Solve the following linear equations for the variable: 3x = 2x + 18 5t − 3 = 3t − 5 4z + 3 = 6 + 2z 2x − 1 = 14 − x x = 4/5(x + 10) 2x/3 + 1 = 7x/15 + 3

$$3x - 2x = 18$$ $$x = 18$$

The solution is $\boxed{x = 18}$.

प्रश्न 2:

$$5t - 3 = 3t - 5$$

$$5t - 3t - 3 = -5$$ $$2t - 3 = -5$$

$$2t = -5 + 3$$ $$2t = -2$$

$$t = \frac{-2}{2}$$ $$t = -1$$

The solution is $\boxed{t = -1}$.

प्रश्न 4:

$$4z + 3 = 6 + 2z$$

$$4z - 2z + 3 = 6$$ $$2z + 3 = 6$$

$$2z = 6 - 3$$ $$2z = 3$$

$$z = \frac{3}{2}$$

The solution is $\boxed{z = \frac{3}{2}}$.

प्रश्न 5:

$$2x - 1 = 14 - x$$

$$2x + x - 1 = 14$$ $$3x - 1 = 14$$

$$3x = 14 + 1$$ $$3x = 15$$

$$x = \frac{15}{3}$$ $$x = 5$$

The solution is $\boxed{x = 5}$.

प्रश्न 7:

$$x = \frac{4}{5}(x + 10)$$

$$x = \frac{4}{5}x + \frac{4}{5} \times 10$$ $$x = \frac{4}{5}x + 8$$

$$x - \frac{4}{5}x = 8$$ $$\frac{5x - 4x}{5} = 8$$ $$\frac{1}{5}x = 8$$

$$x = 8 \times 5$$ $$x = 40$$

The solution is $\boxed{x = 40}$.

प्रश्न 8:

$$\frac{2x}{3} + 1 = \frac{7x}{15} + 3$$

$$15 \left( \frac{2x}{3} \right) + 15(1) = 15 \left( \frac{7x}{15} \right) + 15(3)$$ $$5(2x) + 15 = 7x + 45$$ $$10x + 15 = 7x + 45$$

$$10x - 7x + 15 = 45$$ $$3x + 15 = 45$$

$$3x = 45 - 15$$ $$3x = 30$$

$$x = \frac{30}{3}$$ $$x = 10$$

The solution is $\boxed{x = 10}$.

प्रश्न 10:

$$3m = 5m - \frac{8}{5}$$

$$3m - 5m = -\frac{8}{5}$$ $$-2m = -\frac{8}{5}$$

$$m = \frac{-\frac{8}{5}}{-2}$$ $$m = \frac{8}{5 \times 2}$$ $$m = \frac{8}{10}$$

$$m = \frac{4}{5}$$

The solution is $\boxed{m = \frac{4}{5}}$.

उदाहरण 3:

$$\frac{6x+1}{3} + 1 = \frac{x-3}{6}$$

$$6 \left( \frac{6x+1}{3} + 1 \right) = 6 \left( \frac{x-3}{6} \right)$$ $$2(6x+1) + 6(1) = x-3$$

$$12x + 2 + 6 = x - 3$$ $$12x + 8 = x - 3$$

$$12x - x + 8 = -3$$ $$11x + 8 = -3$$

$$11x = -3 - 8$$ $$11x = -11$$

$$x = \frac{-11}{11}$$ $$x = -1$$

The solution is $\boxed{x = -1}$.