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# Linear Equations in One Variable : Exercise 2.6 (Mathematics NCERT Class 8th)

Solve the following linear equations.

Q.1 ${{8x - 3} \over {3x}} = 2$
Sol. Given, ${{8x - 3} \over {3x}} = 2$
Multiplying by 3x on both the sides, we get,
(8x - 3) = 2 X 3x
(8x - 3) = 6x
8x - 6x = 3
2x = 3
$x = {3 \over 2}$

Q.2 ${{9x} \over {7 - 6x}} = 15$
Sol. Given, ${{9x} \over {7 - 6x}} = 15$
Multiplying by (7 â€“ 6x) on both the sides, we get,
9x = 15 (7 - 6x)
9x =105 -90x
9x +90x = 105
99x =105
$x = {{105} \over {99}}$
$x = {{35} \over {33}}$

Q.3 ${z \over {z + 15}} = {4 \over 9}$
Sol. Given, ${z \over {z + 15}} = {4 \over 9}$
Multiplying by 9(z + 15) on both the sides, we get,
9z = 4 (z + 15)
9z = 4z +60
9z - 4z = 60
5z = 60
$z = {{60} \over 5}$
z=15

Q.4 ${{3y + 4} \over {2 - 6y}} = {{ - 2} \over 5}$
Sol. Given, ${{3y + 4} \over {2 - 6y}} = {{ - 2} \over 5}$
Multiplying by 5(2 â€“ 6y) on both the sides, we get,
5(3y + 4) = -2 (2 - 6y)
15y + 20 = -4 + 12y
15y - 12y = -4 -20
3y = -24
$y = {{ - 24} \over 3}$
y = -8

Q.5 ${{7y + 4} \over {y + 2}} = {{ - 4} \over 3}$
Sol. Given, ${{7y + 4} \over {y + 2}} = {{ - 4} \over 3}$
Multiplying by 3(y + 2) on both the sides, we get,
3 (7y + 4) = -4 (y + 2)
21y + 12 = -4y - 8
21y + 4y = -8-12
25y = -20
$y = {{ - 4} \over 5}$

Q.6 The ages of Hari and Harry are in the ratio 5:7. Four years from now the ratio of their ages will be 3:4. Find their present ages.
Sol. Let the ages of Hari and Harry be 5x and 7x respectively.
After 4 years, the ages of Hari and Harry will be (5x + 4) and (7x + 4) respectively.
Given, ${{5x + 4} \over {7x + 4}} = {3 \over 4}$
4(5x + 4) = 3(7x + 4)
20x +16 = 21x + 12
20 - 21x =12 - 16
-x = -4
x = 4
Therefore, age of Hari = 5x = 5 x 4 = 20
Age of Harry = 7x = 7 x 4 = 28
Thus, the present ages of Hari and Harry are 20 and 28 years respectively.

Q.7 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 . Find the rational number.
Sol. Let the numerator of a rational number be a. Thus, denominator will be a + 8.
Therefore, rational number will be ${a \over {a + 8}}$
Given, ${{a + 17} \over {a + 8 - 1}} = {3 \over 2}$
2 (a + 17) = 3 (a + 7)
2a + 34 = 3a + 21
2a - 3a = 21 - 34
-a = -13
a = 13
Thus, the required rational is ${a \over {a + 8}} = {{13} \over {13 + 8}} = {{13} \over {21}}$

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