Solving Linear Equations Practice

Overview of Solving Linear Equations

• Classifications of Linear Equation Solutions:

  • One Unique Solution: Occurs when algebraic manipulation isolates the variable to a single numerical value (e.g., $x = a$).

  • No Solution: Occurs when algebraic manipulation causes all variable terms to cancel out, leaving a false statement (e.g., $a = b$ where $a \neq b$).

  • Infinitely Many Solutions (Identity): Occurs when algebraic manipulation causes all variable terms to cancel out, leaving a true statement (e.g., $a = a$).

• Problem Set Reference:

Problem 22: $x + 6 = x$

• Original Equation:

  $x + 6 = x$

• Step-by-Step Solution:

  • Subtract $x$ from both sides of the equation:

    $x + 6 - x = x - x$

• Simplify both sides:

    $6 = 0$

• Conclusion:

  • The resulting statement $6 = 0$ is false for all values of $x$.

  • Therefore, the equation has no solution.

Problem 23: $3x - 1 = 1 - 3x$

• Original Equation:

  $3x - 1 = 1 - 3x$

• Step-by-Step Solution:

  • Add $3x$ to both sides of the equation to collect variable terms on one side:

    $3x + 3x - 1 = 1 - 3x + 3x$

    $6x - 1 = 1$

• Add $1$ to both sides of the equation to isolate the constant term:

    $6x - 1 + 1 = 1 + 1$

    $6x = 2$

• Divide both sides by $6$:

    $\frac{6x}{6} = \frac{2}{6}$

    $x = \frac{1}{3}$

• Verification / Check:

  • Substitute $x = \frac{1}{3}$ into the original equation:

    $3\left(\frac{1}{3}\right) - 1 = 1 - 3\left(\frac{1}{3}\right)$

    $1 - 1 = 1 - 1$

    $0 = 0$

• The left side equals the right side, confirming that $x = \frac{1}{3}$ is the correct solution.

Problem 24: $3x + 15 = 3(x + 5)$

• Original Equation:

  $3x + 15 = 3(x + 5)$

• Step-by-Step Solution:

  • Apply the distributive property to expand the right side of the equation:

    $3(x + 5) = 3(x) + 3(5) = 3x + 15$

• Rewrite the equation:

    $3x + 15 = 3x + 15$

• Subtract $3x$ from both sides:

    $15 = 15$

• Conclusion:

  • The resulting statement $15 = 15$ is an identity and is true for all real values of $x$.

  • Therefore, the equation has infinitely many solutions (all real numbers).

Problem 25: $4x - 9 = 3.5x - 9$

• Original Equation:

  $4x - 9 = 3.5x - 9$

• Step-by-Step Solution:

  • Add $9$ to both sides of the equation:

    $4x - 9 + 9 = 3.5x - 9 + 9$

    $4x = 3.5x$

• Subtract $3.5x$ from both sides to collect terms containing $x$:

    $4x - 3.5x = 3.5x - 3.5x$

    $0.5x = 0$

• Divide both sides by $0.5$:

    $x = 0$

• Verification / Check:

  • Substitute $x = 0$ into the original equation:

    $4(0) - 9 = 3.5(0) - 9$

    $-9 = -9$

• The statement is true, confirming that $x = 0$ is the solution.

Problem 26: \$\frac{1}{3}(9x + 3) = 3x + 1$

• Original Equation:

  $\frac{1}{3}(9x + 3) = 3x + 1$

• Step-by-Step Solution:

  • Apply the distributive property to expand the left side of the equation:

    $\frac{1}{3}(9x) + \frac{1}{3}(3) = 3x + 1$

    $3x + 1 = 3x + 1$

• Subtract $3x$ from both sides:

    $1 = 1$

• Conclusion:

  • The statement $1 = 1$ is true for all values of $x$.

  • Therefore, the equation has infinitely many solutions (all real numbers).

Problem 27: $5x - 7 = 4x - 1$

• Original Equation:

  $5x - 7 = 4x - 1$

• Step-by-Step Solution:

  • Subtract $4x$ from both sides to gather variable terms on the left:

    $5x - 4x - 7 = 4x - 4x - 1$

    $x - 7 = -1$

• Add $7$ to both sides to isolate $x$:

    $x - 7 + 7 = -1 + 7$

    $x = 6$

• Verification / Check:

  • Substitute $x = 6$ into the original equation:

    $5(6) - 7 = 4(6) - 1$

    $30 - 7 = 24 - 1$

    $23 = 23$

• The statement is true, confirming that $x = 6$ is the correct solution.

Problem 28: \$\frac{1}{2}x + \frac{1}{2}x = x + 1$

• Original Equation:

  $\frac{1}{2}x + \frac{1}{2}x = x + 1$

• Step-by-Step Solution:

  • Combine like terms on the left side of the equation:

    $\left(\frac{1}{2} + \frac{1}{2}\right)x = 1x = x$

• Rewrite the equation:

    $x = x + 1$

• Subtract $x$ from both sides:

    $x - x = x - x + 1$

    $0 = 1$

• Conclusion:

  • The statement $0 = 1$ is false for all values of $x$.

  • Therefore, the equation has no solution.

Problem 29: $2x + 4 = -(-7x + 4)$

• Original Equation:

  $2x + 4 = -(-7x + 4)$

• Step-by-Step Solution:

  • Distribute the negative sign on the right side of the equation:

    $-(-7x + 4) = 7x - 4$

• Rewrite the equation:

    $2x + 4 = 7x - 4$

• Subtract $2x$ from both sides:

    $4 = 7x - 2x - 4$

    $4 = 5x - 4$

• Add $4$ to both sides:

    $4 + 4 = 5x$

    $8 = 5x$

• Divide both sides by $5$:

    $x = \frac{8}{5}$

• Verification / Check:

  • Substitute $x = \frac{8}{5}$ into the original equation:

    $2\left(\frac{8}{5}\right) + 4 = -\left(-7\left(\frac{8}{5}\right) + 4\right)$

    $\frac{16}{5} + \frac{20}{5} = -\left(-\frac{56}{5} + \frac{20}{5}\right)$

    $\frac{36}{5} = -\left(-\frac{36}{5}\right)$

    $\frac{36}{5} = \frac{36}{5}$

• The statement is true, confirming that $x = \frac{8}{5}$ (or $1.6$) is the correct solution.

Problem 30: $5.5 - x = -4.5 - x$

• Original Equation:

  $5.5 - x = -4.5 - x$

• Step-by-Step Solution:

  • Add $x$ to both sides of the equation:

    $5.5 - x + x = -4.5 - x + x$

    $5.5 = -4.5$

• Conclusion:

  • The statement $5.5 = -4.5$ is false for all values of $x$.

  • Therefore, the equation has no solution.

Problem 31: $-3(2x - 3) = -6x + 9$

• Original Equation:

  $-3(2x - 3) = -6x + 9$

• Step-by-Step Solution:

  • Apply the distributive property to expand the left side of the equation:

    $-3(2x) - 3(-3) = -6x + 9$

• Rewrite the equation:

    $-6x + 9 = -6x + 9$

• Add $6x$ to both sides:

    $9 = 9$

• Conclusion:

  • The statement $9 = 9$ is an identity and is true for all real values of $x$.

  • Therefore, the equation has infinitely many solutions (all real numbers).

Problem 32: $10x - \frac{8}{3} - 4x = 6x$

• Original Equation:

  $10x - \frac{8}{3} - 4x = 6x$

• Step-by-Step Solution:

  • Combine like terms on the left side of the equation:

    $(10x - 4x) - \frac{8}{3} = 6x$

    $6x - \frac{8}{3} = 6x$

• Subtract $6x$ from both sides:

    $-\frac{8}{3} = 0$

• Conclusion:

  • The statement $-\frac{8}{3} = 0$ is false for all values of $x$.

  • Therefore, the equation has no solution.

Problem 33: $6(7x + 7) = 7(6x + 6)$

• Original Equation:

  $6(7x + 7) = 7(6x + 6)$

• Step-by-Step Solution:

  • Apply the distributive property to expand both sides of the equation:

  • Left side:

      $6(7x) + 6(7) = 42x + 42$

- Right side:

      $7(6x) + 7(6) = 42x + 42$

• Rewrite the equation:

    $42x + 42 = 42x + 42$

• Subtract $42x$ from both sides:

    $42 = 42$

• Conclusion:

  • The statement $42 = 42$ is true for all real values of $x$.

  • Therefore, the equation has infinitely many solutions (all real numbers).