Solutions to Equations  Grade 9
The solutions to the questions in solve equations are presented along with all the steps needed and detailed explanations.
Solutions to Questions in solve equations

)
Given
\( \quad \quad 2x + 2 = 6 \)
Subtract \( 2 \) from both sides of the equation
\( \quad \quad 2x + 2  2 = 6  2 \)
Group like terms and simplify
\( \quad \quad 2x = 4 \)
Divide both sides of the equation by \( 2 \) (or multiply both sides by \dfrac{1}{2} \)
\( \quad \quad \dfrac{2x}{2} = \dfrac{4}{2} \)
Simplify
\( \quad \quad x = 2 \)
Check the solution obtained
Left side of equation for \( x = 2 \) : \( \quad 2x + 2 = 2(2)+2 = 6 \)
Left side and right are both equal to \( 6 \) for \( x = 2 \), therefore \( x = 2 \) is a solution to the given equation.

)
Given the equation
\( \quad \quad 5y  2 = 7y  8 \)
Let us arrange the equation so that the variable is on the right side and the constant is on the left side. Subtract \( 5y \) from both sides.
\( \quad \quad 5y  2  5 y = 7y  8  5 y \)
Group like terms and simplify
\( \quad \quad  2 = 2y  8 \)
Add \( 8 \) to both sides.
\( \quad \quad  2 + 8 = 2y  8 + 8 \)
Group like terms and simplify
\( \quad \quad 6 = 2y \)
Divide both sides of the equation by \( 2 \) (or multiply both sides by \( \dfrac{1}{2} \) ).
\( \quad \quad \dfrac{6}{2} = \dfrac{2y}{2} \)
Simplify
\( \quad \quad 3 = y \)
Check the solution obtained
Left side of equation for \( y = 3 \) : \( \quad 5y  2 = 5(3)  2 = 13 \)
Right side of equation for \( y = 3 \) : \( \quad 7y  8 = 7(3)  8 = 13 \)
Hence \( y = 3 \) is a solution to the given equation.

)
Given
\( \quad \quad 2x + 4 + 5x = 7 + 4x  3 \)
Group like terms and simplify
\( \quad \quad 3x + 4 = 4 + 4x \)
Subtract \( 4 \) and \( 3x \) from both sides. (Two opeartions in one step).
\( \quad \quad 3x + 4  4  3x = 4 + 4x  4  3x \)
Group like terms and simplify
\( \quad \quad 0 = x \)
Check the solution obtained
Left side of equation for \( x = 0 \) : \( \quad 2x + 4 + 5x = 2(0) + 4 + 5(0) = 4 \)
Right side of equation for \( x = 0 \) : \( \quad 7 + 4x  3 = 7 + 4(0)  3 = 4 \)
Hence \( x = 0 \) is a solution to the given equation.

)
Given
\( \quad \quad 0.2 d + 4 =  0.1 d  2 \)
Add \( 0.1d \) to both sides and subtract \( 4 \) from both sides. (Two opeartions in one step).
\( \quad \quad 0.2 d + 4 + 0.1 d  4 =  0.1 d  2 + 0.1 d  4 \)
Group like terms and simplify
\( \quad \quad 0.3 d =  6 \)
Divide both sides by \( 0.3 \) and simplify
\( \quad \quad \dfrac{0.3 d}{0.3} =  \dfrac{6}{0.3} \)
\( \quad \quad d =  20 \)
Check the solution obtained
Left side of equation for \( d = 20 \) : \( \quad 0.2 d + 4 = 0.2(20) + 4 = 0 \)
Right side of equation for \( d =  20 \) : \( \quad  0.1 d  2 =  0.1 (20)  2 = 0 \)
Hence \( d = 20 \) is a solution to the given equation.

)
Given the equation
\( \quad \quad 2(2x 6) = (x  4) \)
Use distributive law to expand both sides.
\( \quad \quad 4 x + 12 =  x + 4 \)
Add \( x \) to both sides, group like terms and simplify
\( \quad \quad 4 x + 12 + x =  x + 4 + x \)
\( \quad \quad 3 x + 12 = 4 \)
Subtract \( 12 \) from both sides, group like terms and simplify
\( \quad \quad 3 x + 12  12 = 4  12 \)
\( \quad \quad 3 x =  8 \)
Divide both sides by \(  3 \) and simplify
\( \quad \quad  \dfrac{ 3 x}{3} =  \dfrac{8}{3} \)
\( \quad \quad x = \dfrac{8}{3} \)
Check the solution obtained
Left side of equation for \( x = \dfrac{8}{3} \) : \( \quad 2(2x 6) = 2(2 (\dfrac{8}{3}) 6) = \dfrac{4}{3} \)
Right side of equation for \( x = \dfrac{8}{3} \) : \( \quad (\dfrac{8}{3}  4) = \dfrac{4}{3} \)
Hence \( x = \dfrac{8}{3} \) is a solution to the given equation.

)
Given
\( \quad \quad (x+2)+4 = 2(x+3) + x \)
Use distributive law to expand both sides, group like terms and simplify.
\( \quad \quad x  2 +4 = 2x + 6 + x \)
\( \quad \quad x + 2 = 3x + 6 \)
Add \( x \) to both sides and subtract \( 6 \) from both sides, group like terms and simplify.
\( \quad \quad x + 2 + x  6 = 3x + 6 + x  6 \)
\(\quad \quad  4 = 4 x \)
Divide both sides by \( 4 \) to obtain the solution
\(\quad \quad x =  1 \)
Check the solution obtained
Left side of equation for \( x =  1 \) : \( \quad (x+2)+4 = (1+2)+4 = 3\)
Right side of equation for \( x =  1 \) : \( \quad = 2(x+3) + x = 2(1+3)  1 = 3 \)
Hence \( x =  1 \) is a solution to the given equation.

)
Given
\( \quad \quad \dfrac{x}{5} =  6 \)
In order to get rid of the fraction, multiply both sides of the equation by the denominator \( 5 \)
\( \quad \quad 5 \left( \dfrac{x}{5} \right) = 5( 6) \)
Rearrange as
\( \quad \quad \dfrac{5}{5} x = 5( 6) \)
Simplify to find the solution.
\( \quad \quad x =  30 \)
Check the solution obtained
Left side of equation for \( x =  30 \) : \( \quad \dfrac{x}{5} = \dfrac{30}{5} =  6 \)
Right side of equation is equal to \(  6 \).
Hence \( x =  30 \) is a solution to the given equation.

)
Given the equation
\( \quad \quad  \dfrac{x}{3} = \dfrac{1}{2} \)
Multiply both sides by \(  3 \)
\( \quad \quad  3 \left( \dfrac{x}{3} \right) =  3 \left(\dfrac{1}{2} \right) \)
Simplify
\( \quad \quad \dfrac{3}{3} x = \dfrac{3}{2} \)
\( \quad \quad x =  \dfrac{3}{2} \)
Check the solution obtained
Left side of equation for \( x =  \dfrac{3}{2} \) : \( \quad  \dfrac{x}{3} =  \dfrac{1}{3} x =  \dfrac{1}{3} \left(  \dfrac{3}{2} \right) = \dfrac{1}{2} \)
Both sides of the equation are equal for \( x =  \dfrac{3}{2} \)
Hence \( x =  \dfrac{3}{2} \) is a solution to the given equation.

)
Given
\( \quad \quad  \dfrac{x}{4} = \dfrac{1}{2}  x \)
Multiply both sides by the lowest common multiple (LCM) of the denominators \( 4 \) and \( 2 \) which is equal to \( 4 \).
\( \quad \quad 4 \left( \dfrac{x}{4} \right) = 4 \left(\dfrac{1}{2}  x\right) \)
Use distributive law to expand the right side of the equation
\( \quad \quad 4 \left( \dfrac{x}{4} \right) = 4 \left(\dfrac{1}{2} \right)  4 x \)
Rearrange the fractions as
\( \quad \quad  \dfrac{4}{4} x = \dfrac{4}{2}  4 x \)
Simplify
\( \quad \quad  x = 2  4 x \)
Add \( 4x \) to both sides of the equation and group like terms
\( \quad \quad  x + 4 x = 2  4 x + 4 x \)
\( \quad \quad 3x = 2 \)
Divide boh sides by \( 3 \) and simplify to find the solution.
\( \quad \quad x = \dfrac{2}{3} \)
Check the solution obtained
Left side of equation for \( x = \dfrac{2}{3} \) : \( \quad  \dfrac{x}{4} =  \dfrac{1}{4} (x) =  \dfrac{1}{4} \left( \dfrac{2}{3} \right) =  \dfrac{1}{6} \)
Right side of equation for \( x = \dfrac{2}{3} \) : \( \quad \dfrac{1}{2}  x = \dfrac{1}{2}  \dfrac{2}{3} =  \dfrac{1}{6} \)
Both sides of the equation are equal for \( x = \dfrac{2}{3} \)
Hence \( x = \dfrac{2}{3} \) is a solution to the given equation.

)
Given the equation
\( \quad \quad  \dfrac{x3}{7} = \dfrac{1}{2} ( 2x + 6) \)
Find the lowest common multiple (LCM) of the denominators \( 7 \) and \( 2 \) which is equal to \( 14 \) and multiply both sides of the equation by \( 14 \).
\( \quad \quad 14 \left( \dfrac{x3}{7} \right) = 14 \left(\dfrac{1}{2} ( 2x + 6) \right) \)
Rearrange fractions in order to make it easier to simplify
\( \quad \quad  \dfrac{14}{7} (x  3) = \dfrac{14}{2} ( 2x + 6) \)
Reduce the fractions \( 14/7 = 2 \) and \( 14/2 = 7\)
\( \quad \quad  2 (x  3) = 7( 2x + 6) \)
Use distibutive law to expand both sides
\( \quad \quad 2x + 6 =  14 x + 42 \)
Subtract \( 6 \) from both sides
\( \quad \quad 2x + 6  6 =  14 x + 42  6 \)
Group like terms and simplify
\( \quad \quad  2 x =  14 x + 36 \)
Add \( 14 x \) to both sides
\( \quad \quad  2 x + 14 x =  14 x + 36 + 14 x \)
Group like terms and simplify
\( \quad \quad 12 x = 36 \)
Divide both sides by \( 12 \)
\( \quad \quad \dfrac{12x}{12} x = \dfrac{36}{12} \)
Simplify
\( \quad \quad x = 3 \)
Check the solution obtained
Left side of equation for \( x = 3 \) : \( \quad  \dfrac{x3}{7} =  \dfrac{33}{7} = 0 \)
Right side of equation for \( x = 3 \) : \( \quad \dfrac{1}{2} ( 2x + 6) = \dfrac{1}{2} ( 2(3) + 6) = 0 \)
Both sides of the equation are equal for \( x = 3 \)
Therefore \( x = 3 \) is a solution to the given equation.

)
Given
\( \quad \quad  \dfrac{1}{2}  x + 5 = \dfrac{1}{5} + 2(x2) \)
Find the lowest common multiple (LCM) of the denominators \( 2 \) and \( 5 \) which is equal to \( 10 \) and multiply both sides of the equation by \( 10 \).
\( \quad \quad 10 \left( \dfrac{1}{2}  x + 5 \right) = 10 \left( \dfrac{1}{5} + 2(x2) \right)\)
Use distributive law to exppand both sides of the equation
\( \quad \quad  10 \left( \dfrac{1}{2} \right) 10x + 50 = 10 \left( \dfrac{1}{5} \right) + 20(x  2) \)
Expand \( 20(x  2) = 2x  40 \) and simplify the fractions \( 10 \left( \dfrac{1}{2} \right) = 10/2 = 5 \) and \( 10 \left( \dfrac{1}{5} \right) = 10/5 = 2 \)
\( \quad \quad 5  10x + 50 = 2 + 20 x  40 \)
Group like terms and simplify
\( \quad \quad  10 x + 45 = 20 x  38 \)
Add \( 10 x \) from both sides
\( \quad \quad  10 x + 45 + 10 x = 20 x  38 + 10 x\)
Group like terms and simplify
\( \quad \quad 45 = 30x  38 \)
Add \( 38 \) to both sides
\( \quad \quad 45 + 38 = 30 x\)
Group like terms and simplify
\( \quad \quad 83 = 30x \)
Divide both sides by \( 30 \) and simplify to solve for \( x \).
\( \quad \quad x = \dfrac{83}{30} \)
Check the solution obtained
Left side of equation for \( x = \dfrac{83}{30} \) : \( \quad  \dfrac{1}{2}  x + 5 =  \dfrac{1}{2}  \dfrac{83}{30} + 5 = 26/15 \)
Right side of equation for \( x = \dfrac{83}{30} \) : \( \quad \dfrac{1}{5} + 2(x2) = \dfrac{1}{5} + 2(\dfrac{83}{30}2) = 26/15 \)
Both sides of the equation are equal for \( x = \dfrac{83}{30} \)
Therefore \( x = \dfrac{83}{30} \) is a solution to the given equation.
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
High School Math (Grades 10, 11 and 12)  Free Questions and Problems With Answers
Primary Math (Grades 4 and 5) with Free Questions and Problems With Answers