac{ \not{2}a}{ \not{2}} = \dfrac{18}{2}$$

$$\Rightarrow a = 9$$

#### Solve for ​$$b$$ :   $$\dfrac{6b}{2} = 18$$

A $$18$$

B $$2$$

C $$6$$

D $$3$$

×

Given equation: $$\dfrac{6b}{2} = 18$$

We will use inverse operations to remove $$6$$ and $$2$$ from the given equation.

We will multiply $$2$$ by $$18$$, which is on the right side of the equation.

[Inverse operation of division is multiplication]

$$6b = 2×18$$

$$\Rightarrow 6b = 36$$

Now, we will divide by $$6$$ on both the sides to get only the variable on one side.

[Inverse operation of multiplication is division]

$$\dfrac{ \not{6}b}{ \not{6}} = \dfrac{36}{6}$$

$$\Rightarrow b = 6$$

Hence, option (C) is correct.

### Solve for ​$$b$$ :   $$\dfrac{6b}{2} = 18$$

A

$$18$$

.

B

$$2$$

C

$$6$$

D

$$3$$

Option C is Correct