Linear Equations in One Variable — NCERT Solutions, AI Tutor & Practice

We have $3x - 5 = 16$.

Add $5$ to both sides: $3x = 16 + 5 = 21$

Divide both sides by $3$: $x = \frac{21}{3} = 7$

Verification: $3(7) - 5 = 21 - 5 = 16$ ✓

Hence $x = 7$.

We have $5y + 3 = 2y + 18$.

Subtract $2y$ from both sides: $5y - 2y + 3 = 18$ $3y + 3 = 18$

Subtract $3$ from both sides: $3y = 15$

Divide by $3$: $y = 5$

Verification: LHS $= 5(5)+3 = 28$; RHS $= 2(5)+18 = 28$ ✓

We have $\dfrac{2x}{3} + 1 = \dfrac{7x}{15} + 3$.

The LCM of $3$ and $15$ is $15$. Multiply every term by $15$: $15 \cdot \frac{2x}{3} + 15 \cdot 1 = 15 \cdot \frac{7x}{15} + 15 \cdot 3$ $10x + 15 = 7x + 45$

Subtract $7x$ and $15$ from both sides: $3x = 30$ $x = 10$

Verification: LHS $= \frac{20}{3}+1 = \frac{23}{3}$; RHS $= \frac{70}{15}+3 = \frac{14}{3}+3 = \frac{23}{3}$ ✓

Expand both sides: $3t - 6 = 5t + 5 - 4$ $3t - 6 = 5t + 1$

Bring variable terms to one side and constants to the other: $3t - 5t = 1 + 6$ $-2t = 7$ $t = -\frac{7}{2}$

Verification: LHS $= 3\left(-\frac{7}{2}-2\right) = 3\left(-\frac{11}{2}\right) = -\frac{33}{2}$; RHS $= 5\left(-\frac{7}{2}+1\right)-4 = 5\left(-\frac{5}{2}\right)-4 = -\frac{25}{2}-4 = -\frac{33}{2}$ ✓

Cross-multiply: $3(x - 2) = 4(x + 1)$

Expand: $3x - 6 = 4x + 4$

Rearrange: $3x - 4x = 4 + 6$ $-x = 10$ $x = -10$

Verification: LHS $= \frac{-10-2}{4} = \frac{-12}{4} = -3$; RHS $= \frac{-10+1}{3} = \frac{-9}{3} = -3$ ✓

Let the three consecutive integers be $n$, $n+1$, and $n+2$.

Their sum is $72$: $n + (n+1) + (n+2) = 72$ $3n + 3 = 72$ $3n = 69$ $n = 23$

So the integers are $23$, $24$, and $25$.

Verification: $23 + 24 + 25 = 72$ ✓

Let the son's present age be $x$ years. Then the father's present age is $3x$ years.

After 12 years:

• Son's age $= x + 12$
• Father's age $= 3x + 12$

Given that the father will then be twice the son's age: $3x + 12 = 2(x + 12)$ $3x + 12 = 2x + 24$ $3x - 2x = 24 - 12$ $x = 12$

Son's present age $= 12$ years; father's present age $= 3 \times 12 = 36$ years.

Verification: After 12 years, son $= 24$, father $= 48$, and $48 = 2 \times 24$ ✓

Let the numbers be $5k$ and $3k$.

Their difference is $18$: $5k - 3k = 18$ $2k = 18$ $k = 9$

So the numbers are $5k = 45$ and $3k = 27$.

Verification: $45 : 27 = 5 : 3$ and $45 - 27 = 18$ ✓

The LCM of $3$ and $2$ is $6$. Multiply every term by $6$: $6 \cdot \frac{2x+5}{3} - 6 \cdot \frac{x-1}{2} = 6 \cdot 4$ $2(2x+5) - 3(x-1) = 24$

Expand: $4x + 10 - 3x + 3 = 24$ $x + 13 = 24$ $x = 11$

Verification: $\frac{2(11)+5}{3} - \frac{11-1}{2} = \frac{27}{3} - \frac{10}{2} = 9 - 5 = 4$ ✓

Let the breadth be $b$ cm. Then the length is $(b + 4)$ cm.

Perimeter of a rectangle $= 2(\text{length} + \text{breadth})$: $2\big((b + 4) + b\big) = 52$ $2(2b + 4) = 52$ $4b + 8 = 52$ $4b = 44$ $b = 11$

So breadth $= 11$ cm and length $= 11 + 4 = 15$ cm.

Verification: Perimeter $= 2(15 + 11) = 2(26) = 52$ cm ✓

Let the number be $x$. According to the problem: $\frac{3}{2}x - \frac{2}{5} = \frac{4}{5}$

Add $\frac{2}{5}$ to both sides: $\frac{3}{2}x = \frac{4}{5} + \frac{2}{5} = \frac{6}{5}$

Multiply both sides by $\frac{2}{3}$: $x = \frac{6}{5} \times \frac{2}{3} = \frac{12}{15} = \frac{4}{5}$

Verification: $\frac{3}{2} \cdot \frac{4}{5} - \frac{2}{5} = \frac{12}{10} - \frac{2}{5} = \frac{6}{5} - \frac{2}{5} = \frac{4}{5}$ ✓

Cross-multiply: $3(5x - 3) = 4(2x + 1)$

Expand both sides: $15x - 9 = 8x + 4$

Bring like terms together: $15x - 8x = 4 + 9$ $7x = 13$ $x = \frac{13}{7}$

Verification: LHS $= \frac{5(13/7)-3}{2(13/7)+1} = \frac{65/7 - 21/7}{26/7 + 7/7} = \frac{44/7}{33/7} = \frac{44}{33} = \frac{4}{3}$ ✓