Solve Equations  Grade 9
Grade 9 examples, with a detailed step by step approach, on solving simple equations and equations involving brackets and fractions are presented. Checking the solutions to an equation is also discusssed. More questions and their solutions with detailed explanations are also included.
What is an Equation and its Solution?
We first review the concept of equations and the solution of an equation.
An equation is a statement that expresses the equality of two mathematical expressions. An equation has an equal sign, a right side expression and a left side expression.
Example 1
These are examples of equations with the unknown \( x \)
\( \quad 2 x =  6 \) , \( \quad x + 3 = 7 \) , \( \quad 2(x + 3) =  (2x+4) \)
Each equation has an equal sign that separates the left and right sides of the equation.
The left side of the equation \( \quad \color{red}{2 x  6} = x + 5 \) is \( \quad \color{red}{2 x  6} \).
The right side of the equation \( \quad 2 x  6 = \color{red}{x + 5} \) is \( \quad \color{red}{x + 5} \).
The solution to an equation, with the unknown \( x \), is the set of all values of \( x \) that make the equation a true statement.
Example 2
Which of the following values of \( x \): \(  4, 2\) is/are solution(s) to the equation \( 2 x + 2 = x + 4 \)?
Solution to Example 2
Substitute \( x \) by its numerical value in the left side and the right side of the equation
a) Check \( \color{red}{x =  4} \)
Evaluate left side : \( 2 \color{red}x + 2 = 2 \color{red}{(  4 )} + 2 =  8 + 2 =  6 \) ,
Evaluate right side : \( \color{red}x + 4 = \color{red}{(  4)} + 4 = 0 \)
The numerical values of the left and the right sides are not equal, therefore \( x =  4 \) is NOT a solution to the equation \( 2 x + 2 = x + 4 \).
a) Check \( \color{red}{x = 2} \)
Evaluate left side : \( 2 x + 2 = 2 \color{red}{(2 )} + 2 = 4 + 2 = 6 \) ,
Evaluate right side : \( \color{red}x + 4 = \color{red}{(2)} + 4 = 6 \)
The numerical values of the left and the right sides are equal, therefore \( x = 2 \) is a solution to the equation \( 2 x + 2 = x + 4 \).
Important Properties to Solve Equations
In order to solve an equation, we need mathematical steps that help in getting all terms with the unknown on one side and the constant terms on the other side.
Some of the most important properties used to solve equations are listed below.
1) If we add or subtract the same quantity to both sides of an equation, we obtain an equation having the same solution as the original one.
1) If we multiply or divide both sides of an equation by the same quantity, NOT equal to zero, we obtain an equation having the same solution as the original one.
Solve Simple Equations
Example 3
Solve the equation \( 2x + 1 =  5 \) and check the solution obtained.
Solution to Example 3
The main idea is to have all terms with the unknown \( x \) on one side and all constant terms on the other side of the equation
Let us keep the terms \( 2x \) on the left and have the constant terms on the right side. This can be done by subtracting \( 1 \) from both sides of the equation
\( \quad \quad 2x + 1 \color{red}{ 1} =  5 \color{red}{ 1} \)
Simplify to obtain
\( \quad \quad 2x =  6 \)
To obtain \( x \) from \( 2x \), we divide both sides of the above equation by 2
\( \quad \quad \dfrac{2 x}{\color{red}2} = \dfrac{6}{\color{red}2} \)
Simplify
\( \quad \quad x = 3 \)
Check the solution obtained in the original (given) equation
Evaluate the left side of equation for \( x =  3 \) : \( \quad 2x + 1 = 2(3) + 1 =  5 \)
Evaluate the left side of equation for \( x =  3 \) : \( \quad  5 \)
Left side and right are both equal to \(  5 \) for \( x =  3 \), therefore \( x =  3 \) is a solution to the given equation.
Example 4
Solve the equation \( x  2  3x =  7  x \) and check the solution obtained.
Solution to Example 4
Group like terms in the two sides of the equation. \( x \) and \(  3x \) are like term on the left side and may be grouped to give
\( \quad \quad  2x  2 =  7  x \)
Add \( 2 \) to both sides of the equation in order to eliminate constant terms from the left side.
\( \quad \quad  2x  2 \color{red}{+ 2} =  7  x \color{red}{+ 2} \)
Simplify
\( \quad \quad  2x =  x  5 \)
Add \( x \) to both sides of the equation in order to eliminate terms with \( x \) from the right side.
\( \quad \quad  2x \color{red}{+x } =  x  5 \color{red}{+x }\)
Simplify to obtain
\( \quad \quad  x =  5 \)
If we know \(  x \) and we need \( x \), we multiply both sides of the equation by \(  1 \)
\( \quad \quad \color{red}{(1)}( x) = \color{red}{(1)}( 5) \)
Simplify
\( \quad \quad x = 5 \)
Check the solution obtained in the original (given) equation
Left side of equation for \( x = 5 \) : \( \quad x  2  3x = 5  2 3(5) =  12 \)
Right side of equation for \( x = 5 \) : \( \quad  7  x =  7  (5) =  12 \)
Left side and right are both equal to \(  12 \) for \( x = 5 \), therefore \( x =  3 \) is a solution to the given equation.
Solve Equations Involving Brackets
Example 5
Solve the equation \(  2 (x  2) + 3 = 3 (x + 4)  3 \) and check the solution obtained.
Solution to Example 5
The given equation
\( \quad \quad \color{red}{ 2} (x  2) + 3 = \color{red}3 (x + 4)  3 \)
Use distributive law: \( \quad a(b+c) = ab + ac \quad \), which is one of the basic rules of algebra, to remove the brackets.
Distribute \( \color{red}{2} \) and \( \color{red}3 \).
\( \quad \quad \color{red}{ 2} (x ) \color{red}{ 2} ( 2) + 3 = \color{red}3 (x) + \color{red}3 (4)  3 \)
Simplify
\( \quad \quad  2 x + 4 + 3 =  3 x + 12  3 \)
Group like terms in both sides of the equation.
\( \quad \quad  2 x + 7 =  3 x + 9 \)
Subtract \( 7 \) to both sides of the equation in order to eliminate constant terms from the left side of the equation.
\( \quad \quad  2 x + 7  7 =  3 x + 9  7 \)
Group like terms
\( \quad \quad  2x =  3x + 2 \)
Add \( 3x \) to both sides of the equation in order to eliminate terms in \( x \) from the left right of the equation.
\( \quad \quad  2x + 3 x =  3x + 2 + 3x \)
Group like terms and simplify
\( \quad \quad x = 2 \)
Check the solution obtained in the original (given) equation
Evaluate left side of equation for \( x = 2 \) : \( \quad  2 (x  2) + 3 =  2 ((2)  2) + 3 = 3 \)
Evaluate right side of equation for \( x = 2 \) : \( \quad 3 (x + 4)  3 = 3 ((2) + 4)  3 = 3 \)
Left side and right are both equal to 3 for \( x = 2 \), therefore \( x = 2 \) is a solution to the given equation.
Solve Equations Involving Fractions
The method of solving equations with fractions adopted here is the one where we get rid of fractions first (to avoid dealing with fraction) by multiplication and then solve the equation.
Example 6
Solve the equation \( \quad \dfrac{x}{2} =  3 \) and check the solution obtained.
Solution to Example 6
To eliminate the denominator \( 2 \) in \( \dfrac{x}{2} \), we multiply the two sides of the equation by the denominator \( 2 \)
\( \quad \color{red}2 \left(\dfrac{x}{2} \right) = \color{red}2 ( 3) \)
Simplify
\( \quad x =  6 \)
Check the solution obtained in the original (given) equation
Left side of equation for \( x = 6 \) : \( \quad \dfrac{x}{2} = \dfrac{6}{2} =  3 \)
Left side and right are both equal to \(  3 \) for \( x =  6 \), therefore \( x =  6 \) is a solution to the given equation.
Example 7
Solve the equation \( \quad \dfrac{x}{3}  \dfrac{1}{2} = \dfrac{1}{3} \) and check the solution obtained.
Solution to Example 7
We now have two fractions with denominators \( 2 \) and \( 3 \) in the given equation. In order to get rid of the fractions, we need to multiply both sides of the equation by the LCM (lowest common multiple) of the two different denominators \( 2 \) and \( 3 \).
Find the LCM of \( 2 \) and \( 3 \) which is \( 6 \).
Multiply both sides of the equation by the LCM which is \( 6 \)
\( \quad\quad \color{red}6 \left( \dfrac{x}{3}  \dfrac{1}{2} \right) = \color{red}6 \left(\dfrac{1}{3}\right) \)
Distribute the factor \( 6 \)
\( \quad\quad 6 \left(\dfrac{x}{3} \right)  6 \left(\dfrac{1}{2} \right) = 6 \left(\dfrac{1}{3} \right) \)
Rearrange as
\( \quad\quad \left(\dfrac{6}{3} \right) x  \left(\dfrac{6}{2} \right) = \left(\dfrac{6}{3} \right) \)
Simplify
\( \quad\quad 2x  3 = 2 \)
NOTE: One step, which is the multiplication of both sides of the equation by the LCM of the denominators, is needed to get rid of the fractions because the LCM is a multiple of each denominator.
Solve the above equation by adding \( 3 \) to both sides and simplify to obtain
\( \quad\quad 2 x = 5 \)
Divide both sides by \( 2 \)
\( \quad\quad x = \dfrac{5}{2} \)
Check the solution obtained in the original (given) equation
Left side of equation for \( x = \dfrac{5}{2} \) : \( \quad \dfrac{x}{3}  \dfrac{1}{2} = \dfrac{1}{3} x  \dfrac{1}{2} = \dfrac{1}{3} \left(\dfrac{5}{2}\right)  \dfrac{1}{2} = \dfrac{1}{3} \)
Left side and right are both equal to \( \dfrac{1}{3} \) for \( x = \dfrac{5}{2} \), therefore \( x = \dfrac{5}{2} \) is a solution to the given equation.
Example 8
Solve the equation \( \quad \dfrac{2x + 1}{5} + 2 =  \dfrac{x}{3} \) and check the solution obtained.
Solution to Example 8
We now have fractions with denominators \( 5 \) and \( 3 \) in the given equation. We need to multiply both sides of the equation by the LCM (lowest common multiple) of the two different denominators \( 5 \) and \( 3 \).
Find the LCM of \( 5 \) and \( 3 \) which is \( 15 \).
NOTE: One step, which is the multiplication of both sides of the equation by the LCM of the denominators, is needed to get rid of the fractions because the LCM is a multiple of each denominator.
Multiply both sides of the equation by the LCM \( 15 \)
\( \quad\quad \color{red}{15} \left( \dfrac{2x + 1}{5} + 2 \right) = \color{red}{15} \left(\  \dfrac{x}{3} \right) \)
Distribute the factor \( 15 \)
\( \quad\quad 15 \left(\dfrac{2x+1}{5} \right) + 15 (2) = 15 \left(  \dfrac{x}{3} \right) \)
Rearrange as
\( \quad\quad \dfrac{15}{5}(2x+1) + 15 (2) = \dfrac{15}{3}(x) \)
Simplify
\( \quad\quad 3 (2x+1) + 30 =  5x \)
Distribute factor \(3 \) in the left side and group like terms
\( \quad\quad 6 x + 3 + 30 =  5x \)
\( \quad\quad 6x + 33 =  5 x \)
Subtract \( 33 \) from both sides and add \( 5x \) to both sides. (NOTE: we have carried out two operations in one step.)
\( \quad\quad 6x + 33 \color{red}{ 33 + 5x } =  5 x \color{red}{ 33 + 5x } \)
Group like terms
\( \quad\quad 11 x =  33 \)
Divide both sides by \( 11 \)
\( \quad\quad \dfrac{ 11 x} {11} = \dfrac{33}{11} \)
Simplify
\( \quad\quad x =  3 \)
Check the solution obtained in the original (given) equation
Left side of equation for \( x =  3 \) : \( \quad \dfrac{2x + 1}{5} + 2 = \dfrac{2(3) + 1}{5} + 2 = 1 \)
Right side of equation for \( x = 3 \) : \( \quad  \dfrac{x}{3} =  \dfrac{3}{3} = 1 \) Left side and right are both equal to \( 1 \) for \( x = 3 \), therefore \( x =  3 \) is a solution to the given equation.
Questions

Solve the following equations and check the solution found.
 ) \( 2x + 2 = 6 \)
 ) \( 5y  2 = 7y  8 \)
 ) \( 2x + 4 + 5x = 7 + 4x  3 \)
 ) \( 0.2 d + 4 =  0.1 d  2 \)
 ) \( 2(2x 6) = (x  4) \)
 ) \( (x+2)+4 = 2(x+3) + x \)
 ) \( \dfrac{x}{5} =  6 \)
 ) \(  \dfrac{x}{3} = \dfrac{1}{2} \)
 ) \(  \dfrac{x}{4} = \dfrac{1}{2}  x \)
 ) \(  \dfrac{x3}{7} = \dfrac{1}{2} ( 2x + 6) \)
 ) \(  \dfrac{1}{2}  x + 5 = \dfrac{1}{5} + 2(x2) \)
 ) \( 2x + 2 = 6 \)
More References and Links
Solve Equations, Systems of Equations and InequalitiesFind Lowest Common Multiple
Middle School Math (Grades 6, 7, 8, 9)  Free Questions and Problems With Answers
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