Solving Equations: Questions and Step-by-Step Solutions - Grade 9

Guided Examples

Review these guided examples to see how the properties are applied to isolate variables, handle brackets, and clear fractions.

1. Simple Equation: Solve the equation \( 2x + 1 = -5 \) and check the solution obtained.

   View Step-by-Step Solution

   The main idea is to isolate \( x \) on one side of the equation. Subtract \( 1 \) from both sides:

   \[ \begin{aligned} 2x + 1 - 1 &= -5 - 1 \\ 2x &= -6 \end{aligned} \]

   Divide both sides by 2:

   \[ \begin{aligned} \dfrac{2x}{2} &= \dfrac{-6}{2} \\ x &= -3 \end{aligned} \]

   Check the Solution:
   Left Side (LHS): \( 2(-3) + 1 = -6 + 1 = -5 \)
   Right Side (RHS): \( -5 \)
   Since LHS = RHS, \( x = -3 \) is correct.

2. Variables on Both Sides: Solve the equation \( x - 2 - 3x = -7 - x \) and check the solution.

   View Step-by-Step Solution

   First, group the like terms on the left side (\( x \) and \( -3x \)):

   \[ -2x - 2 = -7 - x \]

   Add \( 2 \) to both sides to move constant terms to the right:

   \[ -2x = -x - 5 \]

   Add \( x \) to both sides to move variable terms to the left:

   \[ \begin{aligned} -2x + x &= -5 \\ -x &= -5 \end{aligned} \]

   Multiply or divide both sides by \( -1 \):

   \[ x = 5 \]

   Check the Solution:
   LHS: \( 5 - 2 - 3(5) = 5 - 2 - 15 = -12 \)
   RHS: \( -7 - (5) = -12 \)
   Since LHS = RHS, \( x = 5 \) is correct.

3. Equations Involving Brackets: Solve the equation \( -2(x - 2) + 3 = 3(-x + 4) - 3 \).

   View Step-by-Step Solution

   Use the distributive law, \( a(b+c) = ab + ac \), to remove the brackets on both sides:

   \[ -2x + 4 + 3 = -3x + 12 - 3 \]

   Combine like terms on each side:

   \[ -2x + 7 = -3x + 9 \]

   Add \( 3x \) to both sides, and subtract \( 7 \) from both sides to isolate \( x \):

   \[ \begin{aligned} -2x + 3x &= 9 - 7 \\ x &= 2 \end{aligned} \]

   Check the Solution:
   LHS: \( -2(2 - 2) + 3 = -2(0) + 3 = 3 \)
   RHS: \( 3(-2 + 4) - 3 = 3(2) - 3 = 6 - 3 = 3 \)
   Since LHS = RHS, \( x = 2 \) is correct.

4. Equations Involving Fractions: Solve the equation \( \dfrac{x}{3} - \dfrac{1}{2} = \dfrac{1}{3} \).

   View Step-by-Step Solution

   To eliminate the fractions, multiply both sides by the Lowest Common Multiple (LCM) of the denominators 3 and 2, which is 6.

   \[ 6 \left( \dfrac{x}{3} - \dfrac{1}{2} \right) = 6 \left( \dfrac{1}{3} \right) \]

   Distribute the 6 to every term:

   \[ \begin{aligned} \dfrac{6x}{3} - \dfrac{6}{2} &= \dfrac{6}{3} \\ 2x - 3 &= 2 \end{aligned} \]

   Add 3 to both sides, then divide by 2:

   \[ \begin{aligned} 2x &= 5 \\ x &= \dfrac{5}{2} \end{aligned} \]

   Check the Solution:
   LHS: \( \dfrac{1}{3} \left(\dfrac{5}{2}\right) - \dfrac{1}{2} = \dfrac{5}{6} - \dfrac{3}{6} = \dfrac{2}{6} = \dfrac{1}{3} \)
   Since LHS = RHS, \( x = \dfrac{5}{2} \) is correct.

Practice Questions

Solve the following equations and verify your answer by checking the solution.

1. \( 2x + 2 = 6 \)

   View Step-by-Step Solution

   Subtract 2 from both sides:

   \[ \begin{aligned} 2x + 2 - 2 &= 6 - 2 \\ 2x &= 4 \end{aligned} \]

   Divide both sides by 2:

   \[ \begin{aligned} \dfrac{2x}{2} &= \dfrac{4}{2} \\ x &= 2 \end{aligned} \]

   Check: LHS: \( 2(2) + 2 = 4 + 2 = 6 \). Both sides are equal.

2. \( 5y - 2 = 7y - 8 \)

   View Step-by-Step Solution

   Subtract \( 5y \) from both sides:

   \[ \begin{aligned} 5y - 2 - 5y &= 7y - 8 - 5y \\ -2 &= 2y - 8 \end{aligned} \]

   Add 8 to both sides:

   \[ \begin{aligned} -2 + 8 &= 2y \\ 6 &= 2y \end{aligned} \]

   Divide by 2:

   \[ y = 3 \]

   Check: LHS: \( 5(3) - 2 = 13 \). RHS: \( 7(3) - 8 = 13 \). Both sides are equal.

3. \( -2x + 4 + 5x = 7 + 4x - 3 \)

   View Step-by-Step Solution

   Combine like terms on each side:

   \[ 3x + 4 = 4x + 4 \]

   Subtract \( 3x \) and 4 from both sides:

   \[ \begin{aligned} 4 - 4 &= 4x - 3x \\ 0 &= x \end{aligned} \]

   Check: LHS: \( -2(0) + 4 + 5(0) = 4 \). RHS: \( 7 + 4(0) - 3 = 4 \). Both sides are equal.

4. \( 0.2d + 4 = -0.1d - 2 \)

   View Step-by-Step Solution

   Add \( 0.1d \) to both sides and subtract 4 from both sides:

   \[ \begin{aligned} 0.2d + 0.1d &= -2 - 4 \\ 0.3d &= -6 \end{aligned} \]

   Divide both sides by 0.3:

   \[ \begin{aligned} d &= \dfrac{-6}{0.3} \\ d &= -20 \end{aligned} \]

   Check: LHS: \( 0.2(-20) + 4 = -4 + 4 = 0 \). RHS: \( -0.1(-20) - 2 = 2 - 2 = 0 \). Both sides are equal.

5. \( -2(2x - 6) = -(x - 4) \)

   View Step-by-Step Solution

   Expand both sides using the distributive law:

   \[ -4x + 12 = -x + 4 \]

   Add \( 4x \) to both sides and subtract 4:

   \[ \begin{aligned} 12 - 4 &= -x + 4x \\ 8 &= 3x \end{aligned} \]

   Divide by 3:

   \[ x = \dfrac{8}{3} \]

   Check: LHS: \( -2\left(2\left(\frac{8}{3}\right) - 6\right) = -2\left(\frac{16}{3} - \frac{18}{3}\right) = -2\left(-\frac{2}{3}\right) = \frac{4}{3} \).
   RHS: \( -\left(\frac{8}{3} - 4\right) = -\left(\frac{8}{3} - \frac{12}{3}\right) = -\left(-\frac{4}{3}\right) = \frac{4}{3} \). Both sides are equal.

6. \( -(x + 2) + 4 = 2(x + 3) + x \)

   View Step-by-Step Solution

   Expand and combine like terms:

   \[ \begin{aligned} -x - 2 + 4 &= 2x + 6 + x \\ -x + 2 &= 3x + 6 \end{aligned} \]

   Add \( x \) and subtract 6 from both sides:

   \[ \begin{aligned} 2 - 6 &= 3x + x \\ -4 &= 4x \\ -1 &= x \end{aligned} \]

   Check: LHS: \( -(-1 + 2) + 4 = -1 + 4 = 3 \). RHS: \( 2(-1 + 3) + (-1) = 2(2) - 1 = 3 \). Both sides are equal.

7. \( \dfrac{x}{5} = -6 \)

   View Step-by-Step Solution

   Multiply both sides by 5 to isolate \( x \):

   \[ \begin{aligned} 5 \left(\dfrac{x}{5}\right) &= 5(-6) \\ x &= -30 \end{aligned} \]

   Check: LHS: \( \dfrac{-30}{5} = -6 \). Both sides are equal.

8. \( -\dfrac{x}{3} = \dfrac{1}{2} \)

   View Step-by-Step Solution

   Multiply both sides by -3 to clear the fraction and the negative sign:

   \[ \begin{aligned} (-3) \left(-\dfrac{x}{3}\right) &= (-3) \left(\dfrac{1}{2}\right) \\ x &= -\dfrac{3}{2} \end{aligned} \]

   Check: LHS: \( -\dfrac{(-3/2)}{3} = \dfrac{3/2}{3} = \dfrac{1}{2} \). Both sides are equal.

9. \( -\dfrac{x}{4} = \dfrac{1}{2} - x \)

   View Step-by-Step Solution

   Multiply the entire equation by 4 to clear the fractions:

   \[ \begin{aligned} 4 \left(-\dfrac{x}{4}\right) &= 4 \left(\dfrac{1}{2} - x\right) \\ -x &= 2 - 4x \end{aligned} \]

   Add \( 4x \) to both sides:

   \[ \begin{aligned} 3x &= 2 \\ x &= \dfrac{2}{3} \end{aligned} \]

   Check: LHS: \( -\dfrac{2/3}{4} = -\dfrac{2}{12} = -\dfrac{1}{6} \).
   RHS: \( \dfrac{1}{2} - \dfrac{2}{3} = \dfrac{3}{6} - \dfrac{4}{6} = -\dfrac{1}{6} \). Both sides are equal.

10. \( -\dfrac{x - 3}{7} = \dfrac{1}{2}(-2x + 6) \)

    View Step-by-Step Solution

    Multiply both sides by the LCM of 7 and 2 (which is 14) to clear the fractions:

    \[ \begin{aligned} 14 \left( -\dfrac{x - 3}{7} \right) &= 14 \left( \dfrac{1}{2}(-2x + 6) \right) \\ -2(x - 3) &= 7(-2x + 6) \end{aligned} \]

    Expand using the distributive law:

    \[ -2x + 6 = -14x + 42 \]

    Add \( 14x \) and subtract 6 from both sides:

    \[ \begin{aligned} 12x &= 36 \\ x &= 3 \end{aligned} \]

    Check: LHS: \( -\dfrac{3 - 3}{7} = 0 \). RHS: \( \dfrac{1}{2}(-2(3) + 6) = 0 \). Both sides are equal.

11. \( -\dfrac{1}{2} - x + 5 = \dfrac{1}{5} + 2(x - 2) \)

    View Step-by-Step Solution

    First, clear the fractions by multiplying every term by the LCM of 2 and 5 (which is 10):

    \[ 10 \left(-\dfrac{1}{2} - x + 5\right) = 10 \left(\dfrac{1}{5} + 2(x - 2)\right) \]

    Distribute the 10 carefully:

    \[ -5 - 10x + 50 = 2 + 20(x - 2) \]

    Combine constant terms on the left and expand the brackets on the right:

    \[ \begin{aligned} -10x + 45 &= 2 + 20x - 40 \\ -10x + 45 &= 20x - 38 \end{aligned} \]

    Add \( 10x \) and add 38 to both sides:

    \[ \begin{aligned} 45 + 38 &= 20x + 10x \\ 83 &= 30x \\ x &= \dfrac{83}{30} \end{aligned} \]

    Check: LHS: \( -\dfrac{1}{2} - \dfrac{83}{30} + 5 = -\dfrac{15}{30} - \dfrac{83}{30} + \dfrac{150}{30} = \dfrac{52}{30} = \dfrac{26}{15} \).
    RHS: \( \dfrac{1}{5} + 2\left(\dfrac{83}{30} - \dfrac{60}{30}\right) = \dfrac{3}{15} + 2\left(\dfrac{23}{30}\right) = \dfrac{3}{15} + \dfrac{46}{30} = \dfrac{3}{15} + \dfrac{23}{15} = \dfrac{26}{15} \). Both sides are equal.

Links and References

• Solve Equations, Systems of Equations and Inequalities
• Find Lowest Common Multiple
• Middle School Math (Grades 6, 7, 8, 9) - Free Questions and Problems
• High School Math (Grades 10, 11 and 12) - Free Questions and Problems
• Primary Math (Grades 4 and 5) - Free Questions and Problems