Exercise 7.3 | Cambridge Math | Class 8 | Linear Equations in One Variable

1. Solve the following:

(a) $\ \frac{5x-7}{3x}=2$
⇒ $\ 5x-7=2×3x$
⇒ $\ -7 = 6x-5x$
⇒ $\ x=-7$


(b) $\ \frac{5-7y}{2+4y}=\frac{-8}{7}$
⇒ $\ 7(5-7y)=(-8)(2+4y)$
⇒ $\ 35-49y=-16-32y$
⇒ $\ 35+16=49y-32y$
⇒ $\ 51=17y ⇒ y=\frac{51}{17}⇒y=3$


(c) $\ \frac{9y}{7-6y}=15$
⇒ $\ 9y=15(7-6y)$
⇒ $\ 9y = 105-90y$
⇒ $\ 9y+90y = 105$
⇒ $\ 99y=105 ⇒ 99y=105$
⇒ $\ y=\frac{105}{99} ⇒ y=\frac{35}{33}$


(d) $\ \frac{2p-4}{3p+2}=\frac{-2}{3}$
⇒ $\ 3(2p-4)=(-2)(3p+2)$
⇒ $\ 6p-12=-6p-4$
⇒ $\ 6p+6p=12-4$
⇒ $\ 12p=8 ⇒ p=\frac{8}{12} ⇒ p=\frac{2}{3}$


(e) $\ \frac{2a-\frac{3}{4}}{9a+\frac{4}{7}}=\frac{1}{4}$
⇒ $\ 4\left(2a-\frac{3}{4}\right) = 1\left(9a+\frac{4}{7}\right)$
⇒ $\ 4(2a)-4\left(\frac{3}{4}\right)=9a+\frac{4}{7}$
⇒ $\ 8a-3=9a+\frac{4}{7}$
⇒ $\ 8a-9a=\frac{4}{7}+3$
⇒ $\ -a=\frac{21+4}{7}=\frac{25}{7}$
⇒ $\ a=-\frac{25}{7}$


(f) $\ \frac{5x-11}{3x+7}=\frac{-2}{1}$
⇒ $\ 1(5x-11)=(-2)(3x+7)$
⇒ $\ 5x-11=-6p-14$
⇒ $\ 5x+6x=11-14$
⇒ $\ 11x=-3 ⇒ x=-\frac{3}{11}$


(g) $\ \frac{5x+2}{2x-5}=\frac{22}{5}$
⇒ $\ 5(5x+2)=22(2x-5)$
⇒ $\ 25x+10=44x-110$
⇒ $\ 10+110=44x-25x$
⇒ $\ 120=19x ⇒ x=\frac{120}{19}$


(h) $\ \frac{\frac{2}{5}x-4}{\frac{3}{4}x+7}=\frac{4}{5}$
⇒ $\ 5\left(\frac{2}{5}x-4\right) = 4\left(\frac{3}{4}x+7\right)$
⇒ $\ \frac{10x}{5}-20=4×\frac{3x}{4}+4×7$
⇒ $\ \frac{10x}{5}-20=3x+28$
⇒ $\ \frac{10x}{5}-3x=28+20$
⇒ $\ \frac{10x-15x}{5}-3x=48 ⇒ -5x=48×5$
⇒ $\ x=-\frac{48×5}{5} ⇒ x=-48$


(i) $\ \frac{2-b}{b+16}=\frac{3}{5}$
⇒ $\ 5(2-b)=3(b+16)$
⇒ $\ 10-5b=3b+48$
⇒ $\ 10-48=3b+5b⇒ -38=8b$
⇒ $\ b=-\frac{38}{8}⇒b=-\frac{19}{4}$


(j) $\ \frac{(5x-7)-(2x+3)}{6x+11}=\frac{8}{3}$
⇒ $\ 3[(5x-7)-(2x+3)]=8(6x+11)$
⇒ $\ 3(5x-7-2x-3)=48x+88$
⇒ $\ 15x-21-6x-9=48x+88$
⇒ $\ 9x-30=48x+88$
⇒ $\ 9x-48x=88+30$
⇒ $\ -39x=118 ⇒ x=-\frac{118}{39}$


(k) $\ \frac{ 0.3+0.7x}{2}=0.85x$
⇒ $\ 0.3+0.7x=2×0.85x$
⇒ $\ 0.3=1.70x-0.7x$
⇒ $\ 0.3 =x$
⇒ $\ x=0.3$


(l) $\ \frac{1}{9x-(5x-1)}=\frac{7}{6x-(2x-1)}$
⇒ $\ 1[6x-(2x-1)]=7[6x-(2x-1)]$
⇒ $\ 6x-(2x-1)=7[6x-2x+1)]$
⇒ $\ 6x-2x+1=42x-14x+7$
⇒ $\ 6x-2x+1=42x-14x+7$
⇒ $\ 4x+1=28x+7$
⇒ $\ 1-7=28x-4x ⇒-6=24x$
⇒ $\ x=-\frac{6}{24}⇒ x=-\frac{1}{4}$



2. The difference between two positive integers is 51. The quotient when one integer is divided by the other is 4. Find the two integers.
Solution:
Let one of the positive integers be $\ x$ and the other be $\ x-51$
According to question,
$\ \frac{x}{x-51}=4$
$\ x=4(x-51) ⇒x=4x-204$
⇒ $\ 204=4x-x⇒ 204=3x$
⇒ $\ \frac{204}{3}=x ⇒x=68$
⸫ The required positive integers are $\ 68$ and $\ 68-51=17$.


3. The numerator of a rational number is less than the denominator by 8. If the denominator is decreased by 1 and numerator is increased by 17, then the number obtained is $\ \frac{3}{2}$. Find the rational number.
Solution:
Let the denominator of the rational number be $\ x$ and so the numerator be $\ x-8$
⸫ The rational number = $\ \frac{x-8}{x}$
According to question,
$\ \frac{(x-8)+17}{x-1}=\frac{3}{2}$
⇒ $\ 2(x-8+17)=3(x-1)$
⇒ $\ 2x-16+34=3x-3$
⇒ $\ 18+3=3x-2x$
⇒ $\ x=21$
⸫ The required rational number is $\ \frac{13}{21}$


4. The denominator of a rational number is greater than its numerator by 3. If numerator and denominator are increased by 1 and 4, respectively, the number obtained is $\ \frac{1}{2}$. Find the rational number.
Solution:
Let the numerator of the rational number be $\ x$ and so the denominator be $\ x+3$
⸫ The rational number =$\ \frac{x-8}{x}$
According to question,
$\ \frac{x+1}{x+3+4}=\frac{1}{2}$
⇒ $\ 2(x+1)=1(x+3+4)$
⇒ $\ 2x+2=x+7 ⇒2x-x=7-2$
⇒ $\ x=5$
⸫ The required rational number is $\ \frac{5}{8}$


5. The ratio of two numbers is 3: 5. If each number is increased by 10, the ratio between the numbers so formed is 5: 7. Find the two original numbers.
Solution:
Let the common ratio between the numbers be $\ x$
So the numbers are $\ 3x$ and $\ 5x$
According to question,
$\ \frac{3x+10}{5x+10}=\frac{5}{7}$
⇒ $\ 21x+70=25x+50$
⇒ $\ 70-50=25x-21x ⇒20=4x$
⇒ $\ \frac{20}{4} ⇒ x= 5$
⸫ The required numbers are $\ 3×5=15$ and $\ 5×5=25$


6. A 2-digit number is such that ones digit is 1 more than the tens digit. The ratio of this number and the number formed by reversing its digits is 5: 6. Find the number.
Solution:
Let the tens digit of the number be $\ x$ and so the ones digit be $\ x+1$
⸫ The number is $\ 10x+(x+1)=10x+x+1=11x+1$
But, when the digits are reversed,
The number becomes $\ 10(x+1)+x=10x+10+x=11x+10$
According to question,
$\ \frac{11x+1}{11x+10}=\frac{5}{6}$
⇒ $\ 5(11x+10)=6(11x+1)$
⇒ $\ 55x+50=66x+6$
⇒ $\ 50-6=66x-55x ⇒ 44=11x$
⇒ $\ \frac{44}{11} =x ⇒ x=4$
⸫ The required number is $\ =11×4+1=44+1=45$


7. A rational number is such that its denominator is twice its numerator. If both the numerator and the denominator are increased by 13, the rational number becomes $\ \frac{2}{3}$ in the simplest form. Find the rational number.
Solution:
Let the numerator of the rational number be $\ x$ and so the denominator be $\ 2x$
⸫ The rational number = $\ \frac{x}{2x}$
According to question,
$\ \frac{x+13}{2x+13}=\frac{2}{3}$
⇒ $\ 3(x+13)=2(2x+13)$
⇒ $\ 3x+13=4x+26$
⇒ $\ 39-26=4x-3x$
⇒ $\ x=13$
⸫ The required rational number is $\ \frac{13}{26}$


8. The length of a rectangle is twice its breadth. If the ratio of the perimeter of the rectangle and its area is1:2, find the dimensions of the rectangle.
Solution:
Let the breadth of the rectangle be $\ x$ and the length be $\ 2x$
Perimeter of the rectangle = $\ 2(x+2x)$ unit
Area of the rectangle = $\ x×2x$ sq. unit
According to question,
$\ \frac{2(x+2x)}{x×2x}=\frac{1}{2}$
⇒ $\ 2[2(x+2x)]=1(x×2x)$
⇒ $\ 2[2x+4x]=x×2x$
⇒ $\ 4x+8x=x×2x ⇒12x=x×2x$
⇒ $\ \frac{12x}{2x}=x ⇒ x=6$
The length= $\ 6$ unit
The length= $\ 2×6=12$ unit
⸫ The required dimension of the rectangle is $\ 12$ unit × $\ 6$ unit.


9. A fraction is such that the sum of its numerator and denominator is 15. If 8 is added to the numerator and12 is added to the denominator, the fraction becomes $\ \frac{3}{4}$. Find the original fraction.
Solution:
Let the numerator of the fraction be $\ x$ and the denominator be $\ 15-x$.
⸫ The fraction = $\ \frac{x}{15-x}$
According to question,
$\ \frac{x+8}{15-x+12}=\frac{3}{4}$
⇒ $\ 4(x+8)=3(15-x+12)$
⇒ $\ 4x+32=45-3x+36$
⇒ $\ 4x+3x=81-32 ⇒ 7x=49$
⇒ $\ x=\frac{49}{7} ⇒ x=7$
⸫ The required original fraction is $\ \frac{7}{15}$