**Equation**: **In mathematics we use the symbol “=” to show that quantities are equal to each other. **Therefore, linear equations are those algebraic expression with the symbol =.

Ex: x+3=8

*The standard form of an equation is a statement that an algebraic expression is equal to 0.*

*Ex: x-3=0*

*The statement need not to be true for any variables in it; or may be true only for certain values of the variables in it.*

*Ex: Consider 3x-5=0. If we look for integer value of x, the statement is not true. On the other hand if we are looking for rational value of x, then x=(5/3). So the statement 3x-5=0 is true for rational values.*

*The value of the statement which makes the statement true is called ***Solution. **

* **Sometimes we may come across statements like 2x-5=x+6. This is not in standard form. However we can bring it to standard form; x-11=0.*

**An equation containing one variable raised to the power of** **1****is called a** *linear equation** in one variable***.**

**For example,** **2x + 3 = 11****, where x≠0.**

Solving the equation: The process of finding the value of the variable(s) in the equation.

**Solving a Linear Equation in One Variable:**

**Axiom 1:** If equals are added to equals, we again get equals.

**Ex 1:** Solve the equation x-15=0

**Solution:**

Use the **Axiom 1:** If equals are added to equals, we again get equals.

Let us add 15 on both the sides of the given problem to get the value of x.

It is given that,

x-15+15 = 15

x-0 = 15

This reduces to, x = 15.

**Axiom 2:** If equals are subtracted from equals, we again get equals.

**Example 2:** Solve the equation x+9 = 20

**Solution:**

We know that, the given problem x+9 = 20 is not in standard form. We can bring this to standard form by subtracting 20 both sides (**Axiom 2:** If equals are subtracted from equals, we again get equals.)

i.e., x+9-20 = 20-20

x-11 = 0

We have got x-11 = 0, to find the value of x, we can make use of **Axiom 1:** If equals are added to equals, we again get equals.

That implies, x-11+11 = 11.

This reduces to x = 11.

**Example 3: Solve 2x-3 = x+8**

**Solution**:

We know that, the given problem 2x-3 = x+8 is not in standard form. We can bring this to standard form by subtracting x-3 both sides. (**Axiom 2:** If equals are subtracted from equals, we again get equals.)

i.e., (2x-3)-(x-3) = (x+8)-(x-3)

x-0 = 0+11

x = 11

**Axiom 3:** If equals are multiplied to equals, we again get equals.

**Example 4: **solve (x/3) = 9

**Solution****:**

By making use of axiom3: if equals are multiplied to equals, we get equals.

Let us multiply by 3 on both the sides of the given equation (x/3) = 9

i.e., (x/3)*3 = 9*3

Thus, x = 27

**Axiom 4:** If equals are divided by non-zero equal quantities, we again get equals.

**Example 5**: Solve (2x/9) = 5

**Solution:**

By making use of axiom 3 and axiom 4, let us multiply the given equation by (9/2)

Note: This is as same as that of multiplying by 9 at first and dividing by 2.

We get,

(2x/9)*(9/2) = 5*(9/2)

x = (45/2)

**Example 6: solve 15x = 120**

**Solution:**

By making use of axiom-4, let us divide both the sides of the equation by 15.

We get,

(15x)/15 = 120/15

Thus, x = 8

**Example 7: Solve 13y=100.**

**Solution**:

We divide both the sides by 13 and we get,

Note:** Axiom 4:** If equals are divided by non-zero equal quantities, we again get equals

i.e., 13y/13 = 100/13

This reduces to y = 100/13.

**Note:** *Another important content of solving an equation is that of verifying the solution. We have to check whether the so called solution we obtain makes the given statement true or not. This is done by substituting the value of the variable we have got in the given equation and by verifying the truth of the statement.*

**Example 8: Is 2 a solution of the equation 3x-5 = 19?**

**Solution:**

In the problem it is said to check whether 2 is a solution of 3x-5 = 19 or not.

Now let us substitute x = 2 in 3x-5 = 19.

We obtain 3x-5 = 3(2)-5 = 6-5 = 1, which is left hand side or LHS. But the right hand side or RHS is 19. Since 1 ≠ 19. We see that LHS is not equal to RHS.

Therefore, x ≠ 2. Thus, x is not a solution.

**Example 9: Is 7 a solution of 2x-4 = 10?**

**Solution:**

In the problem it is said to check whether 7 is a solution of 2x-4 = 10 or not.

Now let us substitute x = 7 in 2x-4 = 10.

We obtain 2x-4 = 2(7)-4 = 14-4 = 10, which is left hand side or LHS. and right hand side or RHS = 10. Since LHS = RHS.

Therefore, x = 7. Thus, x is a solution.

**Example 10: Solve the equation 2x-3 = 7**

**Solution:**

Use **Axiom 1:** If equals are added to equals, we again get equals.

Let us add 3 on both the sides of the given relation 2x-3 = 7.

We get, 2x-3+3 = 7+3

This gives, 2x = 10 ———–(1)

Now again, use **Axiom 4:** If equals are divided by non-zero equal quantities, we again get equals

Divide the equation (1) by 2, that implies,

2x/2 = 10/2

Which reduces to, x = 5.

Now, let us check whether the solution x = 5 is true or not. Substitute the value of x in 2x-3 = 7, such that,

2x-3 = 2(5)-3 = 10-3 = 7. Thus equation satisfied.

Note: We can reduce the number of steps in solving an equation.

Consider the equation 2x-3 = 7. We can simply write this as 2x=7+3.

Actually we are adding 3 on both the sides. But we dropped that separate step here and do the addition mentality to write 2x-3+3 = 2x. We say, we have transposed -3 to the other side. In fact this is the convenient way for doing fast calculations. Again we can divide both the sides by 2 in mind and write x = 5.

**Rule for transposing: When we are transposing an expression from one side to the other side of equality, we have to change the sign of the expression we are transposing. **

Note: Thus -6x becomes +6x and vice versa.

**Example 12: Solve 8x-3 = 9-2x**

**Solution:**

It is given that 8x-3 = 9-2x,

Let us transpose variables on one side and constants on the other side.

Make use of **Rule for transposing: When we are transposing an expression from one side to the other side of equality, we have to change the sign of the expression we are transposing. **

We get, 8x+2x = 9+3

Thus, 10x = 12

x = 12/10

This reduces to, x = 6/5.

**Example 13: Solve 8x+9 = 3(x-1)+7**

**Solution:**

It is given that 8x+9 = 3(x-1)+7,

8x+9 = 3x-3+7

8x+9 = 3x+4

Let us transpose variables on one side and constants on the other side.

Make use of **Rule for transposing: When we are transposing an expression from one side to the other side of equality, we have to change the sign of the expression we are transposing. **

We get, 8x-3x = 4-9

Thus, 5x = -5

x = 5/(-5)

This reduces to, x = -1.

**Example 14: Solve ^{2}/_{3 }x = ^{3}/_{8} x + ^{7}/_{12}**

**Solution:**

Note: Here, the constants in the given problem are rational numbers. for easy simplification, let us multiply both the sides of the given problem by LCM of all coefficients.

LCM of 3, 8 and 12 is 24. So now, let us multiply by 24 on both the sides of the given problem i.e., ^{2}/_{3 }x = ^{3}/_{8} x + ^{7}/_{12}

Use **Axiom 3:** If equals are multiplied to equals, we again get equals.

i.e., (^{2}/_{3 }x*24 = (^{3}/_{8} x + ^{7}/_{12})*24

This simplifies to, 16x = 9x+14

16x-9x = 14

Hence, 7x = 14 or x = 2.

**Example 15: Solve the equation ^{(2x+7)}/_{5 }– ^{(3x+11)}/_{2} = ^{(2x+8)}/_{3 }– 5**

**Solution**:

Note: Here, the constants in the given problem are rational numbers. For easy simplification let us multiply both the sides of the given problem by LCM of all coefficients.

Here 2, 3 and 5 appear in the denominator of various fractions. Their LCM is 30. So now, let us multiply by 30 on both the sides of the given problem i.e.,

^{(2x+7)}/_{5 }– ^{(3x+11)}/_{2} = ^{(2x+8)}/_{3 }– 5

Use **Axiom 3:** If equals are multiplied to equals, we again get equals.

i.e., (^{(2x+7)}/_{5 }– ^{(3x+11)}/_{2})*30 = (^{(2x+8)}/_{3 }– 5)*30

(^{(2x+7)}/_{5})*30 – (^{(3x+11)}/_{2})*30 = (^{(2x+8)}/_{3})*30 – 5*30

This simplifies to, 6(2x+7) – 15(3x+11) = (2x+8)10 – 150

12x+42-45x-165 = 20x+80-150

12x-45x-20x = 165-150-42+80

-53x = 53

Hence, x = -1.

**Example 16: Solve (x+4) ^{2} – (x-5)^{2} = 9**

**Solution:**

This equation apparently looks like an equation which is not linear. But let us expand this using identities, we get,

Identities: (a+b)^{2 }= (a^{2}+2ab+b^{2})

(a-b)^{2} = (a^{2}-2ab+b^{2})

(x+4)^{2} = (x^{2}+2(x)(4)+4^{2}) = (x^{2}+8x+16) ———-(1)

(x-5)^{2} = (x^{2}-2(5)(x)+5^{2}) = (x^{2}-10x+25) ———–(2)

Let us substitute equation(1) and (2) in the given problem (x+4)^{2} – (x-5)^{2} = 9

(x+4)^{2} – (x-5)^{2} = (x^{2}+8x+16) – (x^{2}-10x+25) = 9

x^{2}+8x+16 – x^{2}+10x-25=9

18x-9 = 9

**Use Axiom 1:** If equals are added to equals, we again get equals.

18x = 18

Hence, x = 1

**Linear Equations In One Variable – Exercise 2.3.2**

**Solve the following**

**(i) x+3 = 11**

**Solution**:

Use **Axiom 2:** If equals are subtracted equals, we again get equals.

x+3-3 = 11-3

We get, x = 8

**(ii) y-9 = 21**

**Use Axiom 1:** If equals are added to equals, we again get equals.

**y-**9+9 = 21+9

Hence, y = 30

**(ii) 10 = z+3**

**Solution**:

z+3 = 10

Use **Axiom 2:** If equals are subtracted from equals, we again get equals.

z+3-3 = 10-3

Hence, z = 7

^{(iii) 3}/_{11 }+ x = ^{9}/_{11 }

**Solution**:

Let us multiply the given equation by 11(which is denominator) in the given problem.

(^{3}/_{11 }+ x)*11 = ^{9}/_{11} * 11

3+11x = 9

This reduces to,

11x = 6

Hence, x = 6/11

**(iv) 10x = 30**

**Solution**:

Use **Axiom 4:** If equals are divided by non-zero equal quantities, we again get equals

10x/10 = 30/10

Hence, x = 3

**(v) ^{s}/_{7 }= 4*7**

Solution:

Use **Axiom 3:** If equals are multiplied to equals, we again get equals.

^{s}/_{7 }* 7 = 4*7

Therefore, s = 28

**(vi) ^{3x}/_{6 }= 10**

Solution:

Use **Axiom 3:** If equals are multiplied to equals, we again get equals.

^{3x}/_{6 }* 6 = 10 * 6

3x= 60

Use **Axiom 4:** If equals are divided by non-zero equal quantities, we again get equals

3x/3 = 60/3

Hence, x = 20

**(vii) 1.6 = ^{x}/**

_{1.5 }

**Solution**:

Use **Axiom 3:** If equals are multiplied to equals, we again get equals.

1.6*1.5 = ^{x}/_{1.5 }* 1.5

2.4 = x or x=2.4

**(viii) 8x – 8 = 48**

**Solution**:

Use **Axiom 1:** If equals are added to equals, we again get equals.

8x -8 +8 = 48+8

8x = 56

Use **Axiom 4:** If equals are divided by non-zero equal quantities, we again get equals

8x/8 = 56/8

Thus, x=7

^{(ix) x}/_{3 }+ 1 = ^{7}/_{15 }

**Solution**:

We have 3 and 15 in the denominator of the given problem. LCM of 3 and 15 is 15. Use **Axiom 4:** If equals are divided by non-zero equal quantities, we again get equals

(^{x}/_{3 }+ 1)*15 = ^{7}/_{15} * 15

^{x}/_{3}*15 + 1*15 = ^{7}/_{15}*15

5x +15 = 7

Use

5x = 7-15

5x = -8

Hence, x = -8/5

**(x) ^{x}/_{5 }= 12**

Solution:

Use **Axiom 3:** If equals are multiplied to equals, we again get equals.

^{x}/_{5} * 5 = 12*5

Thus, x = 60.

(xi) ^{3x}/_{5 }= 15

**Solution**:

Use **Axiom 3:** If equals are multiplied to equals, we again get equals.

^{3x}/_{5} *5 = 15*5

3x = 75

3x/3 = 75/3

Thus, x = 25

**(xii) 3(x+6) = 24**

**Solution**:

Given 3(x+6) = 24

3x+18 = 24

3x = 24 -18

3x = 6

x = 6/3

Thus, x = 2

**(xiii) ^{x}/_{4 }-8 = 1**

**Solution**:

(^{x}/_{4 }-8)*4 = 1*4

x – 8*4 = 4

x – 32 = 4

x = 4+32

Thus, x = 36

**(ivx) 3(x+2) – 2(x-1) = 7**

**Solution**:

3(x+2) – 2(x-1) = 7

3x+6 -2x+2 = 7

x+8 = 7

x = 7-8

Thus, x = -1

**Solve the equations:**

**(i) 5x = 3x+24**

**Solution:**

It is given that 5x = 3x+24

Use

5x-3x = 24

That implies, 2x = 24

x = 24/2

Hence, x = 12

**(ii) 8t + 5 = 2t -31**

**Solution:**

It is given that 8t+5 = 2t-31

Use

8t-2t = -31-5

6t = -36

That implies, t=(-36)/6

Hence, t = -6

**(iii) 7x – 10 = 4x +11**

**Solution:**

It is given that 7x -10 = 4x+11

7x-4x = 11+10

That implies, 3x = 21

x = 21/3

Hence, x = 7

**(iv) 4z+3 = 6+2z**

**Solution:**

It is given that 4z+3 = 6+2z

4z-2z = 6-3

That implies, 2z = 3

Hence, z = 3/2

**(v) 2x-1 = 14 – x**

**Solution:**

It is given that 2x-1 = 14-x

2x+x = 14+1

That implies, 3x = 15

x = 15/3

Hence, x = 5

**(vi) 6x+1 = 3(x-1)+7**

**Solution:**

It is given that 6x+1 = 3(x-1)+7

6x+1 = 3x-3+7

6x+1 = 3x+4

6x-3x = 4-1

That implies, 3x = 3

Hence, x = 1

**(vii) ^{2x}/_{5 }–^{ 3}/_{2 }= ^{x}/_{2 }+ 1**

**Solution:**

It is given that ^{2x}/_{5 }–^{ 3}/_{2 }= ^{x}/_{2 }+ 1

LCM of 2 and 5 is 10. So multiply by 10 on both the sides of the given problem.

(^{2x}/_{5 }–^{ 3}/_{2})*10 = (^{x}/_{2 }+ 1 )*10

(^{2x}/_{5})*10 – (^{3}/_{2})*10 = (^{x}/_{2})*10 + 1*10

4x – 15 = 5x + 10

4x – 15 = 5x + 10

4x-5x = 10 +15

-x = 25

Hence, x = -25

**(viii) ^{x-3}/_{5 }– 2 = ^{2x}/_{5 }**

**Solution:**

It is given that ^{x-3}/_{5 }– 2 = ^{2x}/_{5}

Here denominator is 5, so multiply the equation by 5 on both the sides.

This gives,

(^{x-3}/_{5 }– 2)*5 = (^{2x}/_{5})*5

(x-3) – 10 = 2x

x – 2x =10+3

-x = 13

Hence, x = -13

**(ix) 3(x+1) = 12+4(x-1)**

**Solution:**

It is given that 3(x+1) = 12+4(x-1)

3x+3 = 12+4x-4

3x-4x = 12-4-3

That implies, -x = 5

Hence, x = -5

(x) 2x – 5 = 3(x-5)

**Solution:**

It is given that 2x – 5 = 3(x-5)

2x -5 = 3x -15

2x-3x = -15+5

That implies, -x = -10

Hence, x = 10

**(xi) 6(1-4x)+7(2+5x) = 53**

**Solution:**

It is given that 6(1-4x)+7(2+5x) = 53

6-24x+14+35x = 53

11x+20 = 53

11x = 53-20

That implies, 11x = 33

Hence, x = 3

**(xii) 3(x+6)+2(x+3) = 64**

**Solution:**

It is given that 3(x+6)+2(x+3) = 64

3x+18+2x+6 = 64

5x+24 = 64

5x = 64 – 24

5x = 40

That implies, x = 40/5

Hence, x = 8

**(xiii) ^{2m}/_{3 }+8 = ^{m}/_{2} -1**

**Solution:**

It is given that ^{2m}/_{3 }+8 = ^{m}/_{2} -1

Here denominators are 2 and 3. LCM of 2 and 3 is 6. So let us multiply the given problem by 6 on both the sides.

(^{2m}/_{3 }+8)*6 = (^{m}/_{2} -1)*6

(^{2m}/_{3})*6 +(8)*6 = (^{m}/_{2} )*6-(1)*6

4m+48 = 3m – 6

4m – 3m = -6 -48

m = -54

Hence, m = -54

**(ivx) ^{3}/_{4 }(x-1) = x-3**

**Solution:**

It is given that ^{3}/_{4 }(x-1) = x-3

Here denominator is 4. So let us multiply the given problem by 4 on both the sides.

(^{3}/_{4 }(x-1))*4 = (x-3)*4

3x – 3 = 4x – 12

3x – 4x = -12+3

-x = -9

That implies, x = 9.

**Application of Linear Equations – Linear Equations in One Variable**

Let us study the real life applications of linear equations.

**Example 17: Seven times a number if increased by 11, is 81. Find the number.**

**Solution:**

Let us solve the problem in following steps.

**Step 1**: First we convert given data to a appropriate equation.

Let the number be x. (we do not know that value of this x) We formulate an equation including this unknown x, using the given data(given problem).

Seven times a number means 7x. Increasing this data by 11 leads to 7x+11. the problem says that 7x+11 = 81. (now you can see unknown x in the equation.)

**Step 2:** We have to solve the equation 7x+11 = 81. We have learnt methods of solving such equations. Use “Rule of transpose”.

7x = 81 – 11

7x = 70

Then, x= 70/7

Hence, x=10.

**Step 3: **We have to check whether the number 10 satisfies the statement of our problem. Now seven times the number gives 7*10=70. Adding 11 to this gives 81. That is precisely the data says. Thus x = 10.

**Example 18: The present age of the siri’s mother is three times the present of siri’s age. After 5years, their ages add to 66years. Find their present ages.**

**Solution:**

**Step 1**: Suppose siri’s age is x years. Then her mother’s age is 3x. Then after 5 years their ages would be x+5 and 3x+5 respectively. The data says that these two numbers would add up to 66. Thus we get the equation, (x+5)+(3x+5) = 66.

**Step 2: ** We have to solve the equation (x+5)+(3x+5) = 66. Then 4x+10 = 66. Use “rule of transpose” 4x = 66 – 10 = 56

x = 56/4 = 14.

**Step 3: **We have to check whether the number 14 satisfies the statement of our problem. Substitute the value of x, (14+5)+(3*14+5) = 66. This satisfies the condition.

**Example 19: The sum of 3 three consecutive even number is 252. Find them.**

**Solution:**

Let x be the least even number among the three consecutive even numbers. Then the other numbers are x+2 and x+4. This is because any two consecutive numbers differ by 2. The given condition says that

x+(x+2)+(x+4) = 252

Thus, we get, 3x+6 = 252

3x = 252 – 6 = 246

x = 246/3 = 82

Hence, the numbers are 82,84,86.

We check, 82+84+86 = 252, hence condition satisfied.

**Example 20: If the perimeter of a triangle is 14cm and the sides are x+4, 3x+1 and 4x+1. Find x.**

**Solution:**

We know that perimeter of a triangle is sum of its three sides. Since the sides are given to be x+4, 3x+1, and 4x+1.

So the perimeter is (x+4)+(3x+1)+(4x+1) = 14

This reduces to, 8x+6 = 14

8x = 14 – 6

8x = 8

Thus, x = 1.

Hence the sides are x+4 = 5

3x+1 = 4

4x+1 = 5

i.e., 5, 4, 5cm.

**Example 21: Let P be a point on a line AB such that P lies between A and B. AP = 3PB. Given that AB = 10cm, find the length of AP.**

**Solution:**

Since P lies between A and B. We have AB = AP + PB.

Thus 10 = 3PB + PB = 4PB. We can solve for PB and get PB = 10/4.

Therefore, PB = 5/2cm.

Hence, AP = 3PB = 3*(5/2)

AP = 15/2cm.

**Example 22: The sum of two numbers is 75 and they are in the ratio 3:2. Find the numbers.**

**Solution:**

The numbers which are in ratio 3:2 are 3x and 2x. We are given,

3x+2x = 75

5x = 75

x = 15.

Hence the numbers are 3x = 45 and 2x = 30.

We verify: 45/30 = 3/2 and 45 +30 =75.

**Linear Equation in One Variable – Exercise 2.3.3**

**If 4 is added to a number and sum id multiplied be 3, the result is 30. Find the number.**

**Solution:**

Let the number be x. ( x is unknown)

It is given that 4 is added to the number x, i.e., 4+x and the sum is multiplied by 3, this implies, (4+x)*3. It is also mentioned that the result of this will be 30.

Thus, (4+x)*3 = 30

12+3x = 30

3x = 30 – 12

3x = 18

This reduces to, x = 18/3

Therefore, x = 6.

**Find the three consecutive odd numbers whose sum is 219.**

**Solution:**

Let 3 consecutive odd numbers be x, x+2, x+4 (because each consecutive odd numbers differs by 2)

We have, x+(x+2)+(x+4) = 219

3x + 6 = 219

3x = 219 – 6

3x = 213

Therefore, x = 213/3 = 71

The three consecutive odd numbers are 71, 73 and 75

**A rectangle has a length which is 5cm less than twice of its breadth. If the length of the rectangle is decreased by 5cm and breadth is increased by 2cm, then perimeter of the resulting rectangle will be 74cm. Find the length and breadth of the original rectangle.**

**Solution:**

Given data: Let breadth b x an d length is 5cm less than twice of its breadth i.e., length = (2x-5)cm.

If the length is decreased by 5cm then, length = (2x-5-5) = (2x – 10)cm and breadth is increased by 2cm then, breadth = (x+2)cm, thus perimeter of the resulting rectangle will be 74cm.

We know, the perimeter of the rectangle = 2(Length + breadth)

Therefore, 74 = 2(2x-10+ x+2)

(74/2) = 3x – 8

37 + 8 = 3x

45 = 3x

x = 45/3 = 15 cm

Therefore, breadth of the original rectangle is 15cm and length of the original rectangle is (2x-5) = (2*15 – 5) = 30 – 5 = 25cm.

**A number subtracted by 30 gives 14 subtracted by 3 times the number. Find the number.**

**Solution:**

Given Data: Let the number be x. A number subtracted by 30 i.e., x – 30, gives 14 subtracted by 3times the number i.e., x – 30 = 14 – 3x

x + 3x = 14 +30

4x = 44

x = 11

**sristi’s salary is same as 4 times Azar’s salary. If together they earn Rs. 3,750 month, find their individual salary.**

**Solution:**

Let Sristi’s salary be x. It is given that Sristi’s salary is 4 times the Azar’s salary, therefore Azar’s salary is 4x. Sum of their salary is equal to Rs. 3750. i.e., x + 4x = 3,750

x + 4x = 3750

5x = 3750

x = 3750/5

x = 750

**Prakruthi age is 6 times Sahil’s age. After 15 years, Prakruthi will be 3 times old as Sahil. Find their age.**

**Solution:**

Let Sahil’s age be x and Prakruthi’s age is 6times Sahil’s age i.e., Prakruthi’s age is 6x. After 15years Prakruthi’s age will be 3 times sail’s age, i.e.,6x+15 = 3(x+15)

6x+15 = 3(x+15)

6x – 3x = 45 – 15

3x = 30

x = 10

Sahil’s age, x = 10 and Prakruthi’s age, 6x = 6*10 = 60.

**In figure, AB is a straight line. Find x.**

**Solution:**

A straight line = 180◦

x+20+x+40+x = 180◦

3x + 60 = 180◦

3x = 180◦ – 60◦

3x = 120◦

x = 120◦/3

x = 40◦

**8. If 5 is subtracted from three times a number, the result is 16. Find the number.**

**Solution:**

3x – 5 = 6

3x = 6 + 5

3x = 11

x = 11/3

**9. Find two numbers such that one of them exceeds the other by 9 and their sum is 81.**

**Solution:**

Let that number be x. Then the other number be (x+9)

Sum of these two numbers is x+(x+9) = 81

2x = 81 – 9

2x = 72

x = 72/2

x = 36

So, the two numbers be 36 and 45.

**10. The length of a rectangular field is twice its breadth. If the perimeter of the field is 288m, find the dimensions of the field.**

**Solution:**

Let breadth of the rectangular field be x and length be 2x , it is also given that its perimeter is 288m.

We have, 2(x+2x) = 288

3x = 288/2

3x = 144

x = 144/3

x = 48

Hence, breadth = 48

Length, 2x = 48*2 = 96

Therefore, dimensions of the field are 46 and 96.

**11. Ahmed’s father is thrice as old as Ahmed. After 12 years, his age will be twice that of his son. Find their present age.**

**Solution:**

Let Ahmed’s age be x and his father’s age be 3x. After 12 years, his age will be twice that of his son. (3x+12)=2(x+12)

3x+12 = 2(x+12)

3x – 2x = 24 -12

x = 12

Therefore, Ahmed’s age is 12 and is father’s age is 36.

**12. Sanju is 6years older than his brother Nishu. If the sum of their ages is 28years, what is their present age?**

**Solution:**

Let Nishu’s age be x years and Sanju is 6years older than Nishu i.e., (x+6). Sum of their ages is 28years. i.e., x+(x+6) = 28

x+ (x+6) = 28

2x + 6 = 28

2x = 28 – 6

2x = 22

x = 11

**13. Viji is twice as old as is brother Deepu. If the difference of their ages is 11 years, find their present age.**

**Solution:**

Let deepu’s age be x. Viji is twice old as deepu then , viji is age is 2x. Difference is their ages is 11 years. i.e., 2x – x = 11

2x – x = 11

x = 11

Deepu’s age is 11

Viji age is 22

14. Mrs Joseph is 27years older than her daughter Bindu. After 8years she will be twice as old as Bindu. Find their present ages.

**Solution:**

Let bindu’s age be x. Her Mother’s age be x+27. After 8 years she will be twice as old as Bindu i.e., x+27+8 = 2(x+8)

x+27+8 = 2x +16

x – 2x = +16 – 27 -8

-x = -19

x = 19

Bindu’s age is 19 and her mother’s age is x+27 = 19+27 = 46

**15. After 16years, Leena will be three times as old as she is now. Find her present age.**

**Solution:**

Let Leena’s age be x. After 16 years, her age will be three times as old as she is now, i.e., x+16 = 3x

x+16 = 3x

x + 16 = 3x

-2x = -16

x = 8

Therefore, Leena’s age is 8years now.

## 1 thought on “Linear Equations in One Variable – Class VIII”

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