Linear Equations In One Variable

1.0Linear Equations

A statement of equality of two algebraic expressions which involves one or more unknown quantities (called variables) is known as an equation. E.g : $2\mathrm{x}-5=23$.

• $2xy+1=3\to$ This is not an example of linear equation as the degree is two.

2.0Linear equations In One Variable Definition

A linear equation which has only one variable is called a linear equation in one variable. The standard form of a linear equation in one variable is $\mathrm{ax}+\mathrm{b}=\mathrm{c}$, where $\mathrm{a},\mathrm{b}$ and c are constants and x is a variable. E.g: $3x=8;\sqrt{\frac{3}{2}}x+5=125$

3.0Solution Of A Linear Equation

The value of the variable which satisfies the given equation is called a solution or the root of the equation.

You can solve an equation by writing an equivalent equation that has the variable isolated on one side. Linear equations are equivalent equations if they have the same solution(s). To change, or transform, an equation into an equivalent equation, think of an equation as having two sides that need to be "in balance."

4.0Balancing Method

• The following rules are important for solving linear equations.
• Addition rule: Same quantity can be added to both sides of an equation.
• Subtraction rule: Same quantity can be subtracted from both sides of an equation.
• Multiplication rule: Same quantity can be multiplied to both sides of an equation.
• Division rule: Both sides of an equation can be divided by the same non-zero quantity.
• Any term can be transposed from one side of the equation to the other by changing its sign.

5.0Transposition Method

Transposing means moving from one side to the other(i.e., from LHS to RHS or from RHS to LHS). When a term is transposed from one side of the equation, its sign gets changed. Transposition of an expression can be carried out in the same way as the transposition of a term.

Solving Equations With Variables On One Side

Some equations have variables on one side. To solve such equations, collect the numbers on the other side, after simplifying one or both sides of the equation, if needed. Then use inverse operations to isolate the variable.

• Q. Solve: $3\mathrm{x}+7=-8$ Solution: To isolate the variable, undo the addition and then the multiplication. $3x+7=-8$ $\Rightarrow 3\mathrm{x}+7-7=-8-7$ $\Rightarrow 3\mathrm{x}=-15$ $\Rightarrow \frac{3\mathrm{x}}{3}=\frac{-15}{3}$ Subtract 7 from each side to undo the addition. (Subtraction property of equality) (Simplify both sides.) (Divide each side by 3 to undo the multiplication.) (Division property of equality) $\Rightarrow \mathrm{x}=-5$
• Q. SCIENCE LINK: The temperature within Earth's crust increases about $30^{\circ }\mathrm{C}$ for each kilometre beneath the surface. If the temperature at Earth's surface is $24^{\circ }\mathrm{C}$, at what depth would you expect the temperature to be $114^{\circ }\mathrm{C}$ ? Explanation 1:
  
  Explanation 2: The temperature will be $114^{\circ }\mathrm{C}$ at a depth of 3 kilometres.
• Q. Solve: $7\mathrm{x}-3\mathrm{x}-8=24$. (Combine like terms first) Solution: $7\mathrm{x}-3\mathrm{x}-8=24$ $\Rightarrow 4\mathrm{x}-8=24$ (Combine like terms 7x and -3 x ) $\Rightarrow 4\mathrm{x}-8+8=24+8$ (Add 8 to each side to undo the subtraction.) $\Rightarrow 4\mathrm{x}=32$ $\Rightarrow \frac{4\mathrm{x}}{4}=\frac{32}{4}$ (Divide each side by 4 to undo the multiplication.) $\Rightarrow \mathrm{x}=8$
• Q. Solve the equation (Use the distributive property) (i) $8\mathrm{x}-2(\mathrm{x}+7)=16$ (ii) $5x+3(x+4)=28$ Solution: (i) Distribute a negative number. (ii) Distribute a positive number.
  
  $\Rightarrow 8x-2x-14=16$ $\Rightarrow 6x-14=16$ $\Rightarrow 6x-14+14=16+14$ $\Rightarrow 6\mathrm{x}=30$ $\Rightarrow \frac{6x}{6}=\frac{30}{6}$ $\Rightarrow \mathrm{x}=5$
• Q. Solve: $4=\frac{2}{3}(x+3)\cdot ($ Multiply by a reciprocal first) Solution:

$x=2$ If two expressions are equal to each other and add or subtract the exact same thing to both sides, the two sides will remain equal. $4=\frac{2}{3}(x+3)$ $\Rightarrow \frac{3}{2}(4)=\frac{3}{2}\left(\frac{2}{3}\right)(x+3)$ (Multiply each side by $\frac{3}{2}$, the reciprocal of $\frac{2}{3}$.) $\Rightarrow 6=x+3$ $\Rightarrow 6-3=x+3-3$ (Subtract 3 from each side.) $\Rightarrow 3=x$

Solving Equations With Variables On Both Sides

Some equations have variables on both sides. To solve these equations, you can first collect the variable terms on one side of the equation. The examples will show you that collecting the variable terms on the side with the greater coefficient of variable will result in a positive coefficient.

• Q. Solve: $7x+19=-2x+55$ (Collect variables on left side) Solution: Look at the coefficient of the x-terms. Since 7 is greater than -2 , collect the $x$-terms on the left side to get a positive coefficient. $7x+19=-2x+55$ $\Rightarrow 7\mathrm{x}+19+2\mathrm{x}=-2\mathrm{x}+55+2\mathrm{x}$ (Add 2 x to each side.) $\Rightarrow 9x+19=55$ $\Rightarrow 9\mathrm{x}+19-19=55-19$ (Subtract 19 from each side) $\Rightarrow 9\mathrm{x}=36$ $\Rightarrow \frac{9x}{9}=\frac{36}{9}$ (Divide each side by 9.) $x=4$
• Q. Solve: $80-9y=6y$ (Collect variables on right side) Solution: Remember that $80-9y$ is the same as $80+(-9y)$. Since 6 is greater than -9 , collect the $y$ terms on the right side to get a positive coefficient. $80-9y=6y$ $\Rightarrow 80-9y+9y=6y+9y$ (Add $9y$ to each side.) $\Rightarrow 80=15y$ $\Rightarrow \frac{80}{15}=\frac{15y}{15}$ (Divide each side by 15.) $\Rightarrow \frac{16}{3}=y$
• Q. Solve: $3\mathrm{x}-10+4\mathrm{x}=5\mathrm{x}-7$. (Combine like terms first) Solution: $3\mathrm{x}-10+4\mathrm{x}=5\mathrm{x}-7$ $\Rightarrow 7\mathrm{x}-10=5\mathrm{x}-7$ $\Rightarrow 7\mathrm{x}-10-5\mathrm{x}=5\mathrm{x}-7-5\mathrm{x}$ (Subtract 5 x from each side) $\Rightarrow 2\mathrm{x}-10=-7$ $\Rightarrow 2\mathrm{x}-10+10=-7+10$ (Add 10 to each side) $\Rightarrow 2\mathrm{x}=3$ $\Rightarrow \frac{2\mathrm{x}}{2}=\frac{3}{2}$ (Divide each side by 2) $\Rightarrow \mathrm{x}=\frac{3}{2}$
• Q. RATIONAL NUMBERS: A number is as much greater than 84 as it is less than 108. Find it. Explanation: Number $-84=108-$ Number Let the number be x $\mathrm{x}-84=108-\mathrm{x}$ $\mathrm{x}-84=108\Rightarrow 2\mathrm{x}=108+84\Rightarrow 2\mathrm{x}=192$ $\Rightarrow \mathrm{x}=\frac{192}{2}=96$ The number is 96 .

Reducing Equations To Simpler Form

You have learnt several ways to transform an equation into an equivalent equation. As you solve more complicated equations, you will continue to use these same steps to isolate the variable.

• Q. Solve: $4(1-x)+3x=-2(x+1)(A$ more complicated equation) Solution: $4(1-x)+3x=-2(x+1)$ $\Rightarrow 4-4\mathrm{x}+3\mathrm{x}=-2\mathrm{x}-2$ (Use distributive property) $\Rightarrow 4-x=-2x-2$ $\Rightarrow 4-x+2x=-2x-2+2x$ (Add $2x$ to each side) $\Rightarrow 4+x=-2$ $\Rightarrow 4+\mathrm{x}-4=-2-4$ (Subtract 4 from each side) $\Rightarrow \mathrm{x}=-6$
• Q. AGE RELATED TASK: Kanwar is three years older than Anima. Six years ago, Kanwar's age was four times Anima's age. Find the present ages of Kanwar and Anima. Explanation:
  

6.0Fraction In Algebraic Expressions

To solve equations that contain fractional coefficients, you can get rid of all the fractional coefficients by multiplying both sides of the equation by any common multiple of the denominators of the fractions. You don't have to remove fractions, but it can make solving the equation a lot easier. The most efficient thing to multiply by is the least common multiple (LCM).

• Q. Solve and check the root of $\frac{2}{3}x-\frac{5}{6}x-3=\frac{1}{2}x-5$. Solution: LCM of 3,6 and 2 is 6 .

Fractional coefficient in algebraic expressions

Some equations contain fractional coefficients and you can't isolate the variables until you sort out the fractions.

(x-2) is all divided by 5. You need to deal with the fraction before you can get $x$ on its own.

$\frac{x-2}{5}-\frac{x-3}{6}=\frac{3}{10}$

The denominators are all different, so multiply the equation by the LCM of all three denominators.

If there are fractional coefficients in the equation, multiply both sides of the equation by the least common multiple of the denominators of the fractions to remove the fractional coefficients. You need to keep each numerator ( $x-2,x-3$, and 3 ) as a group. Then you can apply the distributive property to eliminate the grouping symbols.

• Q. Solve: $\frac{x-2}{5}-\frac{x-3}{6}=\frac{3}{10}$. Solution: There are three different denominators, so you need the LCM. List the prime factors, and use them to work out the LCM. $5=56=2\times 310=2\times 5$ So, LCM $=2\times 3\times 5=30$ Multiply both sides of the equation by 30 to clear the fractional coefficients. $\frac{30}{1}\times \frac{(x-2)}{5}-\frac{30}{1}\times \frac{(x-3)}{6}=\frac{30}{1}\times \frac{3}{10}$ $\Rightarrow 6(\mathrm{x}-2)-5(\mathrm{x}-3)=3\times 3$ $\Rightarrow 6\mathrm{x}-12-5\mathrm{x}+15=9$ $\Rightarrow 6\mathrm{x}-5\mathrm{x}-12+15=9$ $\Rightarrow x+3=9$ $\Rightarrow \mathrm{x}=6$
• Q. COIN TASK : 70 coins of 10-paise and 50-paise coins are mixed in a purse. If the total value of the money in the purse is $₹19$, find the number of each type of coins. Explanation: Number of 10 paisa coins = x Number of 50 paisa coins $=70-x$ $₹\frac{10}{100}x\times$ $+₹\frac{50}{100}(70-x)\times$ $=₹19$ $\Rightarrow \frac{x}{10}+\frac{70-x}{2}=19$ $\Rightarrow \mathrm{x}+5(70-\mathrm{x})=10\times 19$ (Multiplying both sides by 10) $\Rightarrow \mathrm{x}+350-5\mathrm{x}=190$ $\Rightarrow -4\mathrm{x}=190-350$ $\Rightarrow -4\mathrm{x}=-160$ $\Rightarrow \mathrm{x}=\frac{-160}{-4}\Rightarrow \mathrm{x}=40$ Number of 50 paisa coins $=(70-x)=30$ Number of 10 -paisa coins $=40$

7.0Equations Reducible To Linear Form

Consider the following equation involving ratios of linear expressions. $\frac{x+1}{2x+3}=\frac{3}{8}$ It is not a linear equation, since the expression in LHS is not linear. But we can put it in the form of a linear equation, by multiplying both the sides of the equation by $(2x+3)$. $\left(\frac{x+1}{2x+3}\right)\times (2x+3)=\frac{3}{8}\times (2x+3)\Rightarrow (x+1)=\frac{3(2x+3)}{8}$ Now the equation is in linear form and we know how to solve it. The above step can be directly obtained by cross multiplication. In cross multiplication, we multiply the numerator of LHS by the denominator of RHS and the denominator of LHS by the numerator of RHS. The resultant expressions are equal to each other. Consider an equation, $\frac{\mathrm{ax}+\mathrm{b}}{\mathrm{cx}+\mathrm{d}}=\frac{\mathrm{m}}{\mathrm{n}}$ Using cross multiplication, we get $(ax+b)\times n=(cx+d)\times m$

• Q. FRACTIONS: The denominator of a rational number is greater than it's numerator by 3. If 3 is subtracted from the numerator and 2 is added to it's denominator, the new number becomes $1/5$. Find the rational number. Explanation:
  
  Numerator $=\mathrm{x}$ Denominator $\mathrm{x}+3$ $\frac{\mathrm{x}-3}{(\mathrm{x}+3)+2}=\frac{1}{5}$ $5(x-3)=(x+5)$ (By cross-multiplication) $\Rightarrow 5\mathrm{x}-15=\mathrm{x}+5$ $\Rightarrow 5\mathrm{x}-\mathrm{x}=15+5$ $\Rightarrow 4\mathrm{x}=20$ $\Rightarrow \mathrm{x}=\frac{20}{4}=5$ $\Rightarrow \mathrm{x}+3=5+3=8$ Numerator $=5$, Denominator $=8$, Rational number $=5/8$ $4\times (2x+7)=3\times (x+6)$ Or $8\mathrm{x}+28=3\mathrm{x}+18$ Transposing the terms, we get $8\mathrm{x}-3\mathrm{x}=18-28$ or $5\mathrm{x}=-10$ Dividing both sides by 5 , we get $\mathrm{x}=-2$

8.0Applications

Practical problems often involve some known and some unknown quantities. We convert such problems into equations and then solve them to solve a wordproblem, use the following three-step problem solving plan.

Step-I : Explore the problem: To solve a verbal problem, first read the problem carefully and explore what the problem is about. Identify what information is given. Identify what you are asked to find. Step-II : Plan the solution: One strategy you can use to solve a problem is to write an equation. Choose a variable to represent one of the unspecified numbers in the problem. This is called defining a variable. Then use the variable to write expressions for the other unspecified numbers. Step-III : Solve the problem: Use the strategy you choose in Step-II to solve the problem.

1. MENSURATION: A wire of length 21 cm is bent in the form of a rectangle in such a way that it's length becomes $1\frac{1}{2}\mathrm{cm}$ less than twice its breadth. Find the length and the breadth of the rectangle. Explanation
   

Length $=2x-\frac{3}{2}=2\times 4-\frac{3}{2}=8-\frac{3}{2}=\frac{13}{2}=6\frac{1}{2}\mathrm{cm}$ Breadth $=\mathrm{x}=4\mathrm{cm}$.

Note:

$S=\frac{D}{T};D=S\times T;T=\frac{D}{S}$ Where $\mathrm{S}=$ Speed in km/hr $\mathrm{T}=$ Time in hours $\mathrm{D}=$ Distance in km. $\mathrm{xkm}/\mathrm{hr}=\mathrm{x}\times \frac{5}{18}\mathrm{m}/\sec$ $\mathrm{xm}/\sec =\mathrm{x}\times \frac{18}{5}\mathrm{km}/\mathrm{hr}$

2. SPEED, DISTANCE & TIME : The distance between city A and B is $510\mathbf{km}$. Two cars begin their journey from these cities and move directly towards each other. From city $A$, the car is moving at a speed of $100\mathbf{km}$ per hour and from city $B$, the car is moving at 70 km per hour. Assuming the cars start at the same time, find after how many hours will they meet and how far is their meeting point from city $A$ ?
   
   Explanation:
   
   Time to meet $=\mathrm{t}$ hours Speed of car-1 = 100 km per hour Speed of car-2 $=70\mathrm{km}$ per hour $100t+70t=510$ $\Rightarrow 170\mathrm{t}=510$ $\Rightarrow \mathrm{t}=\frac{510}{170}=3$ hours The two cars will meet after 3 hours from the start. Distance of the meeting point from city $A=$ Speed of car $1\times$ Time to reach meeting point $=100\times 3=300\mathrm{km}$ ( $\mathrm{d}=\mathrm{s}\times \mathrm{t}$ )
3. A gazelle can run 73 feet per second for several minutes. A cheetah can run faster (88 feet per second) but can only sustain its top speed for about 20 seconds before it is worn out. How far away from the cheetah does the gazelle need to stay for it to be safe? Explanation: You can use a diagram to picture the situation. If the gazelle is too far away for the cheetah to catch it within 20 seconds, the gazelle is probably safe.
   
   To find the "safety zone" you can first find the starting distance for which the cheetah reaches the gazelle in 20 seconds. Gazelle's distance $=73\times 20$ (feet) Algebraic model Cheetah's distance $=88\times 20$ (feet) $$\begin{gathered} 73\times 20+x=88\times 20 \\ 1460+x=1760 \\ x=1760-1460 \\ x=300 \end{gathered}$$ Gazelle needs to stay more than 300 feet away from the cheetah to be safe.
4. NUMBERS: Find two consecutive even numbers such that, three-fourth of the first exceeds two-fifth of the other by 9 . Explanation: $\frac{3}{4}$ (An even number) $=\frac{2}{5}$ (Consecutive even number) + $9$ An even number $=2x$. Consecutive even number $=2\mathrm{x}+2$ $\frac{3}{4}(2x)=\frac{2}{5}(2x+2)+9$, $\Rightarrow \frac{3\mathrm{x}}{2}=\frac{4\mathrm{x}+4}{5}+9$ $\Rightarrow 10\left(\frac{3x}{2}\right)=10\left(\frac{4x+4}{5}\right)+90$ (Multiplying by LCM of 2 and 5) $\Rightarrow 15\mathrm{x}=8\mathrm{x}+8+90$ $\Rightarrow 15\mathrm{x}-8\mathrm{x}=98\Rightarrow 7\mathrm{x}=98\Rightarrow \mathrm{x}=14$ The first number $=2x=2\times 14=28$ The next number $=2x+2=28+2=30$
5. DIGIT NUMBER: The sum of the digits of a two digit number is 14 . The number obtained by inter-changing the digits is 36 more than the given number. Explanation: Unit digit $=x$, tens digit $=14-x$ Original number $=10$ (tens digit) + unit digit $10(14-x)+x=140=-10x+x=140-9x$ Number obtained by reversing the digits $=10(x)+(14-x)=9x+14$ A.T.Q $\Rightarrow 9\mathrm{x}+14=36+(140-9\mathrm{x})$ $\Rightarrow 9x+9x=140+36-14$ $\Rightarrow 18\mathrm{x}=162$ $\Rightarrow \mathrm{x}=\frac{162}{18}=9\Rightarrow (14-\mathrm{x})=14-9=5$ Original Number $=10(5)+9=59$
6. ECONOMICS: A video store charges ₹8 to rent a video game for five days. You must be a member to rent from the store, but the membership is free. A video game club in town charges only ₹3 to rent a game for five days. But membership in the club is ₹50 per year. Which rental plan is more economical? Explanation: Find the number of rentals for which the two plans would cost the same. Store rental fee = 8 (rupees per game) Number of games to be rented = x Club rental fee = ₹ 3 (per game) Club membership fee $=₹50$ $8\mathrm{x}=3\mathrm{x}+50$ $\Rightarrow 5x=50$ $\Rightarrow x=10$ If you rent 10 video games in a year, the cost would be the same at either the store or the club. If you rent more than 10 , the club is more economical. If you rent fewer than 10 , the store is more economical.