1) $\frac{3}{4}x = \frac{1}{4}$
Multiply both sides by the reciprocal of $\frac{3}{4}$, which is $\frac{4}{3}$:
$x = \frac{1}{4} \cdot \frac{4}{3} = \frac{1}{3}$
2) $\frac{7}{2} = \frac{c}{5}$
Multiply both sides by 5:
$5 \cdot \frac{7}{2} = c$
$c = \frac{35}{2} = 17.5$
3) $\frac{6}{5} = \frac{n}{\left(\frac{3}{8}\right)}$
This means $\frac{6}{5} = \frac{n}{3/8} = n \cdot \frac{8}{3}$
So, $n = \frac{6}{5} \cdot \frac{3}{8} = \frac{18}{40} = \frac{9}{20}$
4) $-\frac{2}{3} = \frac{1}{2}m$
Multiply both sides by 2:
$-\frac{4}{3} = m$
5) $\frac{7}{6} = \frac{7}{3}p$
Divide both sides by $\frac{7}{3}$, or multiply by $\frac{3}{7}$:
$p = \frac{7}{6} \cdot \frac{3}{7} = \frac{3}{6} = \frac{1}{2}$
6) $\frac{8}{17} = \frac{5}{7}$
This is not a solvable equation as it's not true. However, if it's meant to be $\frac{8}{17} = \frac{5}{7}x$, then:
$x = \frac{8}{17} \cdot \frac{7}{5} = \frac{56}{85}$
7) $\frac{n}{\left(\frac{4}{9}\right)} = \frac{10}{9}$
This means $n \cdot \frac{9}{4} = \frac{10}{9}$
So, $n = \frac{10}{9} \cdot \frac{4}{9} = \frac{40}{81}$
8) $\frac{2}{3}x = \frac{8}{3}$
Multiply both sides by $\frac{3}{2}$:
$x = \frac{8}{3} \cdot \frac{3}{2} = \frac{24}{6} = 4$
9) $\frac{2}{7} = \frac{1}{8}x$
Multiply both sides by 8:
$\frac{16}{7} = x$
10) $-\frac{9}{10} = \frac{1}{3}x$
Multiply both sides by 3:
$-\frac{27}{10} = x$
Parent Tip: Review the logic above to help your child master the concept of solving addition and subtraction equations worksheet.
