When working with fractions built into linear equations, it is often easiest to remove the fraction in the very first step. This generally means finding the LCD of the fraction and then multiplying every term in the entire equation by the LCD.

Questions

For questions 1 to 18, solve each linear equation.

1. ${\displaystyle \frac{3}{5}}(1+p)={\displaystyle \frac{21}{20}}$
2. $-{\displaystyle \frac{1}{2}}={\displaystyle \frac{3k}{2}}+{\displaystyle \frac{3}{2}}$
3. $0=-{\displaystyle \frac{5}{4}}(x-{\displaystyle \frac{6}{5}})$
4. ${\displaystyle \frac{3}{2}}n-8=-{\displaystyle \frac{29}{12}}$
5. ${\displaystyle \frac{3}{4}}-{\displaystyle \frac{5}{4}}m={\displaystyle \frac{108}{24}}$
6. ${\displaystyle \frac{11}{4}}+{\displaystyle \frac{3}{4}}r={\displaystyle \frac{160}{32}}$
7. $2b+{\displaystyle \frac{9}{5}}=-{\displaystyle \frac{11}{5}}$
8. ${\displaystyle \frac{3}{2}}-{\displaystyle \frac{7}{4}}v=-{\displaystyle \frac{9}{8}}$
9. ${\displaystyle \frac{3}{2}}({\displaystyle \frac{7}{3}}n+1)={\displaystyle \frac{3}{2}}$
10. ${\displaystyle \frac{41}{9}}={\displaystyle \frac{5}{2}}(x+{\displaystyle \frac{2}{3}})-{\displaystyle \frac{1}{3}}x$
11. $-a-{\displaystyle \frac{5}{4}}(-{\displaystyle \frac{8}{3}}a+1)=-{\displaystyle \frac{19}{4}}$
12. ${\displaystyle \frac{1}{3}}(-{\displaystyle \frac{7}{4}}k+1)-{\displaystyle \frac{10}{3}}k=-{\displaystyle \frac{13}{8}}$
13. ${\displaystyle \frac{55}{6}}=-{\displaystyle \frac{5}{2}}({\displaystyle \frac{3}{2}}p-{\displaystyle \frac{5}{3}})$
14. $-{\displaystyle \frac{1}{2}}({\displaystyle \frac{2}{3}}x-{\displaystyle \frac{3}{4}})-{\displaystyle \frac{7}{2}}x=-{\displaystyle \frac{83}{24}}$
15. $-{\displaystyle \frac{5}{8}}={\displaystyle \frac{5}{4}}(r-{\displaystyle \frac{3}{2}})$
16. ${\displaystyle \frac{1}{12}}={\displaystyle \frac{4}{3}}x+{\displaystyle \frac{5}{3}}(x-{\displaystyle \frac{7}{4}})$
17. $-{\displaystyle \frac{11}{3}}+{\displaystyle \frac{3}{2}}b={\displaystyle \frac{5}{2}}(b-{\displaystyle \frac{5}{3}})$
18. ${\displaystyle \frac{7}{6}}-{\displaystyle \frac{4}{3}}n=-{\displaystyle \frac{3}{2}}n+2(n+{\displaystyle \frac{3}{2}})$

Answer Key 2.4