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.
\(\dfrac{ \not{3}a}{ \not{3}}=\dfrac{ \not{3}}{ \not{3}}\)
\(\Rightarrow a=1\)
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]
\(\dfrac{ \not{2}b}{ \not{2}} = \dfrac{26}{2}\)
\(\Rightarrow b=13\)
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]
\(\dfrac{d}{3}+ \not{4}- \not{4} = 8-4\)
\(\Rightarrow \dfrac{d}{3} = 4\)
[Inverse operation of division is multiplication]
\(\dfrac{ \not{3}d}{ \not{3}} = 4×3\)
\(\Rightarrow d=12\)
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]
\(\Rightarrow\;x=6\)
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]
\( \dfrac{7y}{7} = 15×7\)
\(\Rightarrow y = 105\)
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]
\(\dfrac{ \not{2}a}{ \not{2}} = \dfrac{18}{2}\)
\(\Rightarrow a = 9\)