Table 1 Assessment problems, with standard algorithms and situationally appropriate methods

1

\(4\left(x-2\right)=24\)

Begin by distributing the parentheses

\(4x-4\cdot 2=24\)

Divide a constant to both sides before distributing

\(x-2=\frac{24}{4}\)

2

\(3\left(x+0.69\right)=15\)

\(3x+3\cdot 0.69=15\)

\(x+ 0.69=\frac{15}{3}\)

3

\(4\left(x+ \frac{3}{5}\right)=12\)

\(4x+4\cdot \frac{3}{5}=12\)

\(x+\frac{3}{5}= \frac{12}{4}\)

4

\(4\left(x+6\right)+3\left(x+6\right)=21\)

Begin by distributing the parentheses

\(4x+4\cdot 6+3x+3\cdot 6=21\)

Change in variable – combine

\(7\left(x+6\right)=21\)

5

\(5\left(x+ \frac{3}{7}\right)+3\left(x+ \frac{3}{7}\right)=16\)

\(5x+5\cdot \frac{3}{7}+3x+3\cdot \frac{3}{7}=16\)

\(8\left(x+\frac{3}{7}\right)=16\)

6

\(2\left(x-0.31\right)+3\left(x-0.31\right)=15\)

\(2x-2\cdot 0.31+3x-3\cdot 0.31=15\)

\(5\left(x-0.31\right)=15\)

7

\(8\left(x-5\right)=3\left(x-5\right)+20\)

Begin by distributing the parentheses

\(8x-8\cdot 5=3x-3\cdot 5+20\)

Change in variable – subtract from both

\(5\left(x-5\right)=20\)

8

\(8\left(x-\frac{2}{5}\right)-11=6\left(x-\frac{2}{5}\right)\)

\(8x- \frac{16}{5} -11=6x-\frac{12}{5}\)

\(2\left(x-\frac{2}{5}\right) =11\)

9

\(5\left(x+ 0.6\right)+3x=5\left(x+ 0.6\right)+7\)

\(5x+3+3x=5x+3+7\)

\(3 x =7\)

10

\(\frac{2x-6}{2}+ \frac{6x-18}{3}=5\)

Begin by obtaining a common denominator for the two expressions

\(\frac{3\left(2x-6\right)}{2\cdot 3}+ \frac{2\left(6x-18\right)}{2\cdot 3}=5\)

Reducing each fraction before combining

\(\left(x-3\right)+\left(2x-6\right)=5\)

11

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

\(\frac{3\cdot \left(x+3\right)}{3\cdot 3}+ \frac{3x+9}{9}=1\)

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

12

\(\frac{5x+5}{5}+ \frac{6x+6}{6}=6\)

\(\frac{6\left(5x+5\right)}{6\cdot 5}+ \frac{5\left(6x+6\right)}{5\cdot 6}=6\)

\(\left(x+1\right)+\left(x+1\right)=6\)