Exercise 6B
Question 1
$\frac{x}{2}+ 5 = 8$
Sol :
$\frac{x}{2}=3$
x=6
Question 2
$\frac{y}{3} – 1 = 7$
Sol :
$\frac{y}{3}=8$
y=24
Question 3
$2 – \frac{m}{8} = 5$
Sol :
$\frac{-m}{8}=5-2$
$\frac{-m}{8}=3$
-m=24
m=24
Question 4
$\frac{m}{40} – 7 = 0$
Sol :
$\frac{m}{40}=7$
m=280
Question 5
$\frac{2x}{5} – 18 = -24$
Sol :
$\frac{2x}{5}=-24+18$
$\frac{2x}{5}=-6$
2x=-30
x=-15
Question 6
$\frac{n}{2} + \frac{n}{3} = 5$
Sol :
L.C.M of 2 and 3 is 6, multiplying both sides by 6
$\frac{n}{2}\times 6 +\frac{n}{3} \times 6=5 \times 6$
3n+2n=30
$n=\frac{30}{5}=6$
Question 7
$6\frac{1}{2} + 1\frac{1}{2} (1 – 2x) = -4$
Sol :
$\frac{13}{2}+\frac{3}{2} (1-2x)=-4$
$\frac{31}{2}+\frac{3}{2}-3x=-4$
$-3x=-4-\frac{13}{2}-\frac{3}{2}$
$-3x=-\left(4+\frac{31}{2}+\frac{3}{2}\right)$
$3x=\frac{8+13+3}{2}$
$3x=\frac{24}{2}$
3x=12
$x=\frac{12}{3}=4$
Question 8
$\frac{4x-3}{5}-\frac{5x-3}{8}=1$
Sol :
L.C.M of 5 and 8 is 40, multiplying both sides by 40
$\left(\frac{4x-3}{5}\times 40\right)-\left(\frac{5x-3}{8}\times 40\right)=1\times 40$
(32x-24)-(25x-15)=40
32x-24-25x+15=40
32x-25x=40+24-15
7x=49
$x=\frac{49}{7}=7$
Question 9
$\dfrac{4 z+3}{3}+\dfrac{1}{3}=\dfrac{3 z-1}{2}$
Sol :
$\dfrac{(4z+3)+(1)}{3}=\dfrac{3z-1}{2}$
$\dfrac{4 z+4}{3}=\dfrac{3 z-1}{2}$
2(4z+4)=3(3z-1)
8z+8=9z-3
8+3=+9z-8z
11=z or z=11
Question 10
$\frac{1}{2}(8-6n) = n$
Sol
4-3n=n
-3n-n=-4
-4n=-4
n=1
It is very helpful for me thanks 😊☺️☺️
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