Linear Equations in One Variable Practice Questions PDF

Okay, here is the markdown conversion of the text in the images you sent. ### Linear Equations in One Variable **8** #### Jog Your Memory 1. The degree of a variable in a linear equation is **1**. 2. If a number added to itself gives 500, the number is **250**. 3. $\frac{x}{2}+5=3$, then $x =...

Okay, here is the markdown conversion of the text in the images you sent. ### Linear Equations in One Variable **8** #### Jog Your Memory 1. The degree of a variable in a linear equation is **1**. 2. If a number added to itself gives 500, the number is **250**. 3. $\frac{x}{2}+5=3$, then $x = $ **-4** 4. Is $x = -3$ a solution of equation $3x + 9 = 0$? **Yes** 5. Two consecutive odd integers can be considered as $x$ and **x+2**. 6. If 4 times a number increased by 3 is 39, the number is **9**. 7. If $\frac{x}{6}+\frac{x}{3} = 6$, then the value of $x =$ **12**. 8. Is $5x-3y + 7 = 0$ a linear equation in one variable? **No**. 9. Two consecutive natural numbers whose sum is 85 are **42** and **43**. 10. The difference between two consecutive even numbers is **2**. #### Introduction So far, we have learnt about algebraic expressions and equations. We know that a linear equation has a variable with highest power 1 or a linear equation in one variable has only one variable with highest power 1. For example: $x + 1 = 3$ is a linear equation, but $x + x^2 + 2 = 0$ is not a linear equation. Also $2x + 1$ is a linear equation in one variable but $x + y = 1$ is not a linear equation in one variable. A linear equation has two sides, LHS or left hand side and RHS or right hand side. For a certain value of the variable, the RHS equals LHS and this value is also called root of the equation. Let us recall the ways to solve a linear equation. #### Solving Linear Equations **When Variable is Only on One Side** * **Example 1**: Solve for $x$, $3x = 18$ * **Solution: $3x = 18$** $\implies \frac{3x}{3} = \frac{18}{3}$ (Dividing both sides by 3) $\implies x = 6$ Hence, $ x = 6$ * **Example 2**: Solve for $y$, $3y - 7 = 5$ * **Solution: $3y - 7 = 5$** $\implies 3y - 7 + 7 = 5 + 7$ (Adding 7 on both sides) $\implies 3y = 12$ $\implies \frac{3y}{3} = \frac{12}{3}$ (Dividing both sides by 3) Hence, $y = 4$ * **Example 3**: Solve for $\frac{x}{5} - \frac{6}{3} = 2$ * **Solution: $\frac{x}{5} - \frac{6}{3} = 2$** LCM of 5 and 3 = 15 $\implies \frac{3x-30}{15} = 2$ $\implies (\frac{3x-30}{15})\times 15 = 2 \times 15$ (Multiplying both sides by 15) $\implies 3x-30 = 30$ $\implies 3x - 30 + 30 = 30 + 30$ (Adding both sides by 30) $\implies 3x = 60$ $\implies \frac{3x}{3} = \frac{60}{3}$ (Dividing both sides by 3) $\implies x = 20$ Hence, $x = 20$ * **Example 4**: Solve for $m$, $\frac{1}{4}-m = -\frac{3}{2}$ * **Solution: $\frac{1}{4}-m = -\frac{3}{2}$** LCM of 4 and 1 is 4 $\implies \frac{1-4m}{4} = -\frac{3}{2}$ $\implies (1-4m) \times 2 = -3 \times 4$ (Cross multiplication) $\implies 2 - 8m = -12$ $\implies 2 - 8m - 2 = -12 - 2$ (Subtracting 2 from both sides) $\implies -8m = -14$ $\implies m = \frac{-14}{-8}$ $\implies m = \frac{7}{4}$ Hence, $m = \frac{7}{4}$ **Try These** Solve for $x$: 1. $\frac{2x}{5} = 15$ 2. $4-2x=9$ 3. $\frac{8}{3}x=-2$ * **Example 5**: What should be added to $\frac{1}{2}$ to make it 6. * **Solution**: Let $x$ be added to $\frac{1}{2}$ to make it 6. Its linear equation will be $\frac{1}{2} + x = 6$ $\implies \frac{1}{2} + x = 6$ $\implies \frac{1}{2} +x - \frac{1}{2} = 6 - \frac{1}{2}$ (Subtracting $\frac{1}{2}$ from both sides) $\implies x = \frac{12-1}{2}$ (LCM of 2 and 1 is 2) $\implies x = \frac{11}{2}$ So, $\frac{11}{2}$ should be added to $\frac{1}{2}$ to get 6. * **Example 6**: The sum of four multiples of 8 is 208. Find the four numbers * **Solution**: Let the four numbers be $x$, $x + 8$, $x + 16$ and $x + 24$ So,$x+x+8+x+16+x+24 = 208$ $\implies 4x + 48 = 208$ $\implies 4x + 48 - 48 = 208 - 48$ (Subtracting 48 from both sides) $\implies 4x = 160$ $\implies \frac{4x}{4} = \frac{160}{4}$ (Dividing both sides by 4) $\implies x = 40$ So, the numbers are 40, 48, 56 and 64. * **Example 7**: Perimeter of a rectangle is $\frac{16}{3} m$, if the length is $\frac{5}{2} m$, find its breadth. * **Solution**: Let breadth of the rectangle be $x$, and length is $\frac{5}{2} m$ So, perimeter of the rectangle $ = 2(l + b) = \frac{16}{3}$ $\implies 2(\frac{5}{2} + x) = \frac{16}{3}$ $\implies 5+2x = \frac{16}{3}$ $\implies 5 + 2x - 5= \frac{16}{3} - 5$ (Subtracting 5 from both sides) $\implies 2x = \frac{16-15}{3}$ $\implies \frac{2x}{2} = \frac{1}{3} \times \frac{1}{2}$ $\implies x = \frac{1}{6}$ Therefore, breadth of the rectangle is $\frac{1}{6} m$. * **Example 8**: Mayank's mother's present age is four times his present age. After 7 years, their ages add up to 64. Find their ages. * **Solution**: Let Mayank's present age be $x$. So, Mayank's mother's present age is $4x$. After 7 years $= x + 7$ After 7 years $= 4x + 7$ Adding their ages after 7 years; $(x + 7) + (4x + 7) = 64$ $\implies 5x + 14 = 64$ $\implies 5x + 14 - 14 = 64 - 14$ (Subtracting 14 from both sides) $\implies 5x = 50$ $\implies \frac{5x}{5} = \frac{50}{5}$ (Dividing both sides by 5) $\implies x = 10$ So, presently Mayank is 10 years old and his mother is 40 years old. #### Solving Linear Equations **When Variable is on Both the Sides** * **Example 9**: Solve $3x + 5 = 2x + 10$. * **Solution**: $3x + 5 = 2x + 10$ $\implies 3x + 5 - 5 = 2x + 10 - 5$ (Subtracting 5 from both the sides) $\implies 3x = 2x + 5$ $\implies 3x - 2x = 5$ (Transposing 2x to LHS) $\implies x = 5$ **Try These** Kajal scored 87% marks in the annual exammination. If she has scored 1305 marks, find the maximum marks. * **Example 10**: If $x = \frac{2}{5} - x \frac{7}{5}$, find $x$. * **Solution**: $x = \frac{2}{5} - x \frac{7}{5}$ $\implies x + \frac{2}{5} x = -\frac{7}{5}$ (Transposing $\frac{2}{5} x$ to LHS) $\implies \frac{5x-2x}{5} = -\frac{7}{5}$ (LCM of 1 and 5 is 5) $\implies \frac{3x}{5} = -\frac{7}{5}$ $\implies \frac{3x}{5} \times 5 = -\frac{7}{5} \times 5$ (Multiplying both sides by 5) $\implies 3x = -7 $ $\implies \frac{3x}{3} = \frac{-7}{3}$ (Dividing both sides by 3) $\implies x = \frac{-7}{3}$ * **Example 11**: Solve for $x$, $2(4-5x)-5 = (3x+1)$. * **Solution**: $2(4-5x)-5 = (3x+1)$ $\implies 8 - 10x - 5 = 3x + 1$ $\implies 3 - 10x = 3x + 1$ $\implies -10x -3x = 1 - 3$ (Transposing 3 to RHS and 3x to LHS) $\implies -13x = -2$ $\implies \frac{-13x}{-13} = \frac{-2}{-13}$ (Dividing both sides by 13) $\implies x = \frac{2}{13}$ * **Example 12**: Find $m$, if $3.9m - 3(1.5-m) = 1.5m - 1.8$ * **Solution**: $3.9m - 3(1.5 -m) = 1.5m - 1.8$ $\implies 3.9m - 4.5 + 3m = 1.5m - 1.8$ $\implies 6.9m - 4.5 = 1.5m - 1.8$ $\implies 6.9m - 1.5m = 4.5 - 1.8$ (Transposing $1.5m$ to LHS and 4.5 to RHS) $\implies 5.4m = 2.7$ $\implies \frac{5.4m}{5.4} = \frac{2.7}{5.4}$ (Dividing both sides by 5.4) $\implies m = \frac{1}{2}$ * **Example 13**: The sum of a two-digit number and the number when its digits are interchanged is 165. If the difference in the two digits is 3, find the number. * **Solution**: Let ones digit be $x$. As the difference in the two digits is 3, then the other digit will be $x - 3$. According to the question, the two numbers are $10(x - 3) + x$ and $10x + (x - 3)$. Also, $[10(x-3)+x] + [10x + (x - 3)] = 165$. $\implies 10x - 30 + x + 10x + x - 3 = 165$ $\implies 22x - 33 = 165$ $\implies 22x = 165 + 33$ (Transposing 33 to RHS) $\implies 22x = 198$ $\implies \frac{22x}{22} = \frac{198}{22}$ (Dividing both side by 22) $\implies x = 9$, other digit is $9 - 3 = 6$ So, the number is 69. * **Example 14**: Akshit's age after 15 years is four times his present age. What is his age? * **Solution**: Let Akshit's present age be $x$. Akshit's age after 15 years $= x + 15$ According to the questions: $x + 15 = 4x$ $\implies x + 15 - 15 = 4x - 15$ (Subtracting 15 from both sides) $\implies x = 4x - 15$ $\implies x - 4x = -15$ (Transposing 4x to LHS) $\implies -3x = -15$ $\implies \frac{-3x}{-3} = \frac{-15}{-3}$ (Dividing (-3) on both sides) $\implies x = 5$ So, Akshit is 5 years old at present. **Try These** A positive number is 4 times the other number. If 5 is added to both, one number is thrice the other. Find the numbers. ### EXERCISE 8.1 **Understanding Based Questions** 1. Solve the following equations and verify the answers: a. $9x - 7 = 20$ b. $\frac{y}{2} + 6 = 34$ c. $13a = 78$ d. $\frac{4t}{5} = 12$ e. $28 + 10p = 72$ f. $t + 1.5 = 7$ g. $47 - \frac{3m}{7} = 50 $ h. $-2.5 (x+3) = 10$ 2. 3. Solve the following equations: a. $5x + 7 = 12x + 28$ b. $12z - 11 = 3z-38$ c. $5(2d - 6) = −3(4d + 12)$ d. $-0.4(0.5a + 7 ) = 1.2 − 0.2a$ e. $\frac{5y}{8} + 12 = 52 - \frac{5y}{16}$ f. $\frac{5s}{9} + \frac{s}{12} - 159 = \frac{5s}{6}$ g. $\frac{2}{5}m+\frac{1}{2}(m-5)=3$ h. $\frac{1}{4}(y-5) = \frac{2}{7}(14-3y)$ i. $\frac{5}{3}(1.5n + 20) = -0.5n + 6$ **Application Based Questions** 3. Two equal sides of an isosceles triangle are each 5 cm more than the third side. If the perimeter of the triangle is 100 cm. Find the length of the three sides. 4. Two numbers are in the ratio 7:2. If the difference between them is 100, find both the numbers. 5. The sum of three consecutive multiples of 6 is 666. Find the multiples. 6. $\frac{1}{6}$ is subtracted from a rational number and the difference is multiplied by 5, the result is twice the given rational number. Find the rational number. 7. Sum of the digits of a 2-digit number is 8. When the digits of the number are interchanged, the resulting new number is smaller than the original number by 36. What is the original number ? **Real-life Application Based Questions** 8. Ankita has ₹255 in her piggy bank in the form of coins of ₹5 and ₹2. If the number of ₹5 coins is thrice the number of ₹2 coins, find the total number of coins of each denomination that, Ankita has. 9. Harsh's mother is 30 years older than him. In 12 years, Harsh's mother will be three times as old as Harsh. What are the present ages of Harsh and his mother? 10. Ishan sold a shirt for ₹1085 and gained 7.5%. Find the cost price the shirt. 11. In an entrance examination. a student requires 60% of the total marks pass. If Umesh got 200 marks, which is 20 marks more than the passing marks, fond the total marks. 12. Length of length of a rectangular field is 7 less than twice of its breath If the perimeter of the rectange is 196n, find the area the held. 13. Yash has triangular plot The of in the 5% is ₹147. 14. The third of security The ### Reducing Equations to Linear Equations * **Example 15**: Solve the following: a. $\frac{x+2}{3x-5} = \frac{5}{4}$ b. $\frac{x+7}{x-6} = \frac{x-5}{x+2} = 0$ c. $\frac{7}{m-10} = \frac{2}{m+4}$ d. $\frac{1}{z-1}+\frac{2}{z-2}=\frac{3}{z-4}$ * **Solution** a. $\frac{x+2}{3x-5}=\frac{5}{4}$ $\implies 4(x+2) = 5(3x-5) \quad \text{(By cross-multiplication)}$ $\implies 4x+8=15x-25$ $\implies 4x - 15x = -25 - 8 \quad \text{(Transposing 15x to LHS and 8 to RHS)}$ $\implies -11x=-33$ $\implies x = \frac{-33}{-11} \quad \text{(Transposing the coefficient\(-11\) to RHS)}$ $\therefore x=3$ b. $\frac{x+7}{x-6}=\frac{x-5}{x+2} = 0$ $\implies \frac{x+7}{x-6}=\frac{x-5}{x+2} $ $\implies (x+7)(x+2)=(x-5)(x-6)$ $\implies x^{2}+2x+7x+14=x^{2}-6x-5x+30$ $\implies x^{2}+9x+14=x^{2}-11x+30$ $\implies x^{2}+9x+14-x^{2}=-11x+30$$\quad\text{(Transposing x² to LHS)}$ $9x+14=-11x+30$