Simple Equations

1.0Introduction

Variables

In a mathematical equation, a variable is a letter or alphabet used in place of an unspecified number in expressions, equations or formulas. That is, in the given expression a variable acts as a placeholder for the unknown number. Generally, a single letter is used to represent a variable. Ex- $\mathrm{x},\mathrm{y},\mathrm{z},\mathrm{t}$ etc.

Expression

An expression or algebraic expression is any mathematical statement which consists of numbers, variables and an arithmetic operation (,,$+-\times ,÷$ ) between them. For e.g, $4\mathrm{m}+5$ is an expression where 4 m and 5 are the terms and $m$ is the variable of the given expression separated by the arithmetic sign + .

Equations

An equation is a statement of equality which contains one or more unknown quantities or variables.

2.0Linear equations or simple equations

An equation involving only a linear polynomial is called a linear equation.

• Example: $4x+2=-2$ $\frac{5}{4}x+9=\frac{1}{2}$

Solution of linear equation:

The value of the variable which makes the equation a true statement is called the solution or root of the equation.

• Example: The solution of the equation $9+x=13$ is 4 . In other words, the value of $x$, called the variable, which satisfies the given equation is called the solution or root of the equation.

• An expression does not have equality sign, but an equation always has an equality sign.

3.0Balancing an equation

The first step in solving an equation involves balancing the equation on both sides. There are four rules which must be followed while balancing an equation. (i) In an equation, to maintain the balance or equality, any number added to one side must also be added to the other side. (ii) The same number can be subtracted from both sides of an equation. (iii) If one side of an equation is multiplied by a number, the other side must also be multiplied by the same number. (iv) Both sides of an equation can be divided by the same non zero number.

Transposing Method

In this method, we transpose the numbers from one side of the equation to the other side so that all the terms with variable come on one side and all the constants come on another side. If a positive number is transposed, it becomes a negative number and vice-versa and the multiplication will become the division and vice-versa. Let's see an example to understand this method better.

Transposition helps in isolating the variable and solving simple equations efficiently.

An equation remains the same if the LHS and the RHS are interchanged.

4.0From Solution to Equation

We cannot only solve an equation, but also can make equations by following the reverse path. Also, given an equation, we can get one solution but with the given solution we can make many equations.

5.0Applications of Linear Equations to Practical Situations

We have learnt how to convert daily life situations expressed in statement to simple equations. We also know how to solve linear equations to find the solution of practical problems and puzzles. The steps to solve these word problems are as follows: Step 1: Read the problem carefully. Step 2: Figure out what we have to find and what is given. Step 3: Assume the unknown quantity by variables $x$ or $y$. Step 4: Form an equation and solve it. Step 5: Verify whether the solution satisfies the equation.

6.0Numerical Ability

Q. Check whether the value given in the bracket is a solution (root) of the given equation or not? (i) $y-11=4$ (at $y=16$ ) (ii) $5x-2=13$ (at $x=2$ ) (iii) $3m=21$ (at $m=7$ )

• Explanation (i) Substituting $y=16$ in the given equation, we get L.H.S. $=16-11=5\ne$ R.H.S. $\therefore y=16$ is not a solution of the equation $y-11=4$ (ii) Substituting $\mathrm{x}=2$ in the given equation, we get L.H.S. $=5\times 2-2=8\ne$ R.H.S. $\therefore \mathrm{x}=2$ is not a solution of the equation $5\mathrm{x}-2=13$ (iii) Substituting $m=7$ in the given equation, we get L.H.S. $=3\times 7=21=$ R.H.S. $\therefore \mathrm{m}=7$ is a solution of the equation $3\mathrm{m}=21$.

Q. Solve the following equations without transposing: (i) $2x+14=26$ (ii) $\frac{x}{2}-4=\frac{x}{3}+5$ (iii) $5\mathrm{y}+10=4\mathrm{y}-10$

• Explanation (i) $2\mathrm{x}+14=26$ Subtract 14 from both sides $\Rightarrow 2\mathrm{x}+14-14=26-14$ $\Rightarrow 2\mathrm{x}=12$ Divide both the sides by 2 $\Rightarrow \frac{2\mathrm{x}}{2}=\frac{12}{2}$ $\Rightarrow \mathrm{x}=6$ (ii) $\frac{x}{2}-4=\frac{x}{3}+5$ Subtract $\frac{x}{3}$ from both sides $\Rightarrow \frac{x}{2}-4-\frac{x}{3}=\frac{x}{3}+5-\frac{x}{3}$ $\Rightarrow \frac{x}{2}-\frac{x}{3}-4=5$ Adding 4 to both sides $\frac{x}{2}-\frac{x}{3}-4+4=5+4$ $\Rightarrow \frac{x}{2}-\frac{x}{3}=9$ $\Rightarrow \frac{x}{6}=9$ Multiply both the sides by 6 $\mathrm{x}=54$ (iii) $5y+10=4y-10$ Subtract $4y$ from both the sides $\Rightarrow 5y+10-4y=4y-10-4y$ $\Rightarrow y+10=-10$ Subtract 10 from both the sides $\Rightarrow y+10-10=-10-10$ $\Rightarrow y=-20$ (iv) $\frac{5x}{6}=x-2$ Subtract $\frac{5x}{6}$ from both the sides $\Rightarrow \frac{5x}{6}-\frac{5x}{6}=x-2-\frac{5x}{6}$ $\Rightarrow 0=\mathrm{x}-\frac{5\mathrm{x}}{6}-2$ Adding 2 to both the sides $\Rightarrow 0+2=x-\frac{5x}{6}-2+2$ $\Rightarrow 2=\frac{x}{6}$ Multiply both sides by 6 $\Rightarrow 2\times 6=\frac{x}{6}\times 6$ $\Rightarrow \mathrm{x}=12$

Q. Solve the equation (i) $3+2(p-7)=9$ (ii) $6(2+x)=12$ (iii) $3x+\frac{1}{5}=2-x$ (iv) $\frac{1}{4}x+\frac{1}{6}x=\frac{1}{2}x+\frac{3}{4}$

• Explanation (i) $3+2(p-7)=9$ $2(p-7)=9-3$ $2(p-7)=6$ $\mathrm{p}-7=\frac{6}{2}$ p-7 = 3 $\mathrm{p}=3+7$ $\mathrm{p}=10$ (ii) $6(2+x)=12$ The equation remains the same if all expressions on L.H.S. are inter changed with those on R.H.S. and vice versa. $2+x=\frac{12}{6}$ $2+x=2$ $\mathrm{x}=2-2$ $\mathrm{x}=0$ (iii) $3x+\frac{1}{5}=2-x$ $3\mathrm{x}+\mathrm{x}=2-\frac{1}{5}$ [transposing -x to L.H.S. and $\frac{1}{5}$ to R.H.S.] $4\mathrm{x}=\frac{9}{5}$ $\mathrm{x}=\frac{1}{4}\times \frac{9}{5}=\frac{9}{20}$ (iv) $\frac{1}{4}x+\frac{1}{6}x=\frac{1}{2}x+\frac{3}{4}$ The LCM of the denominators $4,6,2$ and 4 is 12 . Multiplying both sides by 12 , we get $3\mathrm{x}+2\mathrm{x}=6\mathrm{x}+9$ $\Rightarrow 6\mathrm{x}-5\mathrm{x}=-9$ $\Rightarrow \mathrm{x}=-9$. Thus, $x=-9$ is a solution of the given equation.

q. Solve : $\frac{5x-4}{8}-\frac{x-3}{5}=\frac{x+6}{4}$

• Solution We have $\frac{5x-4}{8}-\frac{x-3}{5}=\frac{x+6}{4}$ Multiplying both sides by 40 , the LCM of 8,5 and 4 we get, $5(5\mathrm{x}-4)-8(\mathrm{x}-3)=10(\mathrm{x}+6)$ $\Rightarrow 25\mathrm{x}-20-8\mathrm{x}+24=10\mathrm{x}+60$ $\Rightarrow 17\mathrm{x}+4=10\mathrm{x}+60$ $\Rightarrow 17\mathrm{x}-10\mathrm{x}=60-4$ [transposing 10x to L.H.S and 4 to R.H.S] $\Rightarrow 7\mathrm{x}=56$ $\Rightarrow \mathrm{x}=\frac{1}{7}\times 56=8$ [multiplying both sides by $\frac{1}{7}$ ] Thus, $x=8$ is a solution of the given equation.

q. Solve: $x-\left(2x-\frac{3x-4}{7}\right)=\frac{4x-27}{3}-3$ Explanation: We have : $x-\left(2x-\frac{3x-4}{7}\right)=\frac{4x-27}{3}-3$ Removing the brackets, we get $x-2x+\frac{3x-4}{7}=\frac{4x-27}{3}-3$ $-x+\frac{3x-4}{7}=\frac{4x-27}{3}-3$ Multiplying both sides by 21, the LCM of 7 and 3 , we get $-21x+3(3x-4)=7(4x-27)-63$ $\Rightarrow -21\mathrm{x}+9\mathrm{x}-12=28\mathrm{x}-189-63$ $\Rightarrow -12\mathrm{x}-12=28\mathrm{x}-252$ $\Rightarrow -12\mathrm{x}-28\mathrm{x}=-252+12$ [by transposing 28x on LHS & -12 on RHS] $\Rightarrow -40\mathrm{x}=-240$ $\Rightarrow \mathrm{x}=6$ [on dividing both sides by -40]. $\therefore \mathrm{x}=6$ is a solution of the given equation.

Q. If $x=8$, then find the equation.

• Explanation We have, $x=8$ $\Rightarrow \mathrm{x}+5=8+5$ (Adding 5 to both sides) $\Rightarrow \mathrm{x}+5=13$ Hence, $x+5=13$ is an equation for $x=8$ Another way, $\mathrm{x}=8$ $\Rightarrow \mathrm{x}\times 4=8\times 4$ (Multiply 4 to both sides) $\Rightarrow 4\mathrm{x}=32$ Hence, $4\mathrm{x}=32$ is also an equation for $\mathrm{x}=8$. In this way, we can find as many equations as we want for the solution $x=8$.

Q. Write equations for the following statements: (i) Six added to thrice of a number gives 33 . (ii) Subtracting 6 from three times a number gives 9 .

• Explanation (i) Let the number be $x$ $6+3x=33$ (ii) Let the number be $x$ $3x-6=9$

Q. Write the following equations in statements form: (i) $x+9=2$ (ii) $4x-3=9$

• Explanation (i) 9 added to a number gives 2 . (ii) Subtracting 3 from four times a number gives 9

Q. Set up equations and solve them to find the unknown number in the following cases: (i) Rahul says that he has $3$ marbles less than four times the marbles Rohit has. Rahul has 29 marbles. (ii) Meena's father is 43 years old. He is 19 years older than twice the Meena's age

• Explanation (i) Let Rohit have x marbles. Rahul has 29 marbles. $\therefore 4\mathrm{x}-3=29$ $4\mathrm{x}=32$ $x=8$ (ii) Let Meena's age be y years. Meena's father's age $=43$ years $2\mathrm{y}+19=43$ $2y=24$ $y=12$

Q. The sum of a number and 7 is equal to 15 . What is the number?

• Solution Let the number be x . $\therefore \mathrm{x}+7=15$ (given) $\mathrm{x}=15-7$ or $\mathrm{x}=8$

Q. The sum of three consecutive integers is 10 more than twice the smallest of the integers. Find the integers.

• Solution Let the smallest integer be x . Then the three integers are $\mathrm{x},\mathrm{x}+1,\mathrm{x}+2$. Their sum is $x+x+1+x+2=3x+3$. Ten more than twice the smallest integer $=2\mathrm{x}+10$. $\therefore 3\mathrm{x}+3=2\mathrm{x}+10$ $3\mathrm{x}=2\mathrm{x}+10-3=2\mathrm{x}+7$ $3x-2x=7$ $\therefore \mathrm{x}=7$ $\therefore$ The three integers are 7, 8 and 9 .

7.0Memory Map