Solutions to Equations - Grade 9

The solutions to the questions in solve equations are presented along with all the steps needed and detailed explanations.

Note: In what follows LHS means the evaluation of left hand side of the equation and RHS means evaluation of the right hand side of the equation.

Solutions to Questions in solve equations


  1. Given \[ 2x + 2 = 6 \] Subtract 2 from both sides: \[ 2x + 2 - 2 = 6 - 2 \] Simplify: \[ 2x = 4 \] Divide both sides by 2: \[ \dfrac{2x}{2} = \dfrac{4}{2} \] \[ x = 2 \] Check the solution obtained \[ \text{LHS: } 2x + 2 = 2(2) + 2 = 6 \] Both sides are equal, so \(x = 2\) is a solution.

  2. Given \[ 5y - 2 = 7y - 8 \] Subtract \(5y\) from both sides: \[ 5y - 2 - 5y = 7y - 8 - 5y \] Simplify: \[ -2 = 2y - 8 \] Add 8 to both sides: \[ -2 + 8 = 2y - 8 + 8 \] \[ 6 = 2y \] Divide both sides by 2: \[ y = 3 \] Check the solution obtained \[ \text{LHS: }5y - 2 = 5(3) - 2 = 13, \quad \text{RHS: } 7y - 8 = 7(3) - 8 = 13 \] Hence \(y = 3\) is a solution.

  3. Given \[ -2x + 4 + 5x = 7 + 4x - 3 \] Combine like terms: \[ 3x + 4 = 4 + 4x \] Subtract \(3x + 4\) from both sides: \[ 0 = x \] Check the solution obtained \[ \text{LHS: } -2(0) + 4 + 5(0) = 4, \quad \text{RHS: } 7 + 4(0) - 3 = 4 \] So \(x = 0\) is a solution.

  4. Given \[ 0.2d + 4 = -0.1d - 2 \] Add \(0.1d\) and subtract 4 from both sides: \[ 0.3d = -6 \] Divide both sides by 0.3: \[ d = -20 \] Check the solution obtained \[ \text{LHS: } 0.2(-20) + 4 = 0, \quad \text{RHS: } -0.1(-20) - 2 = 0 \] Hence \(d = -20\) is a solution.

  5. Given \[ -2(2x - 6) = -(x - 4) \] Expand using distributive law: \[ -4x + 12 = -x + 4 \] Add \(x\) to both sides: \[ -3x + 12 = 4 \] Subtract 12: \[ -3x = -8 \] Divide by -3: \[ x = \dfrac{8}{3} \] Check the solution obtained \[ \text{LHS: } -2(2(8/3)-6) = 4/3, \quad \text{RHS: } -(8/3-4) = 4/3 \] Hence \(x = 8/3\) is a solution.

  6. Given \[ -(x+2)+4 = 2(x+3)+x \] Expand and simplify: \[ -x - 2 + 4 = 2x + 6 + x \quad \Rightarrow \quad -x + 2 = 3x + 6 \] Add \(x\) and subtract 6: \[ -4 = 4x \] \[ x = -1 \] Check the solution obtained \[ \text{LHS: } -( -1 + 2 ) + 4 = 3, \quad \text{RHS: } 2(-1+3) + (-1) = 3 \] Hence \(x = -1\) is a solution.

  7. Given \[ \dfrac{x}{5} = -6 \] Multiply both sides by 5: \[ x = -30 \] Check the solution obtained \[ \dfrac{-30}{5} = -6 \] Hence \(x = -30\) is a solution.

  8. Given \[ - \dfrac{x}{3} = \dfrac{1}{2} \] Multiply both sides by -3: \[ x = -\dfrac{3}{2} \] Check the solution obtained \[ \text{LHS: } -(-3/2)/3 = 1/2 \] Hence \(x = -3/2\) is a solution.

  9. Given \[ - \dfrac{x}{4} = \dfrac{1}{2} - x \] Multiply both sides by 4: \[ -x = 2 - 4x \] Add \(4x\) to both sides: \[ 3x = 2 \] \[ x = \dfrac{2}{3} \] Check the solution obtained \[ \text{LHS: } -2/12 = -1/6, \quad \text{RHS: } 1/2 - 2/3 = -1/6 \] Hence \(x = 2/3\) is a solution.

  10. Given \[ - \dfrac{x-3}{7} = \dfrac{1}{2}(-2x + 6) \] Multiply both sides by 14: \[ -2(x-3) = 7(-2x+6) \] Expand: \[ -2x + 6 = -14x + 42 \] Add 14x to both sides: \[ 12x + 6 = 42 \] Subtract 6: \[ 12x = 36 \] \[ x = 3 \] Check the solution obtained \[ -(3-3)/7 = 0, \quad \frac{1}{2}(-6+6) = 0 \] Hence \(x = 3\) is a solution.

  11. Given \[ -\dfrac{1}{2} - x + 5 = \dfrac{1}{5} + 2(x-2) \] Multiply both sides by 10: \[ 10(-1/2 - x + 5) = 10(1/5 + 2(x-2)) \] Simplify: \[ -5 -10x + 50 = 2 + 20x - 40 \] Combine like terms: \[ -10x + 45 = 20x - 38 \] Add 10x: \[ 45 = 30x - 38 \] Add 38: \[ 83 = 30x \] \[ x = \dfrac{83}{30} \] Check the solution obtained \[ \text{LHS: } -1/2 - 83/30 + 5 = 26/15, \quad \text{RHS: } 1/5 + 2(83/30-2) = 26/15 \] Hence \(x = 83/30\) is a solution.

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