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Solving Equations Involving Multiple Operations (inverse Operation)

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.

 \(\dfrac{ \not{3}a}{ \not{3}}=\dfrac{ \not{3}}{ \not{3}}\)  

\(\Rightarrow a=1\)

Illustration Questions

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] 

\(\dfrac{ \not{2}b}{ \not{2}} = \dfrac{26}{2}\)

\(\Rightarrow b=13\)

Illustration Questions

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]

\(\dfrac{ \not{3}d}{ \not{3}} = 4×3\)

\(\Rightarrow d=12\)

Illustration Questions

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]

\(\Rightarrow\;x=6\)

Illustration Questions

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]

\( \dfrac{7y}{7} = 15×7\)

\(\Rightarrow y = 105\)

Illustration Questions

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]

\(\dfrac{ \not{2}a}{ \not{2}} = \dfrac{18}{2}\)

\(\Rightarrow a = 9\)

Illustration Questions

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

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