Solve Literal Equations and Formulas

Questions on solving literal equations and fromulas are presented along with their detailed solutions. These questions may be used to practice for math placement tests to colleges.

  1. Solve the literal equation \( \quad 2 x + y = 5 + 3x \quad \) for \( x \).
    A) \( \quad x = 5 - y \)
    B) \( \quad x = y + 5 \)
    C) \( \quad x = y - 5 \)
    D) \( \quad x = - 5 - y \)
    E) \( \quad x = 5 y \)



  2. Solve the literal equation \( \quad 2 s n = s + n \quad \) for \( s \).
    A) \( \quad s = 2 - \dfrac{1}{n} \)
    B) \( \quad s = n(2n - 1) \)
    C) \( \quad s = \dfrac{n}{2 n + 1} \)
    D) \( \quad s = n \)
    E) \( \quad s = \dfrac{n}{2 n - 1} \)



  3. Solve the literal equation \( \quad \dfrac {2 + x}{y} = x + 1 \quad \) for \( y \).

    A) \( \quad \dfrac {x+2}{x+1} \)

    B) \( \quad \dfrac {x+1}{x+2} \)

    C) \( \quad \dfrac {x+2}{x-1} \)

    D) \( \quad \dfrac {x-2}{x-1} \)

    E) \( \quad \dfrac {x+2}{1-x} \)



  4. Solve the literal equation \( \quad n = \dfrac {6x + 11}{2x + 3} \quad \) for \( x \)

    A) \( \quad x = \dfrac{3n + 11}{6 - 2n} \)

    B) \( \quad x = 3 n \)

    C) \( \quad x = \dfrac{3n + 11}{6 + 2n} \)

    D) \( \quad x = \dfrac{ 11 - 3 n}{2 n - 6} \)

    E) \( \quad x = \dfrac{3n - 11}{6 + 2n} \)



  5. Solve Solve the literal equation \( \quad \dfrac{2 y + t}{y - t} = 4 \quad \) for \( y \)

    A) \( \quad y = (2 / 5) t \)
    B) \( \quad y = - (3 / 2) t \)
    C) \( \quad y = - (2 / 3) t \)
    D) \( \quad y = - (5 / 2) t \)
    E) \( \quad y = (5 / 2) t \)



  6. Solve Solve the literal equation \( \quad 2 M G + G / M = 2 / M \quad \) for \( G \).

    A) \( \quad G = 2 M^2 + 1 \)

    B)\( \quad G = \dfrac{2}{2 M^2 + 1} \)

    C)\( \quad G = \dfrac{1}{2 M^2 + 1} \)

    D)\( \quad G = \dfrac{2}{2 M^2 - 1} \)

    E)\( \quad G = 2 M^2 \)



  7. The perimeter \( P \) of a rectangle of length \( L \) and width \( W \) is given by \( \quad P = 2L + 2W \). Express \( L \) in terms of \( P \) and \( W \).

    A) \( \quad L = P - W \)
    B)\( \quad L = P - 2 W \)
    C)\( \quad L = P + 2 W \)
    D)\( \quad L = P/2 - W \)
    E)\( \quad L = P/2 + W \)



  8. The formula to convert temperatures from Fahrenheit \( F \) to Celsius \( C \) is given by: \[ C = \dfrac{5}{9} (F - 32) \] Solve the above formula for \( F \) to obtain a formula to convert Celsius \( C \) to Fahrenheit \( F \).

    A) \( \quad F = (9 / 5) C + 32 \)
    B) \( \quad F = (9 / 5) C - 32 \)
    C) \( \quad F = C + 32 \)
    D) \( \quad F = (5 / 9) C + 32 \)
    E) \( \quad F = (5 / 9) C - 32 \)



  9. The total surface area \( A \) of a cylinder of radius \( r \) and height \( H \) is given by \[ A = 2 \pi r^2 + 2 \pi r H \] Find a formula for \( H \) in terms of \( A \) and \( r \).

    A) \( \quad H = \dfrac{A}{2 \pi r} \)

    B) \( \quad H = A - 2 \pi r^2 - 2 \pi r \)

    C) \( \quad H = \dfrac{A}{2 \pi r} - r \)

    D) \( \quad H = \dfrac{A}{2 \pi r^2} - \pi r \)

    E) \( \quad H = \dfrac{A}{2 \pi r} r \)



  10. Solve the equation \( \quad \dfrac{z+y}{x y z } = 2 x \) for \( y \).

    A) \( \quad y = \dfrac{z}{2 x^2 + 1} \)

    B) \( \quad y = \dfrac{z}{2 x^2 - 1} \)

    C) \( \quad y = \dfrac{ - z}{2 x^2 + 1} \)

    D) \( \quad y = \dfrac{z}{2 x^2 - x} \)

    E) \( \quad y = \dfrac{z}{2 x^2 z - 1} \)



  11. Solve the literal equation \( \quad e^{x + 2 y -1 } = 2 \) for \( y \).

    A) \( \quad y = \dfrac{1 + \ln 2 - x}{2} \)

    B) \( \quad y = \dfrac{2 - e^{x-1}}{2} \)

    C) \( \quad y = \dfrac{3 - x}{2} \)

    D) \( \quad y = \dfrac{ - 1 + \ln 2 - x}{2} \)

    E) \( \quad y = (1/2) \dfrac{2} {e^{x -1 }} \)



  12. Solve the literal equation \( \quad \arcsin\left(e^x+y\right) = a \) for \( x \).

    A) \( \quad x = \arcsin(\ln ( a - y)) \)

    B) \( \quad x = \ln (\sin a - y) \)

    C) \( \quad x = \ln (\sin a + y) \)

    D) \( \quad x = \ln (\arcsin a - y) \)

    E) \( \quad x = \ln (a + y) \)





Answers and Solutions to the Above Questions

  1. Answer C
    Solution
    Given the equation: \( \quad 2 x + y = 5 + 3x \quad \)
    Rewrite the equation with the unknown \( x \) on one side: \( \quad y - 5 = 3x - 2x \quad \)
    Simplify and solve for \( x \): \( x = y - 5 \)

  2. Answer E
    Solution
    Given the equation: \( \quad 2 s n = s + n \quad \)
    Rewrite the equation with the unknown \( s \) on one side: \( \quad 2 s n - s = n \quad \)
    Factor and solve for \( s \): \( s = \dfrac{n}{2 n - 1} \)

  3. Answer A
    Solution
    Given the equation: \( \quad \dfrac {2 + x}{y} = x + 1 \)
    Multiply all terms by \( y \): \( \quad y \dfrac {2 + x}{y} = y( x + 1) \)
    Simplify: \( \quad 2 + x = y ( x + 1) \)
    Divide both sides of the equation by \( x + 1 \) and solve for \( y \): \( \quad y = \dfrac{x + 2}{x + 1} \)

  4. Answer D
    Solution
    Given the equation: \( \quad n = \dfrac {6x + 11}{2x + 3} \)
    Multiply all terms by \( 2x + 3 \): \( \quad (2x + 3) n = (2x + 3) \dfrac {6x + 11}{3x+3} \)
    Simplify and expand: \( \quad 2 x n + 3 n = 6x + 11 \)
    Rewrite the equation with all terms in \( x \) on one side: \( \quad 2 x n - 6x = 11 - 3 n \)
    Factor \( x \) on the left side: \( \quad x (2 n - 6) = 11 - 3 n \)
    Divide both sides of the equation by \( 2 n - 6 \):\( \quad \dfrac{x (2 n - 6)}{2 n - 6} = \dfrac{ 11 - 3 n}{2 n - 6} \)
    Simplify and solve for \( x \): \( \quad x = \dfrac{ 11 - 3 n}{2 n - 6} \)

  5. Answer E
    Solution
    Given the equation: \( \quad \dfrac{2 y + t}{y - t} = 4 \)
    Multiply both sides of the equation by \( (y - t) \): \( \quad (y - t) \dfrac{2 y + t}{y - t} = 4 (y - t) \)
    Simplify and expand: \( \quad 2 y + t = 4 y - 4 t \)
    Rewrite the equation with all terms in \( y \) on one side: \( \quad 5t = 2y \)
    Divide both sides of the equation by \( 2 \) and solve for y:\( \quad y = \dfrac{5 t}{2} \)

  6. Answer B
    Solution
    Given the equation: \( \quad 2 M G + G / M = 2 / M \)
    Multiply all terms in the equation by \( M \): \( \quad M (2 M G + G / M) = 2 M / M ) \)
    Simplify: \( \quad 2 M^2 G + G = 2 \)
    Factor \( G \) out on the left side: \( \quad G (2 M^2 + 1) = 2 \)
    Divide both sides of the equation by \( 2 M^2 + 1 \):\( \quad \dfrac{x (G (2 M^2 + 1))}{2 M^2 + 1} = \dfrac{ 2 }{2 M^2 + 1} \)
    Simplify and solve for \( G \): \( \quad G = \dfrac{ 2 }{2 M^2 + 1} \)

  7. Answer D
    Solution
    Given the formula: \( \quad P = 2L + 2W \)
    Rewrite the equation with the term in \( L \) on one side: \( \quad 2 L = P - 2 W\)
    Divide both sides of the equation by \( 2 \): \( \quad \dfrac{2 L }{2} = \dfrac{P - 2 W}{2}\)
    Simplify and solve for \( L \): \( \quad L = P/2 - W \)

  8. Answer A
    Solution
    Given the formula: \( \quad C = \dfrac{5}{9} (F - 32) \)

    Multiply both sides the equation by \( \dfrac{9}{5} \): \( \quad \dfrac{9}{5} C = \dfrac{9}{5} \dfrac{5}{9} (F - 32) \)
    Simplify: \( \quad \dfrac{9}{5} C = F - 32 \)
    Solve for \( F \): \( \quad F = \dfrac{9}{5} C + 32 \)

  9. Answer C
    Solution
    Given the formula: \( \quad A = 2 \pi r^2 + 2 \pi r H \)

    Rewrite the equation with the term with \( H \) on one side: \( \dfrac{9}{5} \): \( \quad A - 2 \pi r^2 = 2 \pi r H \)
    Divide both sides the equation by \( 2 \pi r \): \( \quad \dfrac{A - 2 \pi r^2}{ 2 \pi r} = \dfrac{ 2 \pi r H}{ 2 \pi r} \)
    Simplify and solve for \( H \): \( \quad H = \dfrac{A - 2 \pi r^2}{ 2 \pi r} = \dfrac{A} { 2 \pi r} - r\)

  10. Answer E
    Solution
    Given the literal equation: \( \quad \dfrac{z+y}{x y z } = 2 x \)
    Multiply both sides of the equation by \( x y z \): \( \quad \dfrac{z+y}{x y z } (x y z) = 2 x (x y z) \)
    Simplify and rewrite the equation as: \( \quad z + y = 2 x^2 y z \)
    Rewrite the above equation with all terms in \( y \) on one side: \( \quad y - 2 x^2 y z = - z\)
    Factor \( y \) out: \( \quad y(1 - 2 x^2 z) = - z\)
    Divide both sides by \( (1 - 2 x^2 z) \): \( \quad \dfrac{y(1 - 2 x^2 z)}{(1 - 2 x^2 z)} = \dfrac{- z}{(1 - 2 x^2 z)}\)

    Simplify and solve for \( y \): \( \quad y = \dfrac{- z}{(1 - 2 x^2 z)} = \dfrac{z}{(2 x^2 z - 1)}\)
  11. Answer A
    Solution
    Given the literal equation: \( \quad e^{x + 2 y -1 } = 2 \)
    Take the \( \ln \) of both sides of the equation: \( \quad \ln e^{x + 2 y -1 } = \ln 2 \)
    Simplify using the fact that \( \ln (e^u) = u \): \( \quad x + 2 y -1 = \ln 2 \)
    Rewrite the above equation with all terms in \( y \) on one side: \( \quad 2 y = 1 + \ln 2 - x\)
    Divide both sides by \( 2 \) and solve for \( y \): \( \quad y = \dfrac{1 + \ln 2 - x}{2} \)

  12. Answer B
    Solution
    Given the literal equation: \( \quad \arcsin\left(e^x+y\right) = a \)
    Take the \( \sin \) of both sides of the equation: \( \quad \sin (\arcsin\left(e^x+y\right)) = \sin a \)
    Simplify using the fact that \( \sin (\arcsin u) = u \): \( \quad e^x + y = \sin a \)
    Rewrite the above equation with all terms in \( x \) on one side: \( \quad e^x = \sin a - y\)
    Take the \( \ln \) of both sides: \( \quad \ln e^x = \ln (\sin a - y) \)
    Simplify and solve for \( x \): \( \quad x = \ln (\sin a - y) \)

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