Rearranging Equations higher

This article explores rearranging much complex equations than the previous article. Prior knowledge of what is covered in the previous article is essential.

Rearranging 5x + a = 3x + b

Example: Rearrange the following equation by making x the subject.

5x + a = 3x + b
Answer:

    Explanation:

    Notice that in this equation x appears twice on both sides. We need to make sure it appears once to make it the subject. There is a few basic steps that we have to follow.

    Step 1: Subtract 3x from both sides

    x has to appear on one side of the equation. Therefore we get rid of any terms which contain x on the right hand side of the equation. There is a 3x on the right hand side of the equation, to get rid of it we subtract 3x from both sides of the equation.

    5x – 3x + a= 3x – 3x + b

    This operation should provide the following result;

    2x + a = b
    Step 2: Subtract a from both sides

    The term with x is not a lone yet. We still have to get rid of a on the left hand side of the equation. Subtract a from both sides to get;

    2x = b -a
    Step 3: Divide both sides by 2

    To make sure that the letter x is left a alone without a coefficient we divide both sides of the equation by 2

    x = b – a/2

    x is now the subject of the equation.

    This section explores rearranging equations of the form px + a = qx + b

    Rearrange px + a = qx + b

    Example: Rearrange the following equation to by making x the subject.

    px + a = qx + b
    Answer:

      Explanation:

      To work this out we have to follow a few basic steps just like in the previous example. We have to make sure that x is the only letter left standing on one side.

      Step 1: Subtract a from both sides

      First subtract a from both sides of the equation

      px + a – a = qx + b – a

      The operation should provide the following result;

      px= qx + b – a
      Step 2: Subtract qx from both sides

      x still appears on both sides of the equation to transfer the term with from the right hand side we subtract qx from both sides.

      px – qx = qx – qx + b – a

      By subtracting we’re making sure that x appears on one side. This operation should provide;

      px – qx = qx + b – a
      Step 3: Factorise the left hand side

      Factorisation can be used here to take out factors of x in the left hand side of the equation. This will ensure that only 1 letter of x appears in the left hand side.

      x(p – q) = b – a
      Step 4: Divide both sides by (p – q)

      The multiplication by (p – q) can be undone by dividing both sides by (p – q). This operation should provide the final results.

      x = b – a/p – q

      Rearrange p = (ab + c)(b – d)

      This section explores rearranging equations of the form p=(ab + c)(b -d).

      Example: Rearrange the following equation by making b the subject.

      p = ab + c/b – d
      Answer:

        Explanation:

        The equation looks complicated and difficult to transform. We simply just keep undoing certain sections in the equation for example the fraction. We then continue through by following very basic steps.

        Step 1: Multiply both sides by (b – d)

        It is always easier to undo the fractions first. To undo the fraction multiply both sides by (b – d). We can use cancelling to get rid of a similar numerator with denominator

        p(b – d) = ab + c/(b – d) × (b – d)

        The operation provides a simplified version of the equation.

        p(b – d) = ab + c
        Step 2: Expand brackets

        We have to get rid of the brackets by expanding the left hand side so that b is outside.

        pb – pd = ab + c
        Step 3: Collect like terms

        We must collect like terms to make sure that the terms with the letter b appear on one side of the equation. We will have to add pd on both sides of the equation furst. This should get rid of pd on the left hand side of the equation.

        pb – pd + pd = ab + pd+ c

        Next subtract ab from both sides of the equation. This will make sure that ab appears on the left hand side of the equation.

        pb – ab = ab + pd – ab + c

        These two operations should provide the following equation

        pb – ab = pd + c
        Step 4: Factorise the left hand side

        Factorising should help make the letter b stand out on the left hand side.

        b(p – a) = pd + c
        Step 5: Divide both sides by (p – a)

        The operation of dividing both sides both sides by (p-a) will make sure that b is the only letter left standing on the left hand side of the equation.

        b = pd + c/p – a

        This section explores rearranging equations in the form a = b + bc/d.

        Rearranging a = b + bc/3

        Example: Make b the subject in the following equation;

        a = b + bc/3
        Answer:

          Explanation:

          Step 1:Multiply both sides by 3

          It is always a good idea when changing the subject of an equation to get rid of the fraction first. In this equation to get rid of the fraction We multiply both sides by 3;

          a×3 = 3b×3 + bc/3×3

          The operation provides the following equation.

          3a = 3b + bc
          Step 2: Factorise the right hand side

          We have to make sure that the new subject stands out in the right hand side of the equation. This can be done by using factorisation.

          3a = b(3 + c)
          Step 3: Divide both sides by (3 + c)

          b has to be the only letter left standing on the right hand side of the equation. We must divide both sides by (3 + c) for this to happen;

          3a/3 + c = b(3 + c)/(3 + c)

          This operation should provide the following results.

          b = 3a/3 + c

          Rearranging 1/f = 1/u + 1/v

          Example:Make u the subject in the following equation.

          1/f = 1/u + 1/v
          Answer:

            Explanation:

            The equation appears to be tricky since there is three functions to work with. We multiply in turn by f then u then v to get rid of the fraction. First we multiply through by f

            Step 1: Mutiply both sides by f

            To get rid of the fraction that contains f we multiply both sides of the equation by f.

            1/f × f = 1/u × f + 1/v × f

            This operation should provide the following equation

            1 = f/u + f/v
            Step 2: Multiply through by u

            To get rid of the fraction that contains the letter u in the numerator we must multiply through by the letter u.

            1 × u = f/u × u + f/v × u

            This operation should provide the following equation.

            uv = vf + uf
            Step 3: Subtract uf from both sides

            All the terms that contain the letter u must appear on one side of the equation. To do this we must subtract uf from both sides of the equation,

            uv – uf = vf + uf – uf
            Step 4: Factorise the left hand side

            The letter u has to stand out on the left hand side. To make it stand out we must use factorisation.

            u(v – f) = vf
            Step 5: Divide both sides by (v – f)

            To make sure that u is the only letter left standing on one side of the equation we must divide both sides by (v – f)

            u(v – f)/(v – f) = vf/(v – f)

            This operation should provide the following answer.

            u = vf/(v -f)

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            2 Responses

            1. kate says:

              How would you solve this equation 2/t = t/5t-12?

            1. August 2, 2013

              […] The following shows an equation which involves quadratics; Make x the subject… In the above example we have a square. First we have to get rid of the fraction as we did before. We do this by dividing both sides by e or cross multipliying. We multiply e with b and ax2 + f with 1 to get the following result. Now we must leave x2 on its own since we’re trying to make x the subject. First we subtract f from both sides. Now we divide both sides by a as shown below; That leaves; To make x2 x we must square root it, this means we must square root both sides if they’re to remain equal. This results in; That concludes this chapter on rearranging equations. If you would like to explore further with much harder example click here. […]

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