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Solve the below linear equation problems
$\frac{1}{z + 10} = \frac{2}{z - 4}$
$\frac{1}{z + 4} = \frac{2}{z - 4}$
$\frac{1}{z + 5} = \frac{2}{z - 3}$
$\frac{z}{3} = \frac{z}{9} - \frac{8}{9}$
$\frac{x}{3} + 1 = \frac{x}{10} + 9$
$2(x-1) +3(x-2) +4(x-3) = 16$
$\frac{1}{z + 8} = \frac{2}{z - 7}$
$\frac{z}{6} = \frac{z}{8} - \frac{5}{8}$
$\frac{x}{6} + 2 = \frac{x}{10} + 10$
$\frac{4 x + 10}{7 x + 8} = \frac{2}{5}$
$ 10x + 3 =12 +2x$
$\frac{1}{z + 9} = \frac{2}{z - 10}$
$\frac{1}{q + 7} = \frac{2}{q - 8}$
$\frac{y}{5} = \frac{y}{10} - \frac{1}{10}$
$x + 1 = \frac{x}{6} + 7$
$\frac{1}{z + 4} = \frac{2}{z - 9}$
$\frac{x}{3} + 9 = \frac{x}{6} + 5$
$\frac{1}{z + 8} = \frac{2}{z - 1}$
$\frac{1}{z + 9} = \frac{2}{z - 3}$
$\frac{1}{z + 4} = \frac{2}{z - 2}$
$\frac{7 x + 1}{5 x + 5} = 7$
$\frac{1}{z + 9} = \frac{2}{z - 1}$
$x + 7 = \frac{x}{10} + 7$
$\frac{x + 5}{4 x + 3} = \frac{1}{5}$
$\frac{2 x + 3}{5 x + 3} = \frac{2}{3}$
$ x+1 = 2x+ 6$
$(x-1)(x-2) =(x+3)(x+4)$
$\frac{5 x + 1}{8 x + 1} = 5$
$\frac{1}{z + 5} = \frac{2}{z - 1}$
Answers
$z = -24$
$z = -12$
$z = -13$
$z = -4$
$x = \frac{240}{7}$
$x=4$
$z = -23$
$z = -15$
$x = 120$
$x = - \frac{17}{3}$
$x=1$
$z = -28$
$q = -22$
$y = -1$
$x = \frac{36}{5}$
$z = -17$
$x = -24$
$z = -17$
$z = -21$
$z = -10$
$x = - \frac{17}{14}$
$z = -19$
$x = 0$
$x = -22$
$x = \frac{3}{4}$
$x=-5$
$x=-1$
$x = - \frac{4}{35}$
$z = -11$
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