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Complete the addition and subtraction boxes:

Mathematics

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Step-by-step solution

15 steps

Here are the completed addition and subtraction boxes:

Part a· 5 steps

  1. Calculate the sum for the first row.

    54+74=5+74=124=3\frac{5}{4} + \frac{7}{4} = \frac{5+7}{4} = \frac{12}{4} = 3

  2. Calculate the sum for the second row.

    34+54=3+54=84=2\frac{3}{4} + \frac{5}{4} = \frac{3+5}{4} = \frac{8}{4} = 2

  3. Calculate the difference for the first column.

    54−34=5−34=24=12\frac{5}{4} - \frac{3}{4} = \frac{5-3}{4} = \frac{2}{4} = \frac{1}{2}

  4. Calculate the difference for the second column.

    74−54=7−54=24=12\frac{7}{4} - \frac{5}{4} = \frac{7-5}{4} = \frac{2}{4} = \frac{1}{2}

  5. Calculate the value for the bottom-right cell by adding the results from Step 3 and Step 4.

    12+12=1+12=22=1\frac{1}{2} + \frac{1}{2} = \frac{1+1}{2} = \frac{2}{2} = 1

    The completed box for (a) is:

    [547433454212121]\boxed{ \begin{bmatrix} \frac{5}{4} & \frac{7}{4} & 3 \\ \frac{3}{4} & \frac{5}{4} & 2 \\ \frac{1}{2} & \frac{1}{2} & 1 \end{bmatrix} }

Part b· 5 steps

  1. Simplify 520\frac{5}{20} to 14\frac{1}{4}.

    Calculate the sum for the first row.

    15+310=210+310=510=12\frac{1}{5} + \frac{3}{10} = \frac{2}{10} + \frac{3}{10} = \frac{5}{10} = \frac{1}{2}

  2. Calculate the sum for the second row.

    310+14=620+520=1120\frac{3}{10} + \frac{1}{4} = \frac{6}{20} + \frac{5}{20} = \frac{11}{20}

  3. Calculate the difference for the first column.

    15−310=210−310=−110\frac{1}{5} - \frac{3}{10} = \frac{2}{10} - \frac{3}{10} = -\frac{1}{10}

  4. Calculate the difference for the second column.

    310−14=620−520=120\frac{3}{10} - \frac{1}{4} = \frac{6}{20} - \frac{5}{20} = \frac{1}{20}

  5. Calculate the value for the bottom-right cell by adding the results from Step 3 and Step 4.

    −110+120=−220+120=−120-\frac{1}{10} + \frac{1}{20} = -\frac{2}{20} + \frac{1}{20} = -\frac{1}{20}

    The completed box for (b) is:

    [1531012310141120−110120−120]\boxed{ \begin{bmatrix} \frac{1}{5} & \frac{3}{10} & \frac{1}{2} \\ \frac{3}{10} & \frac{1}{4} & \frac{11}{20} \\ -\frac{1}{10} & \frac{1}{20} & -\frac{1}{20} \end{bmatrix} }

Part c· 5 steps

  1. Calculate the sum for the first row.

    23+43=2+43=63=2\frac{2}{3} + \frac{4}{3} = \frac{2+4}{3} = \frac{6}{3} = 2

  2. Calculate the sum for the second row.

    13+23=1+23=33=1\frac{1}{3} + \frac{2}{3} = \frac{1+2}{3} = \frac{3}{3} = 1

  3. Calculate the difference for the first column.

    23−13=2−13=13\frac{2}{3} - \frac{1}{3} = \frac{2-1}{3} = \frac{1}{3}

  4. Calculate the difference for the second column.

    43−23=4−23=23\frac{4}{3} - \frac{2}{3} = \frac{4-2}{3} = \frac{2}{3}

  5. Calculate the value for the bottom-right cell by adding the results from Step 3 and Step 4.

    13+23=1+23=33=1\frac{1}{3} + \frac{2}{3} = \frac{1+2}{3} = \frac{3}{3} = 1

    The completed box for (c) is:

    [234321323113231]\boxed{ \begin{bmatrix} \frac{2}{3} & \frac{4}{3} & 2 \\ \frac{1}{3} & \frac{2}{3} & 1 \\ \frac{1}{3} & \frac{2}{3} & 1 \end{bmatrix} }
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