Class 8 Linear Equations In One Variable CBSE Questions & Answers

Solve: \(\frac{{2x}}{3} + 1 = \frac{{7x}}{{15}} + 3\)

Solve: \(2y + \frac{5}{3} = \frac{{26}}{3} - y\)

Solve: \(3m = 5m - \frac{8}{5}\)

Solve: \(\frac{x}{2} - \frac{1}{5} = \frac{x}{3} + \frac{1}{4}\)

Solve: \(\frac{n}{2} - \frac{{3n}}{4} + \frac{{5n}}{6} = 21\)

Solve: \(x + 7 - \frac{{8x}}{3} = \frac{{17}}{6} - \frac{{5x}}{2}\)

Solve: \(\frac{{x - 5}}{3} = \frac{{x - 3}}{5}\)

Solve: \(\frac{{3x - 2}}{4} - \frac{{2x + 3}}{3} = \frac{2}{3} - x\)

Solve: \(a - \frac{{a - 1}}{2} = 1 - \frac{{a - 2}}{3}\)

Solve: \({\text{3 }}\left( {t--{\text{ 3}}} \right){\text{ }} = {\text{ 5 }}\left( {{\text{2}}t + {\text{ 1}}} \right)\)

Solve: \({\text{15 }}\left( {y--{\text{ 4}}} \right){\text{ }}--{\text{2 }}\left( {y--{\text{ 9}}} \right){\text{ }} + {\text{ 5 }}\left( {y + {\text{ 6}}} \right){\text{ }} = {\text{ }}0\)

Solve: \({\text{3 }}\left( {{\text{5}}z--{\text{ 7}}} \right){\text{ }}--{\text{ 2 }}\left( {{\text{9}}z--{\text{ 11}}} \right){\text{ }} = {\text{ 4 }}\left( {{\text{8}}z--{\text{ 13}}} \right){\text{ }}--{\text{ 17}}\)

Solve: \(0.{\text{25 }}\left( {{\text{4}}m--{\text{ 3}}} \right){\text{ }} = {\text{ }}0.0{\text{5 }}\left( {{\text{1}}0m--{\text{ 9}}} \right)\)

Solve: \(\frac{{3x + 4}}{{2 - 6x}} = \frac{{ - 2}}{5}\).

The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is \(\frac{3}{2}\). Find the rational number.