chapter-3 PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

Explanation:

Let x be Aftab's current age.

And his daughter's present age is 'y'.

Aftab's age is x-7.

His daughter is y-7 years old.

Now we have to solve is 

x−7 = 7(y−7)

⇒x−7 = 7y−49

⇒x−7y = −42 I (i)

Also, three years from now, or three years later,

Age of Aftab equals x+3 years.

His daughter's age will increase by y+3.

Now  the described situation,

x+3 = 3(y+3)

⇒x+3 = 3y+9

⇒x−3y = 6 …………..…………………(ii) 

By deducting equation (i) from equation (ii), we will get to the following equations: 

(x-3y)-(x-7y) = 6-(-42) 

-3y+7y = 6+42 

4y = 48

y = 12

By deducting equation (i) from equation (ii), we will get the following equations: 

The algebraic equation is expressed by the expressions x-7y = 42 and x-3y = 6.

For, x−7y = −42 or x = −42+7y

The answer grid is

For,  x−3y = 6   or     x = 6+3y

The solution table is

The graphical representation is:

Explanation:

A bat costs "Rs x."

And a ball costs "Rs y"

The question states that the algebraic representation is

3x+6y = 3900

Therefore x+3y = 1300

Hence, 3x+6y = 3900

Or x = (3900-6y)/3

The answer grid is

For, x+3y = 1300

Or x = 1300-3y

The answer grid is

The graphical representation is as follows.

Explanation:

The cost of 1kg of apples is "Rs. x."

And one kg of grapes will cost "Rs. y"

The question states that the algebraic representation is

2x+y = 160

Therefore 4x+2y = 300

The solution table for 2x+y = 160 or y = 160-2x is as follows:

The solution table is as follows for 4x+2y = 300 or y = (300-4x)/2:

The following is the graphical representation:

Explanation:

(i) If there are x number of girls and y number of boys, then The algebraic expression can be represented as follows in accordance with the query.

x +y = 10

x– y = 4

Now, the answers to x+y = 10 or x = 10-y are:

For x – y = 4 or x = 4 + y, the solutions are;

The following is the graphical representation:

The graph demonstrates that the given lines intersect at the position which (7, 3). Hence, there are 3 boys and 7 girls in the class.

(ii). A pencil should cost Rs. x, while a pen should cost Rs. y.

The algebraic expression can be represented; according to the query.

5x + 7y = 50

7x + 5y = 46

The answers to the equations 5x + 7y = 50 or x = (50-7y)/5 are:

The answers to 7x + 5y = 46 or x = (46-5y)/7 are;

The following is the graphical representation:

Explanation:

(i) given expressions 

5x−4y+8 = 0

7x+6y−9 = 0

These equations are compared to a1x+b1y+c1 = 0

 a2x+b2y+c2 = 0 as well

We get,

a1 = 5, b1 = -4, c1 = 8

a2 = 7, b2 = 6, c2 = -9

(a1/a2) = 5/7

(b1/b2) = -4/6 = -2/3

(c1/c2) = 8/-9

Due to this, (a1/a2) ≠  (b1/b2).

As a result, each pair of equations in the question has a single solution, and the lines cross exactly once.

(ii) Given expressions 

9x + 3y + 12 = 0

18x + 6y + 24 = 0

These equations are compared to a1x+b1y+c1 = 0.

a2x+b2y+c2 = 0 as well.

We get,

a1 = 9, b1 = 3, c1 = 12

a2 = 18, b2 = 6, c2 = 24

(a1/a2) = 9/18 = 1/2

(b1/b2) = 3/6 = 1/2

(c1/c2) = 12/24 = 1/2

Considering that (a1/a2) = (b1/b2) = (c1/c2)

The lines coincide because there are an endless number of solutions to the equation pairs provided in the question.

(iii) Given Expressions 

6x – 3y + 10 = 0

2x – y + 9 = 0

These equations are compared to a1x+b1y+c1 = 0.

These equations are compared to a1x+b1y+c1 = 0. 2x+b2y+c2 = 0 as well.

We get,

a1 = 6, b1 = -3, c1 = 10

a2 = 2, b2 = -1, c2 = 9

(a1/a2) = 6/2 = 3/1

(b1/b2) = -3/-1 = 3/1

(c1/c2) = 10/9

Considering that (a1/a2) = (b1/b2) ≠  (c1/c2)

As a result, the pairs of equations provided in the question are parallel to one another, the lines never cross, and the given pair of equations cannot have a solution.

Explanation:

(i) Given : 3x + 2y = 5 or 3x + 2y -5 = 0

and 2x – 3y = 7 or 2x – 3y -7 = 0

Comparing these equations with a1x+b1y+c1 = 0

And a2x+b2y+c2 = 0

We get,

a1 = 3, b1 = 2, c1 = -5

a2 = 2, b2 = -3, c2 = -7

(a1/a2) = 3/2

(b1/b2) = 2/-3

(c1/c2) = -5/-7 = 5/7

Since, (a1/a2) ≠ (b1/b2)

The preceding equations have only one feasible solution since they intersect at a single point. The equations hold true.

(ii) Given 2x – 3y = 8 and 4x – 6y = 9

Therefore,

a1 = 2, b1 = -3, c1 = -8

a2 = 4, b2 = -6, c2 = -9

(a1/a2) = 2/4 = 1/2

(b1/b2) = -3/-6 = 1/2

(c1/c2) = -8/-9 = 8/9

Since , (a1/a2) = (b1/b2) ≠ (c1/c2)

The equations cannot be solved since they are parallel to one another. As a result, the equations are flawed.

(iii)Given (3/2)x + (5/3)y = 7 and 9x – 10y = 14

Therefore,

a1 = 3/2, b1 = 5/3, c1 = -7

a2 = 9, b2 = -10, c2 = -14

(a1/a2) = 3/(2×9) = 1/6

(b1/b2) = 5/(3× -10)= -1/6

(c1/c2) = -7/-14 = 1/2

Since, (a1/a2) ≠ (b1/b2)

The equations have just one feasible solution because they cross at one location. The equations are therefore consistent.

(iv) Given, 5x – 3y = 11 and – 10x + 6y = –22

Therefore,

a1 = 5, b1 = -3, c1 = -11

a2 = -10, b2 = 6, c2 = 22

(a1/a2) = 5/(-10) = -5/10 = -1/2

(b1/b2) = -3/6 = -1/2

(c1/c2) = -11/22 = -1/2

Since (a1/a2) = (b1/b2) = (c1/c2)

These coinciding lines have an endless number of solutions to these linear equations. The equations are therefore consistent.

(v)Given, (4/3)x +2y = 8 and 2x + 3y = 12

a1 = 4/3 , b1= 2 , c1 = -8

a2 = 2, b2 = 3 , c2 = -12

(a1/a2) = 4/(3×2)= 4/6 = 2/3

(b1/b2) = 2/3

(c1/c2) = -8/-12 = 2/3

Since (a1/a2) = (b1/b2) = (c1/c2)

These coinciding lines have an endless number of solutions to these linear equations. The equations are therefore consistent.

Explanation:

(i)Given, x + y = 5 and 2x + 2y = 10

(a1/a2) = 1/2

(b1/b2) = 1/2

(c1/c2) = 1/2

Since (a1/a2) = (b1/b2) = (c1/c2)

The equations are coincidental and can be solved in an endless number of ways.

The equations are therefore consistent.

For, x + y = 5 or x = 5 – y

For 2x + 2y = 10 or x = (10-2y)/2

So, the equations are represented in graphs as follows:

The figure shows that the lines are crossing one another.

The equations can therefore have an endless number of solutions.

(ii) Given, x – y = 8 and 3x – 3y = 16

(a1/a2) = 1/3

(b1/b2) = -1/-3 = 1/3

(c1/c2) = 8/16 = 1/2

Since, (a1/a2) = (b1/b2) ≠ (c1/c2)

The equations have no solutions and run parallel to one another. Consequently, the pair of linear equations is inconsistent.

(iii) Given, 2x + y – 6 = 0 and 4x – 2y – 4 = 0

(a1/a2) = 2/4 = ½

(b1/b2) = 1/-2

(c1/c2) = -6/-4 = 3/2

Since, (a1/a2) ≠ (b1/b2)

The above linear equations have a single point of intersection and a single solution. The two linear equations are hence consistent.

Now, for 2x + y – 6 = 0 or y = 6 – 2x

And for 4x – 2y – 4 = 0 or y = (4x-4)/2

So, the equations are represented in the form of graphs as follows:

These lines only cross each other at one point in the graph, as can be observed (2,2).

(iv) Given, 2x – 2y – 2 = 0 and 4x – 4y – 5 = 0

(a1/a2) = 2/4 = ½

(b1/b2) = -2/-4 = 1/2

(c1/c2) = 2/5

Since, a1/a2 = b1/b2 ≠ c1/c2

These linear equations, therefore, have parallel solutions and cannot be solved. The two linear equations are hence incoherent.

Explanation:

Let's think about it.

The garden has a width of x and a length of y.

Now, in response to the question, we can state the stated condition as;

y – x = 4

and,

y + x = 36

Taking either y = x + 4 or y = x - 4 now

For y + x = 36, y = 36 – x

The two equations' graphical depiction is as follows:

You can see from the graph that the lines meet at a point (16, 20). The garden is therefore 16 inches wide and 20 inches long.

Explanation:

(i) Assuming the equation 2x + 3y - 8 = 0 is linear.

The following requirement must be met in order to find another linear equation with two variables such that the pair thus produced geometrically represented by intersecting lines:

(a1/a2) ≠ (b1/b2)

Another equation that might be used is 2x - 7y + 9 = 0, which results in;

b1/b2 = 3/-7 and (a1/a2) = 2/2 = 1.

It is obvious that the requirement is satisfied by another equation.

(ii) Given the equation 2x + 3y - 8 = 0, which is linear.

The following requirement must be met in order to find another linear equation with two variables such that the pair thus formed can be represented geometrically by parallel lines:

(a1/a2) = (b1/b2) ≠ (c1/c2)

Another equation that might be used is 6x + 9y + 9 = 0, which results in;

(a1/a2) = 2/6 = 1/3

(b1/b2) = 3/9= 1/3

(c1/c2) = -8/9

It is obvious that the requirement is satisfied by another equation.

(iii) The linear equation 2x + 3y - 8 = 0 is given 

To find another linear equation with two variables that conforms to the following requirement and represents the pair so produced geometrically as coincident lines;

(a1/a2) = (b1/b2) = (c1/c2)

Another equation that might be used is 4x + 6y - 16 = 0, which results in;

(a1/a2) equals 2/4, (b1/b2) equals 3/6, and (c1/c2) equals -8/-16=1/2.

It is obvious that the requirement is satisfied by another equation.

Explanation:

Here are two graph equations: 3x + 2y - 12 = 0 and x - y + 1 = 0.

For instance, x – y + 1 = 0 or x = -1+y

For, 3x + 2y – 12 = 0 or x = (12-2y)/3

Hence, the graphical representation of these equations is as given;

;

The graphic demonstrates how these lines cross the x-axis at (-1, 0) and point (2, 3), respectively (4, 0). As a result, the triangle's vertices are (2, 3), (-1, 0), and (4, 0).

Explanation:

(i) Given,

The equations are x + y = 14 and x - y = 4.

The first equation gives us x = 14 - y.

Replace the value of x in the second equation to obtain,

(14 – y) – y = 4

14 – 2y = 4

2y = 10

Or y = 5

We can now determine the precise value of x using the value of y;

∵ x = 14 – y

∴ x = 14 – 5

Or x = 9

Hence, x = 9 and y = 5.

(ii) Given,

The two equations are s – t = 3 and (s/3) + (t/2) = 6

Inferred from the first equation is s = 3 + t ________________(1)

Replace the value of s in the second equation to obtain,

(3+t)/3 + (t/2) = 6

⇒ (2(3+t) + 3t )/6 = 6

⇒ (6+2t+3t)/6 = 6

⇒ (6+5t) = 36

⇒5t = 30

⇒t = 6

In equation (1), change the value of t to s = 3 + 6 = 9.

Hence, t = 6 and s = 9.

(iii) Given,

The two equations are 9x - 3y = 9 and 3x - y = 3.

The first equation gives us x = (3+y)/3.

Now, change the value of x in the second equation provided to get,

9(3+y)/3 – 3y = 9

⇒9 +3y -3y = 9

⇒ 9 = 9

Because x = (3+y) /3 and y has infinite values, x also has infinite values.

(iv) Given,

The two equations are 0.4x + 0.5y = 2.3 and 0.2x + 0.3y = 1.3.

The first equation gives us,

x = (1.3- 0.3y)/0.2 _________________(1)

Now, change the value of x in the second equation provided to get,

0.4(1.3-0.3y)/0.2 + 0.5y = 2.3

⇒2(1.3 – 0.3y) + 0.5y = 2.3

⇒ 2.6 – 0.6y + 0.5y = 2.3

⇒ 2.6 – 0.1 y = 2.3

⇒ 0.1 y = 0.3

⇒ y = 3

When we now change the value of y in equation (1), we obtain,

x = (1.3-0.3(3))/0.2 = (1.3-0.9)/0.2 = 0.4/0.2 = 2

Hence, y = 3 and x = 2.

(v) The two equations are √2 x + √3 y = 0 and √3 x – √8 y = 0

The first equation yields x = – (√3/√2)y, which is as follows:

Using the second equation to find the value of x,

√3(-√3/√2)y – √8y = 0 ⇒ (-3/√2)y- √8 y = 0

⇒ y = 0

The result of substituting the value of y in equation (1) is x = 0.

Thus, x = 0 and y = 0.

(vi) Provided are the two equations: (x/3) + (y/2) = 13/6 and (3x/2)-(5y/3) = -2.

According to the first equation, x = 2(-6+5y)/9 = (-12+10y)/9 and (3/2)x = -2 + (5y/3) ……………………… (1) 

Entering x's value into the second equation to obtain,

((-12+10y)/9)/3 + y/2 = 13/6

⇒y/2 = 13/6 –( (-12+10y)/27 ) + y/2 = 13/6

Now that we have changed the value of y in equation (1), we obtain 

(3x/2) - 5(3)/3 = -2

 (3x/2) - 5 = -2 

x = 2.

Hence, y = 3 and x = 2.

Explanation:

2x + 3y = 11…………………………..(I) 

2x – 4y = -24………………………… (II) 

Equation (II) gives us

x = (11-3y)/2 ………………….(III) 

When the value of x is substituted in equation (II), we obtain

2(11-3y)/2 – 4y = 24

11 – 3y – 4y = -24

-7y = -35

y = 5……………………………………..(IV) 

Equation (III) is completed when the value of y is entered.

x = (11-3×5)/2 = -4/2 = -2

Hence, x = -2 and y = 5.

Also,

y = mx + 3

5 = -2m +3

-2m = 2

m = -1

M thus has a value of -1.

Explanation:

(i) The difference between two numbers is 26 and one number is three times the other. Find them.

Explanation:

Let x and y be the two numbers, with y being greater than x.

The answer to the query is

y = 3x ……………… (1) 

y – x = 26 …………..(2) 

When we enter the value of (1) into (2), we obtain

3x – x = 26

x = 13 ……………. (3) 

When we replace (3) in (1), we obtain y = 39.

Hence, the figures are 13 and 39.

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Explanation:

Give the larger angle the symbol xo and the smaller angle yo.

We are aware that the product of two additional angles is always 180 degrees.

In response to the query,

x + y = 180o……………. (1) 

x – y = 18o ……………..(2) 

From (1), we obtain x = 180° - y. (3)

If we replace (2) with (3), we obtain

180o – y – y =18o

162o = 2y

y = 81o ………….. (4) 

By applying the y value from (3), we obtain

x = 180o – 81o

= 99o

Hence, 99o and 81o are the angles.

(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs.3800. Later, she buys 3 bats and 5 balls for Rs.1750. Find the cost of each bat and each ball.

Explanation:

Let the price of a ball be y and the price of a bat be x.

The answer to the query is

7x + 6y = 3800 ………………. (I) 

3x + 5y = 1750 ………………. (II) 

In (I), we obtain

y = (3800-7x)/6………………..(III) 

replacing (III) in (II). we get,

3x+5(3800-7x)/6 =1750

⇒3x+ 9500/3 – 35x/6 = 1750

⇒3x- 35x/6 = 1750 – 9500/3

⇒(18x-35x)/6 = (5250 – 9500)/3

⇒-17x/6 = -4250/3

⇒-17x = -8500

x = 500 ……………………….. (IV) 

When we replace the value of x in (III), we obtain

y = (3800-7 ×500)/6 = 300/6 = 50

As a result, a bat costs Rs. 500, and a ball costs Rs. 50

iv. The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for traveling a distance of 25 km?

Explanation:

Let the fixed fee be Rs. x and the per-kilometer fee be Rs. y.

The answer to the query is

x + 10y = 105 …………….. (1) 

x + 15y = 155 …………….. (2) 

We obtain x = 105 - 10y  ………………. (3)from (1)

When we change the value of x in (2), we obtain

105 – 10y + 15y = 155

5y = 50

y = 10 …………….. (4)

Adding the value of y to (3) gives us

x = 105 – 10 × 10 = 5

Hence, the set fee is Rs 5 and the per-kilometer fee is Rs 10.

Fee for 25 km: 25 km = x + 25y = 5 + 250 = Rs 255

(v) A fraction becomes 9/11 if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.

Explanation:

Let x/y be the fraction.

The answer to the query is

(x+2) /(y+2) = 9/11

11x + 22 = 9y + 18

11x – 9y = -4 …………….. (1)

(x+3) /(y+3) = 5/6

6x + 18 = 5y +15

6x – 5y = -3 ………………. (2) (2)

We obtain x = (-4+9y)/11  …………….. (3) from  (1)

When we change the value of x in (2), we obtain

6(-4+9y)/11 -5y = -3

-24 + 54y – 55y = -33

-y = -9

y = 9 ………………… (4) 

When we change the value of y in (3), we obtain

x = (-4+9×9 )/11 = 7

Hence, the ratio is 7/9.

(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Explanation:

Let x and y be the ages of Jacob and his son, respectively.

The answer to the query is

(x + 5) = 3(y + 5)

x – 3y = 10 …………………………………….. (1)

(x – 5) = 7(y – 5)

x – 7y = -30 ………………………………………. (2)

From (1), we obtain x = 3y + 10... (3)

When we change the value of x in (2), we obtain

3y + 10 – 7y = -30

-4y = -40

y = 10 ………………… (4)

When we change the value of y in (3), we obtain

x = 3 x 10 + 10 = 40

As a result, Jacob and his son are currently 40 and 10 years old, respectively.

Explanation:

(i) x + y = 5 and 2x – 3y = 4

By the method of elimination.

x + y = 5 ……………………………….. (i)

2x – 3y = 4 ……………………………..(ii)

When the equation (i) is multiplied by 2, we get

2x + 2y = 10 ……………………………(iii)

When equation (ii) is subtracted from (iii) we get,

5y = 6

y = 6/5 ………………………………………(iv)

Substituting the value of y in eq. (i) we get,

x=5−6/5 = 19/5

∴x = 19/5 , y = 6/5

By the method of substitution.

From equation (i), we get:

x = 5 – y………………………………….. (v)

When the value is put in equation (ii) we get,

2(5 – y) – 3y = 4

-5y = -6

y = 6/5

When the values are substituted in equation (v), we get:

x =5− 6/5 = 19/5

∴x = 19/5 ,y = 6/5

(ii) 3x + 4y = 10 and 2x – 2y = 2

By the method of elimination.

3x + 4y = 10……………………….(i)

2x – 2y = 2 ………………………. (ii)

When the equation (i) and (ii) is multiplied by 2, we get:

4x – 4y = 4 ………………………..(iii)

When the Equations (i) and (iii) are added, we get:

7x = 14

x = 2 ……………………………….(iv)

Substituting equation (iv) in (i) we get,

6 + 4y = 10

4y = 4

y = 1

Hence, x = 2 and y = 1

By the method of Substitution

From equation (ii) we get,

x = 1 + y……………………………… (v)

Substituting equation (v) in equation (i) we get,

3(1 + y) + 4y = 10

7y = 7

y = 1

When y = 1 is substituted in equation (v) we get,

A = 1 + 1 = 2

Therefore, A = 2 and B = 1

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7

By the method of elimination:

3x – 5y – 4 = 0 ………………………………… (i)

9x = 2y + 7

9x – 2y – 7 = 0 …………………………………(ii)

When the equation (i) and (iii) is multiplied we get,

9x – 15y – 12 = 0 ………………………………(iii)

When equation (iii) is subtracted from equation (ii) we get,

13y = -5

y = -5/13 ………………………………………….(iv)

When equation (iv) is substituted in equation (i) we get,

3x +25/13 −4=0

3x = 27/13

x =9/13

∴x = 9/13 and y = -5/13 

By the method of Substitution:

From equation (i) we get,

x = (5y+4)/3 …………………………………………… (v)

Putting the value (v) in equation (ii) we get,

9(5y+4)/3 −2y −7=0

13y = -5

y = -5/13

Substituting this value in equation (v) we get,

x = (5(-5/13)+4)/3

x = 9/13

∴x = 9/13, y = -5/13

(iv) x/2 + 2y/3 = -1 and x-y/3 = 3

By the method of Elimination.

3x + 4y = -6 …………………………. (i)

x-y/3 = 3

3x – y = 9 ……………………………. (ii)

When equation (ii) is subtracted from equation (i) we get,

5y = -15

y = -3 ………………………………….(iii)

When equation (iii) is substituted in (i) we get,

3x – 12 = -6

3x = 6

x = 2

Hence, x = 2 , y = -3

By the method of Substitution:

From equation (ii) we get,

x = (y+9)/3…………………………………(v)

Putting the value obtained from equation (v) in equation (i) we get,

3(y+9)/3 +4y =−6

5y = -15

y = -3

When y = -3 is substituted in equation (v) we get,

x = (-3+9)/3 = 2

Therefore, x = 2 and y = -3

Explanation:

(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes if we only add 1 to the denominator. What is the fraction?

Explanation:

Let the fraction be a/b

According to the given information,

(a+1)/(b-1) = 1

=> a – b = -2 ………………………………..(i)

a/(b+1) = 1/2

=> 2a-b = 1…………………………………(ii)

When equation (i) is subtracted from equation (ii) we get,

a = 3 …………………………………………………..(iii)

When a = 3 is substituted in equation (i) we get,

3 – b = -2

-b = -5

b = 5

Hence, the fraction is 3/5.

(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

Solution:

Let us assume, the present age of Nuri is x

And the present age of Sonu is y.

According to the given condition, we can write as;

x – 5 = 3(y – 5)

x – 3y = -10…………………………………..(1)

Now,

x + 10 = 2(y +10)

x – 2y = 10…………………………………….(2)

Subtract eq. 1 from 2, to get,

y = 20 ………………………………………….(3)

Substituting the value of y in eq.1, we get,

x – 3.20 = -10

x – 60 = -10

x = 50

Therefore,

Age of Nuri is 50 years

The age of Sonu is 20 years.

(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Solution:

Let the unit digit and tens digit of a number be x and y respectively.

Then, Number (n) = 10B + A

N after reversing the order of the digits = 10A + B

According to the given information, A + B = 9…………………….(i)

9(10B + A) = 2(10A + B)

88 B – 11 A = 0

-A + 8B = 0 ………………………………………………………….. (ii)

Adding the equations (i) and (ii) we get,

9B = 9

B = 1……………………………………………………………………….(3)

Substituting this value of B, in equation (i) we get A= 8

Hence the number (N) is 10B + A = 10 x 1 +8 = 18

(iv) Meena went to a bank to withdraw Rs.2000. She asked the cashier to give her Rs.50 and Rs.100 notes only. Meena got 25 notes in all. Find how many notes of Rs.50 and Rs.100 she received.

Solution:

Let the number of Rs.50 notes be A and the number of Rs.100 notes be B

According to the given information,

A + B = 25 ……………………………………………………………………….. (i)

50A + 100B = 2000 ………………………………………………………………(ii)

When equation (i) is multiplied by (ii) we get,

50A + 50B = 1250 …………………………………………………………………..(iii)

Subtracting equation (iii) from equation (ii) we get,

50B = 750

B = 15

Substituting in the equation (i) we get,

A = 10

Hence, Meena has 10 notes of Rs.50 and 15 notes of Rs.100.

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs.27 for a book kept for seven days, while Susy paid Rs.21 for a book she kept for five days. Find the fixed charge and the charge for each extra day.

Solution:

Let the fixed charge for the first three days be Rs. A and the charge for each day extra be Rs.B.

According to the information given,

A + 4B = 27 …………………………………….…………………………. (i)

A + 2B = 21 ……………………………………………………………….. (ii)

When equation (ii) is subtracted from equation (i) we get,

2B = 6

B = 3 …………………………………………………………………………(iii)

Substituting B = 3 in equation (i) we get,

A + 12 = 27

A = 15

Hence, the fixed charge is Rs.15

And the Charge per day is Rs.3

Explanation:

(i) Given, x – 3y – 3 =0 and 3x – 9y -2 =0

a1/a2=1/3 ,     b1/b2= -3/-9 =1/3, c1/c2=-3/-2 = 3/2

(a1/a2) = (b1/b2) ≠ (c1/c2)

Since the given set of lines are parallel to each other they will not intersect each other and therefore there is no solution for these equations.

(ii) Given, 2x + y = 5 and 3x +2y = 8

a1/a2 = 2/3 , b1/b2 = 1/2 , c1/c2 = -5/-8

(a1/a2) ≠ (b1/b2)

Since they intersect at a unique point these equations will have a unique solution by cross multiplication method:

x/(b1c2-c1b2) = y/(c1a2 – c2a1) = 1/(a1b2-a2b1)

x/(-8-(-10)) = y/(-15-(-16)) = 1/(4-3)

x/2 = y/1 = 1

∴ x = 2 and y =1

(iii) Given, 3x – 5y = 20 and 6x – 10y = 40

(a1/a2) = 3/6 = 1/2

(b1/b2) = -5/-10 = 1/2

(c1/c2) = 20/40 = 1/2

a1/a2 = b1/b2 = c1/c2

Since the given sets of lines are overlapping each other there will be an infinite number of solutions for this pair of equations.

(iv) Given, x – 3y – 7 = 0 and 3x – 3y – 15 = 0

(a1/a2) = 1/3

(b1/b2) = -3/-3 = 1

(c1/c2) = -7/-15

a1/a2 ≠ b1/b2

Since this pair of lines are intersecting each other at a unique point, there will be a unique solution.

By cross multiplication,

x/(45-21) = y/(-21+15) = 1/(-3+9)

x/24 = y/ -6 = 1/6

x/24 = 1/6 and y/-6 = 1/6

∴ x = 4 and y = 1.

Explanation:

(i) 3y + 2x -7 =0

(a + b)y + (a-b)y – (3a + b -2) = 0

a1/a2 = 2/(a-b) ,       b1/b2 = 3/(a+b) ,           c1/c2 = -7/-(3a + b -2)

For infinitely many solutions,

a1/a2 = b1/b2 = c1/c2

Thus 2/(a-b) = 7/(3a+b– 2)

6a + 2b – 4 = 7a – 7b

a – 9b = -4  ……………………………….(i)

2/(a-b) = 3/(a+b)

2a + 2b = 3a – 3b

a – 5b = 0 ……………………………….….(ii)

Subtracting (i) from (ii), we get

4b = 4

b =1

Substituting this eq. in (ii), we get

a -5 x 1= 0

a = 5

Thus at a = 5 and b = 1, the given equations will have infinite solutions.

(ii) 3x + y -1 = 0

(2k -1)x  +  (k-1)y – 2k -1 = 0

a1/a2 = 3/(2k -1) ,       b1/b2 = 1/(k-1), c1/c2 = -1/(-2k -1) = 1/( 2k +1)

For no solutions

a1/a2 = b1/b2 ≠ c1/c2

3/(2k-1) = 1/(k -1)   ≠ 1/(2k +1)

3/(2k –1) = 1/(k -1)

3k -3 = 2k -1

k =2

Therefore, for k = 2, the given pair of linear equations will have no solution.

Explanation:

8x + 5y = 9 …………………..(1)

3x + 2y = 4 ……………….….(2)

From equation (2) we get

x = (4 – 2y )/ 3  ……………………. (3)

Using this value in equation 1, we get

8(4-2y)/3 + 5y = 9

32 – 16y +15y = 27

-y = -5

y = 5 ……………………………….(4)

Using this value in equation (2), we get

3x + 10 = 4

x = -2

Thus, x = -2 and y = 5.

Now, Using Cross Multiplication method:

8x +5y – 9 = 0

3x + 2y – 4 = 0

x/(-20+18) = y/(-27 + 32 ) = 1/(16-15)

-x/2 = y/5 =1/1

∴ x = -2 and y =5.

Explanation:

(i) Let x be the fixed charge and y be the charge of food per day.

According to the question,

x + 20y = 1000……………….. (i)

x + 26y = 1180………………..(ii)

Subtracting (i) from  (ii) we get

6y = 180

y = Rs.30

Using this value in equation (ii) we get

x = 1180 -26 x 30

x= Rs.400.

As a result, the daily price is Rs. 30 and the fixed charge is Rs. 400.

(ii) Let the fraction be x/y.

So, as per the question given,

(x -1)/y = 1/3 => 3x – y = 3…………………(1)

x/(y + 8) = 1/4  => 4x –y =8 ………………..(2)

Subtracting equation (1) from (2), we get

x = 5 ………………………………………….(3)

Using this value in equation (2), we get,

(4×5)– y = 8

y= 12

Hence, 5/12 is the fraction.

(iii) Let the number of right answers be x and the number of wrong answers be y

According to the given question;

3x−y=40……..(1)

4x−2y=50

⇒2x−y=25…….(2)

Subtracting equation (2) from equation (1), we get;

x = 15 ….….(3)

Putting this in equation (2), we obtain;

30 – y = 25

Or y = 5

Therefore, number of right answers = 15 and number of wrong answers = 5

Hence, the total number of questions = is 20

(iv) Let x km/h be the speed of the car from point A and y km/h be the speed of the car from point B.

If the car travels in the same direction,

5x – 5y = 100

x – y = 20 …………………………………(i)

If the car travels in the opposite direction,

x + y = 100………………………………(ii)

Solving equations (i) and (ii), we get

x = 60 km/h………………………………………(iii)

Using this in equation (i), we get,

60 – y = 20

y = 40 km/h

Therefore, the speed of the car from point A = 60 km/h

Speed of car from point B = 40 km/h.

(v) Let,

The length of the rectangle = x unit

And the breadth of the rectangle = y unit

Now, as per the question given,

(x – 5) (y + 3) = xy -9

3x – 5y – 6 = 0……………………………(1)

(x + 3) (y + 2) = xy + 67

2x + 3y – 61 = 0…………………………..(2)

Using cross multiplication method, we get,

x/(305 +18) = y/(-12+183) = 1/(9+10)

x/323 = y/171 = 1/19

Therefore, x = 17 and y = 9.

Hence, the length of the rectangle = 17 units

And the breadth of the rectangle = 9 units

Explanation:

Let us assume 1/x = m and 1/y = n, then the equation will change as follows.

m/2 + n/3 = 2

⇒ 3m+2n-12 = 0…………………….(1)

m/3 + n/2 = 13/6

⇒ 2m+3n-13 = 0……………………….(2)

Now, using the cross-multiplication method, we get,

m/(-26-(-36) ) = n/(-24-(-39)) = 1/(9-4)

m/10 = n/15 = 1/5

m/10 = 1/5 and n/15 = 1/5

So, m = 2 and n = 3

1/x = 2 and 1/y = 3

x = 1/2 and y = 1/3

(ii) 2/√x + 3/√y = 2

4/√x + 9/√y = -1

Solution:

Substituting 1/√x = m and 1/√y = n in the given equations, we get

2m + 3n = 2 ………………………..(i)

4m – 9n = -1 ………………………(ii)

Multiplying equation (i) by 3, we get

6m + 9n = 6 ………………….…..(iii)

Adding equations (ii) and (iii), we get

10m = 5

m = 1/2…………………………….…(iv)

Now by putting the value of ‘m’ in equation (i), we get

2×1/2 + 3n = 2

3n = 1

n = 1/3

m =1/√x

½ = 1/√x

x = 4

n = 1/√y

1/3 = 1/√y

y = 9

Hence, x = 4 and y = 9

(iii) 4/x + 3y = 14

3/x -4y = 23

Solution:

Putting in the given equation we get,

So, 4m + 3y = 14 => 4m + 3y – 14 = 0  ……………..…..(1)

3m – 4y = 23 => 3m – 4y – 23 = 0  ……………………….(2)

By cross-multiplication, we get,

m/(-69-56) = y/(-42-(-92)) = 1/(-16-9)

-m/125 = y/50 = -1/ 25

-m/125 = -1/25 and y/50 = -1/25

m = 5 and b = -2

m = 1/x = 5

So , x = 1/5

y = -2

(iv) 5/(x-1) + 1/(y-2) = 2

6/(x-1) – 3/(y-2) = 1

Solution:

Substituting 1/(x-1) = m and 1/(y-2) = n  in the given equations, we get,

5m + n = 2 …………………………(i)

6m – 3n = 1 ……………………….(ii)

Multiplying equation (i) by 3, we get

15m + 3n = 6 …………………….(iii)

Adding (ii) and (iii) we get

21m = 7

m = 1/3

Putting this value in equation (i), we get

5×1/3 + n = 2

n = 2- 5/3 = 1/3

m = 1/ (x-1)

⇒ 1/3 = 1/(x-1)

⇒ x = 4

n = 1/(y-2)

⇒ 1/3 = 1/(y-2)

⇒ y = 5

Hence, x = 4 and y = 5

(v) (7x-2y)/ xy = 5

(8x + 7y)/xy = 15

Solution:

(7x-2y)/ xy = 5

7/y – 2/x = 5…………………………..(i)

(8x + 7y)/xy = 15

8/y + 7/x = 15…………………………(ii)

Substituting 1/x =m in the given equation we get,

– 2m + 7n = 5 => -2 + 7n – 5 = 0  ……..(iii)

7m + 8n = 15 => 7m + 8n – 15 = 0 ……(iv)

By cross-multiplication method, we get,

m/(-105-(-40)) = n/(-35-30) = 1/(-16-49)

m/(-65) = n/(-65) = 1/(-65)

m/-65 = 1/-65

m = 1

n/(-65) = 1/(-65)

n = 1

m = 1 and n = 1

m = 1/x = 1    n = 1/x = 1

Therefore, x = 1 and y = 1

(vi) 6x + 3y = 6xy

2x + 4y = 5xy

Solution:

6x + 3y = 6xy

6/y + 3/x = 6

Let 1/x = m and 1/y = n

=> 6n +3m = 6

=>3m + 6n-6 = 0…………………….(i)

2x + 4y = 5xy

=> 2/y + 4/x = 5

=> 2n +4m = 5

=> 4m+2n-5 = 0……………………..(ii)

3m + 6n – 6 = 0

4m + 2n – 5 = 0

By cross-multiplication method, we get

m/(-30 –(-12)) = n/(-24-(-15)) = 1/(6-24)

m/-18 = n/-9 = 1/-18

m/-18 = 1/-18

m = 1

n/-9 = 1/-18

n = 1/2

m = 1 and n = 1/2

m = 1/x = 1 and n = 1/y = 1/2

x = 1 and y = 2

Hence, x = 1 and y = 2

(vii) 10/(x+y) + 2/(x-y) = 4

15/(x+y) – 5/(x-y) = -2

Solution:

Substituting 1/x+y = m and 1/x-y = n in the given equations, we get,

10m + 2n = 4  =>  10m + 2n – 4 = 0  ………………..…..(i)

15m – 5n = -2 =>   15m – 5n + 2 = 0 ……………………..(ii)

Using the cross-multiplication method, we get,

m/(4-20) = n/(-60-(20)) = 1/(-50 -30)

m/-16 = n/-80 = 1/-80

m/-16 = 1/-80 and n/-80 = 1/-80

m = 1/5 and n = 1

m = 1/(x+y) = 1/5

x+y = 5 …………………………………………(iii)

n = 1/(x-y) = 1

x-y = 1……………………………………………(iv)

Adding equations (iii) and (iv), we get

2x = 6   => x = 3 …….(v)

Putting the value of x = 3 in equation (3), we get

y = 2

Hence, x = 3 and y = 2

(viii) 1/(3x+y) + 1/(3x-y) = 3/4

1/2(3x+y) – 1/2(3x-y) = -1/8

Solution:

Substituting 1/(3x+y) = m and 1/(3x-y) = n in the given equations, we get,

m + n = 3/4 …………………………….…… (1)

m/2 – n/2 = -1/8

m – n = -1/4  …………………………..…(2)

Adding (1) and (2), we get

2m = 3/4 – 1/4

2m = 1/2

Putting in (2), we get

1/4 – n = -1/4

n = 1/4 + 1/4 = 1/2

m = 1/(3x+y) = 1/4

3x + y = 4  …………………………………(3)

n = 1/( 3x-y) = 1/2

3x – y = 2 ………………………………(4)

Adding equations (3) and (4), we get

6x = 6

x = 1 ……………………………….(5)

Putting in (3), we get

3(1) + y = 4

y = 1

Hence, x = 1 and y = 1

Explanation:

(i) Let us consider,

Speed of Ritu in still water = x km/hr

Speed of Stream = y km/hr

Now, the speed of Ritu during,

Downstream = x + y km/h

Upstream = x – y km/h

As per the question given,

2(x+y) = 20

Or x + y = 10……………………….(1)

And, 2(x-y) = 4

Or x – y = 2………………………(2)

Adding both the eq.1 and 2, we get,

2x = 12

x = 6

Putting the value of x in eq.1, we get,

y = 4

Therefore,

Speed of Ritu rowing in still water = 6 km/hr

Speed of Stream = 4 km/hr

(ii) Let us consider,

Number of days taken by women to finish the work = x

Number of days taken by men to finish the work = y

Work done by women in one day = 1/x

Work done by women in one day = 1/y

As per the question given,

4(2/x + 5/y) = 1

(2/x + 5/y) = 1/4

And, 3(3/x + 6/y) = 1

(3/x + 6/y) = 1/3

Now, put 1/x=m and 1/y=n, we get,

2m + 5n = 1/4 => 8m + 20n = 1…………………(1)

3m + 6n =1/3 => 9m + 18n = 1………………….(2)

Now, by cross multiplication method, we get here,

m/(20-18) = n/(9-8) = 1/ (180-144)

m/2 = n/1 = 1/36

m/2 = 1/36

m = 1/18

m = 1/x = 1/18

or x = 18

n = 1/y = 1/36

y = 36

Therefore,

Number of days taken by women to finish the work = 18

The number of days taken by men to finish the work = 36.

(iii) Let us consider,

Speed of the train = x km/h

Speed of the bus = y km/h

According to the given question,

60/x + 240/y = 4 …………………(1)

100/x + 200/y = 25/6 …………….(2)

Put 1/x=m and 1/y=n, in the above two equations;

60m + 240n = 4……………………..(3)

100m + 200n = 25/6

600m + 1200n = 25 ………………….(4)

Multiply eq.3 by 10, to get,

600m + 2400n = 40 ……………………(5)

Now, subtract eq.4 from 5, to get,

1200n = 15

n = 15/1200 = 1/80

Substitute the value of n in eq. 3, to get,

60m + 3 = 4

m = 1/60

m = 1/x = 1/60

x = 60

And y = 1/n

y = 80

Therefore,

Speed of the train = 60 km/h

Speed of the bus = 80 km/h

Explanation:

Three years separate Ani and Biju in age.

Either Ani is three years older than Biju or Biju is three years older than Ani.

We learn from both examples that Cathy is 30 years older than Ani's father.

Let Ani and Biju be A and B years old, respectively.

Dharam's age is therefore 2 x A or 2A years.

And Cathy, Biju's sister, is B/2 years old.

Using the provided information,

Case (i)

When Ani is older than Biju by 3 yrs, then A – B = 3 …..(1)

2A − B/2 = 30

4A – B = 60 ….(2)

By subtracting the equation (1) from (2), we get;

3A = 60 – 3 = 57

A = 57/3 = 19

Therefore, the age of Ani = 19 yrs

And the age of Biju is 19 – 3 = 16 yrs.

Case (ii)

When Biju is older than Ani,

B – A = 3 ….(1)

2A − B/2 = 30

4A – B = 60 ….(2)

Adding the equations (1) and (2), we get;

3A = 63

A = 21

Therefore, the age of Ani is 21 yrs

And the age of Biju is 21 + 3 = 24 yrs.

Explanation:

Let's say that two buddies each contribute Rs. x and Rs. y in the capital.

According to what has been stated,

x + 100 = 2(y − 100)…..(i) 

And,

6(x − 10) = (y + 10)….(ii) 

Think about equation (i),

x + 100 = 2(y − 100)

x + 100 = 2y − 200

x − 2y = −300…..(iii) 

in equation (ii),

6x − 60 = y + 10

6x − y = 70…..(iv) 

(iv) × 2 – (iii) 

12x – 2y – (x – 2y) = 140 – (-300)

11x = 440

x = 40

When we change x in equation (iii) to 40, we get;

40 – 2y = -300

2y = 340

y = 170

The two pals, therefore, carried Rs. 40 and Rs. 170.

Explanation:

Let the train's speed be x km/hr, the time it takes to travel a distance be t hours, and the distance be d km.

Train speed is calculated as the product of the distance travelled and the time required to cover that distance, or x = d/t  d = xt.

(i)

Case 1: The train would have taken 2 hours less time if it had been travelling at a speed of 10 km/h higher.

(x + 10) = d/(t – 2)

(x + 10)(t – 2) = d

xt + 10t – 2x – 20 = d

d + 10t – 2x = 20 + d [From (i)]

10t – 2x = 20…..(ii)

Case 2: It would have taken 3 hours longer than expected if the train had been moving at a 10 km/h slower speed.

So, (x – 10) = d/(t + 3)

(x – 10)(t + 3) = d

xt – 10t + 3x – 30 = d

d – 10t + 3x = 30 + d [From (i)]

-10t + 3x = 30…..(iii)

Adding (ii) and (iii), we get;

x = 50

As a result, the train is moving at 50 km/h.

When we change x in equation (ii) to 50, we get;

10t – 100 = 20

10t = 120

t=12 hours

Train journey distance, d = xt = 50 x 12 = 600 km

As a result, the train travels a total of 600 kilometres.

Explanation:

Let y represent the number of pupils in a row and x represent the number of rows.

Students in a row divided by the number of rows equals the total number of pupils in the class.

= xy

Case 1:

Student total is equal to (x - 1) (y + 3).

xy = (x − 1) (y + 3)

xy = xy − y + 3x − 3

3x − y − 3 = 0

3x − y = 3…..(i) 

Case 2:

Student total is equal to (x + 2) (y- 3)

xy = xy + 2y − 3x − 6

3x − 2y = −6…..(ii) 

Taking equation (ii) from the equation I we obtain;

(3x − y) − (3x − 2y) = 3 − (−6)

− y + 2y = 9

y = 9

When we change y = 9 in the equation I we obtain;

3x − 9 = 3

3x = 12

x=4

As a result, xy = 4x9 = 36 is the total number of students in the class.

Explanation:

Given,

∠C = 3 ∠B = 2(∠B + ∠A)

3∠B = 2 ∠A + 2 ∠B

∠B = 2 ∠A

2∠A – ∠B= 0- – – – – – – – – – – – (i)

We know that the sum of a triangle’s interior angles is 180°.

Thus, ∠ A +∠B + ∠C = 180°

∠A + ∠B +3 ∠B = 180°

∠A + 4 ∠B = 180°– – – – – – – – – – – – – – -(ii)

Multiplying equation (i) by 4, we get;

8 ∠A – 4 ∠B = 0- – – – – – – – – – – – (iii)

Adding equations (iii) and (ii), we get;

9 ∠A = 180°

∠A = 20°

Using this in equation (ii), we get;

20° + 4∠B = 180°

∠B = 40°

And 

∠C = 3∠B = 3 x 40 = 120°

Therefore, ∠A = 20°, ∠B = 40°, and ∠C = 120°.

Explanation:

5x – y = 5

⇒ y = 5x – 5

Its solution table will be.

Also given,3x – y = 3

y = 3x – 3

The graphical representation of these lines will be as follows:

As seen in the graph above, the vertices of the triangle formed by the lines and the y-axis have the coordinates (1, 0), (0, -5), and (0, -3).

Explanation:

(i) px + qy = p – q……………(i)

qx – py = p + q……………….(ii)

Multiplying equation (i) by p and equation (ii) by q, we get;

p2x + pqy = p2 − pq ………… (iii)

q2x − pqy = pq + q2 ………… (iv)

Adding equations (iii) and (iv), we get;

p2x + q2 x = p2 + q2

(p2 + q2) x = p2 + q2

x = (p2 + q2)/ (p2 + q2) = 1

Substituting x = 1 in equation (i), we have;

p(1) + qy = p – q

qy = p – q – p

qy = -q

y = -1

(ii) ax + by= c…………………(i)

bx + ay = 1+ c………… ..(ii)

Multiplying equation (i) by a and equation (ii) by b, we get;

a2x + aby = ac ………………… (iii)

b2x + aby = b + bc…………… (iv)

Subtracting equation (iv) from equation (iii),

(a2 – b2) x = ac − bc– b

x = (ac − bc – b)/ (a2 – b2)

x = c(a – b) – b / (a2 – b2)

From equation (i), we obtain

ax + by = c

a{c(a − b) − b)/ (a2 – b2)} + by = c

{[ac(a−b)−ab]/ (a2 – b2)} + by = c

by = c – {[ac(a − b) − ab]/(a2 – b2)}

by = (a2c – b2c – a2c + abc + ab)/ (a2 – b2)

by = [abc – b2c + ab]/ (a2 – b2)

by = b(ac – bc + a)/(a2 – b2)

y = [c(a – b) + a]/(a2 – b2)

(iii) x/a – y/b = 0

ax + by = a2 + b2

x/a – y/b = 0

⇒ bx − ay = 0 ……. (i)

And

ax + by = a2 + b2…….. (ii)

Multiplying equations (i) and (ii) by b and a, respectively, we get;

b2x − aby = 0 …………… (iii)

a2x + aby = a3 + ab2 …… (iv)

Adding equations (iii) and (iv), we get;

b2x + a2x = a3 + ab2

x(b2 + a2) = a(a2 + b2) 

⇒ x = a

Substituting x = 1 in equation (i), we get;

b(a) − ay = 0

ab − ay = 0

ay = ab

⇒ y = b

(iv) (a – b)x + (a + b) y = a2 – 2ab – b2

(a + b)(x + y) = a2 + b2

(a + b) y + (a – b) x = a2 − 2ab − b2 …………… (i)

(x + y)(a + b) = a2 + b2

(a + b) y + (a + b) x = a2 + b2 ………………… (ii)

Subtracting equation (ii) from equation (i), we get;

(a − b) x − (a + b) x = (a2 − 2ab − b2) − (a2+ b2)

x(a − b − a − b) = − 2ab − 2b2

− 2bx = − 2b (a + b)

x = a + b

Substituting x = a + b in equation (i), we get;

y (a + b) + (a + b)(a − b) = a2 − 2ab – b2

a2 − b2 + y(a + b) = a2− 2ab – b2

(a + b)y = −2ab

y = -2ab/(a + b)

(v) 152x – 378y = – 74 

–378x + 152y = – 604

152x – 378y = – 74 ….(i)

–378x + 152y = – 604….(ii)

From equation (i),

152x + 74 = 378y

y = (152x + 74)/378

Or

y = (76x + 37)/189…..(iii)

Substituting the value of y in equation (ii), we get;

-378x + 152[(76x + 37)/189] = -604

(-378x)189  + [152(76x) + 152(37)] = (-604)(189)

-71442x + 11552x + 5624 = -114156

-59890x = -114156 – 5624 = -119780

x = -119780/-59890

x = 2

Substituting x = 2 in equation (iii), we get;

y = [76(2) + 37]/189

= (152 + 37)/189

= 189/189

= 1

Therefore, x = 2 and y = 1

Explanation:

Provided that the quadrilateral ABCD is cyclic.

As is common knowledge, a cyclic quadrilateral's opposing angles are additional.

So,

∠A + ∠C = 180

4y + 20 + (-4x) = 180

-4x + 4y = 160

⇒ -x + y = 40….(i)

And

∠B + ∠D = 180

3y – 5 + (-7x + 5) = 180

⇒ -7x + 3y = 180…..(ii)

Equation (ii) – 3 × (i),

-7x + 3y – (-3x + 3y) = 180 – 120

-4x = 60

x = -15

Substituting x = -15 in equation (i), we get;

-(-15) + y = 40

y = 40 – 15 = 25

Therefore, x = -15 and y = 25.