Chapter-3, PAIR OF LINEAR EQUATIONS IN TWO VARIABLES Extra Questions

Explanation:

The two equations are as follows: 6x - 3y + 10 = 0.

2x - y + 9 = 0

We must discover the graphic solution.

We are aware that for two linear equations, a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, respectively.

If a₁/a₂ = b₁/b₂ c₁/c₂, the chart would comprise of two straight lines, the pair of linear equations is erratic, and there is no solution to the problem.

Here, a₁=6, b₁=-3,

 c₁=10, and a₂=2,

 b₂=-1, c₂=9.

A₁/A₂ = 6/2 = 3, 

B₁/B₂ = -3/-1 = 3,

 and C₁/C₂ = 10/9.

a₁/a₂ = b₁/b₂c₁/c₂.

As a result, the chart of the pair of formulas shows two lines that are parallel.

Explanation:

Both x + 2y + 5 = 0 and -3x - 6y + 1 = 0 are provided.

In this section a₁=1 also b₁=2 also c₁=5 also a₂=3, 

b₂=1 also a₁/a₂=13 also b₁/b₂=26 also c₁/c₂=51 also a₁/a₂ =b₁/b₂  also c₁/c₂ 

Therefore, there is no answer to the pair of problems.

Explanation:

The right response is coincident or C intersecting.

The framework must possess at least one singular solution or infinitely many solutions in order for a pair of linear equations to be consistent.

ii. We find a singular solution when two lines cross at a single place.

iii. Every point on two lines that correspond has the same value. This indicates that there are an endless number of sites where the line can connect, leading to an infinite number of solutions.

iv. Lines expressing a pair of linear equations can only be continuous when they are either parallel with one another or intersect.

Choice(C ) is therefore the appropriate response if a set of linear equations is consistent.

Explanation:

The right response is C No Solution.

Considering that the x-axis (y=0 solution) never intersects y=-7. There is no answer to the given pair of problems.

Explanation:

D Intersecting at (a,b) is the right answer.

the justification for the ideal choice.

The pair of equations  x=a and y=b must initially be drawn on the chart, as illustrated below.

Both lines overlap at (a,b) as can be seen from the chart.

So, the answer (D) is the proper one.

Explanation:

A₁/a₂=B₁/b₂=C₁/c₂ is the prerequisite for parallel crosses. (i) Provided line:

3x – y + 8 = 0

6x – ky + 16 = 0

In this case, a₁=3,b₁=1,c₁=8, 

and a₂=6,b₂=k,

c₂=16 from Eq. (i) 36=1/k=8/16 1/k=1/2 k=2.

Explanation:

If the lines are parallel, the formulas in the solution a₁/a₂ = b₁/b₂ c₁/c₂ are inconsistent.

The correct answer is now 3/2 = 2k/5

=> k = 15/4.

Then, 2k/5 = 30/20 = 3/2 is obtained, satisfying the requirement of constancy.

Explanation:

need for infinitely many solutions

a₁/a₂=b₁/b₂=c₁/c₂

Two lines are provided: 6x - 2y = 3 and cx - y = 2.

Here, a=c, b=1, c=2, and a=6, b=2, c=3 from equation (i). c/6=−1−2=−2−3

⇒c/6=1/2 and c/6=2/3⇒ c=3 and c=4

Because two distinct connections are taken into account, c has two different values.

Therefore, the pair of equations will have an infinite number of answers for any value of c.

Explanation:

A₁/a2=b₁/b2=c₁/c2=1k is a condition for dependent linear equations.

The line's equation is 5x + 7y - 2 = 0.

Here, a₁=5 and b₁=7 and c₁=2.

From Eq. (i),  5/a₂,= 7/b₂,  and = 2/ c₂ =1/k

when any random stable, k, is used.

If k = 2, then a₂ = 10, b₂ = 14, and c₂ = 4. Line's necessary equation equals a₂x+b₂y+c₂=0.

In other words, 10x - 14y = -4 or -10x + 14y - 4 = 0.

Explanation:

D x - 4y -14 = 0 and 5x - y - 13 = 0 are the right answers.

For every problem pair, replace x=2 and y=3.

It is noted that only these equations have this issue as an approach: x4y14=0; 5xy13=0.

Explanation:

The formulas are as follows: x - y = 2. -------------- (1)

x + y = 4 ------------- (2)

We must ascertain the answers x = a and y = b.

2x = 6

x = 6/2

x = 3 after adding (1) and (2).

In (1), enter x = 3, 

3 - y = 2, and 3 - 2 y = 1

As a result, the answers to the equations x = a and y = b are x = 3 and b = 1, respectively.

Explanation:

The right response is D 25 and 25.

Allow x to be the amount of Re1 coins.

Let y be the amount of Rs. 2 coins.

presently, assuming the circumstances

x+y=50 ....(i)

Additionally, x+2y=75....(ii)

When we take formula (i) out of formula (ii), we have

(x+2y)−(x+y)=75−50

⇒y=25

When we replace y=25 in (i), we get

x+25=50⇒x=25.

Explanation:

Assume that the dad is currently x years old and the son is currently y years old.

The relation between the ages 4 years from now is established by the formula (x + 4) = 4(y + 4) x - 4y = 12.....(i)

x = 6y.....(ii) (Given)

Upon entering the value of x from Equation (ii), we obtain the following: 6y - 4y = 12 2y = 12 y = 6

If y = 6, then x must be 36.

Dad is currently 36 years old, and son is 6 years old.

Explanation:

We are aware that the equation for the absence of a solution is a1/a 2= b1/b 2 c1/ c2 (parallel lines).

The values provided in the equations are 2 + 4 -3=0.

6x+12y–6=0

When the equations are compared to ax+by +c =0, a1=2, b1=4, c1=-3; 2=6, 2=12, 2=-6; a1/ 2=2/6=b1/3; c1/b2=4/12=1/3; and 1/ 2=-3/6=-1/2

Here, a1/a 2 equals b1/b 2 c1/ c2, or parallel lines.

As a result, there is no solution to the provided pair of linear equations.

Explanation:

Knowing that a1/a2=b1/b2=c1/c2 is the prerequisite for coinciding lines, the following linear formulas are provided:

3x+1/7y=3

7x+3y=7

Considering the ax+by+c=0, a1=3, b1=17, c1=-3, a2=7, b2=3, c2=-7 formulas, we see that a1/a2=3/7, b1/b2=1/21, and c1/c2=-3/-7=3/7.

Here,a1/a2≠b1/b2.

A set of coinciding lines cannot be represented by the provided set of linear formulas.

Explanation:

The two equations are as follows: -3x - 4y = 12.

4y + 3x = 12

Checking for consistency between the linear formulas is necessary.

We are aware that for 2 linear formulas, a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, respectively.

The pair of linear equations is incorrect if a1/a2=b1/b2c1/c2.

As 3x + 4y = 12, the equation 4y + 3x = 12 can be written.

Here, a2 = 3, b2 = 4, and c2 = 12 where a1 = -3, b1 = -4, and c1 =

Consequently, a1/a2 = -3/3 = -1, b1/b2 = -4/4 = -1, c1/c2 = 12/12 = 1 -1, and c1/c2 = 1 1

Explanation:

The two linear equations are as follows: λx + 3y = -7 2x + 6y = 14.

We must determine whether the pair of equations has an unlimited number of solutions if λ is 1.

We are aware that for two linear equations, a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, respectively.

i) The pair of linear equations are conditional and logical if a1/a2=b1/b2=c1/c2.

ii) A pair of coincident lines will form the graph. The two problems can have an endless number of solutions since each point on the lines will be a solution.

There, a2 = 2, b2 = 6, and c2 = 14, with a1 =λ, b1 = 3, and c1 = -7.

As a result, a1/a2 = /2, b1/b2 = 3/6 = 1/2, and c1/c2 = -7/14 = 1/2.

Case a) λ/2 = 1/2 λ = 2/2 λ = 1

Case b) λ/2 = -1/2

λ = -2/2

λ = -1

Its worth changes over time.

Thus, the number can not be 1 for the pair of equations that have an infinite number of answers.

Explanation:

The two linear formulas are as follows: x - 2y = 8 5x - 10y = c.

To ensure that the equations have a singular answer, we must determine the amount of c.

If a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, respectively, for a pair of linear equations in two variables, and if a1/a2b1/b2, then the graph will be two lines converging at a single point, that is a couple of problems' answer.

In this case, a1 = 1, b1 = 2, and c1 = 8; a2 = 5, b2 = 10, and c2 = c.

As a result, a1/a2 = 1/5, b1/b2 = -2/-10 = 1/5, and c1/c2 = 8/c.

1/5 = 1/5 = 8/c

So, 1/5 = 8/c

c = 8(5) c = 40

When c = 40, the pair of equations a1/a2=b1/b2=c1/c2 contain an unlimited number of answers.

Consequently, the pair holds true for all real values of neither x - 2y = 8 nor 5x - 10y = c has a single answer.

Explanation:

Given that each location along the line has a fixed x manage, the path denoted by the value of x = 7 is parallel to the y-axis. It can also be represented visually.

Consequently, the assertion is untrue.

Explanation:

The two linear formulas are as follows: λx + y = λ2 x + y = 1

We need to figure out what number of λ causes the pair of linear equations to have an unlimited number of answers.

We are aware that for 2 linear equations, a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, respectively.

i) The pair of linear equations are conditional and linear if a1/a2=b1/b2=c1/c2.

ii) A pair of coincident lines will form the graph. The two equations will have an endless number of solutions since each point on the lines will be a solution.

Here, a1 =λ, b1 = 1, and c1 =λ 2.

a2 = 1, b2 =λ, and c2 =.

Therefore, a1/a2 =λ /1 = b1/b2 = 1/λ c1/c2 = 2/1 = 

If there are infinitely many answers, then a1/a2=b1/b2=c1/c2 Case

1) λ = 1/λ λ2 = 1

So, λ = ±1

Case 2) λ = λ2 λ2 - 

λ = 0 λ(λ - 1) = 0

So, λ = 0 

λ - 1 = 0 

λ = 1

So, both situations are satisfied by = 1.

As a result, the pair of linear equations have an infinite number of answers for the value of = 1.

Explanation:

A1 = k, A2 = 12, B1 = 3, B2 = k, C1 = (k - 3) and C2 = -k in the formula above.

If a problem has a limitless number of answers, then a1/a2 = b1/b2 = c1/c2 

In the calculations above, k / 12 = 3 / k = k - 3 / k,

k / 12 = 3/ k,

k2 = 36, and k = 6.

 Additionally, 3k = k(k - 3)/k,

3k = k2 -3k,

k2 = 6k, and 6k = k.

The answer to both formulas' requirements for k is 6.

Consequently, the pair of linear equations have an infinite number of answers with k = 6.

Explanation: 

The two linear formulas are as follows: x + 2y = 1 

(a - b)x + (a + b)y = a + b - 2.

The values of a and b for which the pair of linear equations has an unlimited number of answers must be found.

We are aware that for 2 linear formulas, a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, respectively.

i) The pair of linear equations are contingent and logical if a1/a2=b1/b2=c1/c2.

ii) A pair of coincident lines will form the graph. The two equations will have an endless number of solutions since each point on the lines will be a solution.

Here, a1=1, b1=2, and c1=1.

A2 = A-B, B2 = A+B, 

and C2 = A+B+2

Therefore, a1/a2 = 1/(a - b) b1/b2

= 1/(a+b-2) = 2/(a + b) c1/c2

A1/A2=B1/B2=C1/C2 for all possible answers infinity

2/(a + b) = 1/(a + b - 2) = 1/(a - b)

Case 1) 1/(a - b) = 2/(a + b)

1(a + b) = 2(a - b)

a + b = 2a - 2b

A = 3b and a = 2a + b + b = 0 and a + 3b = 0 -------------------------- (1)

Case 2) 2/(a + b) = 1/(a + b - 2)

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

2a + 2b - 4 = a + b

A + B = 4 --------------------- (2) 2a - a + 2b - a

Replace (1) with (2), resulting in 3b + b = 4 

4b = 4 

b = 4/4

b = 1

b = 1 in (1),

a = 3 in (1), 

and a = 3

Therefore, the pair of linear equations have an infinite number of solutions for the given parameters of a = 3 and b = 1.

Explanation:

The two linear equations are 3x - y - 5 = 0 as is.

6x - 2y - p = 0

The amount of p where the paths depicted by these equations are parallel must be found.

We are aware that for two linear equations, a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, respectively.

If a1/a2=b1/b2c1/c2, the chart is made up of two intersecting lines, the pair of linear formulas are incompatible, and there is no answer to the equations.

Here, a1 = 3, b1 = 1, and c1 = -5,

while a2 = 6, b2 = 2, 

and c2 = -p

Therefore, a1/a2 = 3/6 = 1/2, 

b1/b2 = -1/-2 = 1/2, 

and c1/c2 = -5/-p = 5/p.

For lines that are parallel, a1/a2=b1/b2+c1/c2 1/2 = 1/2 5/p.

So, p ≠ 5(2) p ≠ 10

As a result, for all actual The pair of linear equations show parallel lines for all p values save p = 10.

Explanation:

The provided linear formulas are x - 3y = 2. ------------------- (1)

-2x + 6y = 5 ---------------- (2)

When the two formulas are compared to the basic shape of a line that is straight, axe + by + c = 0, we get the following results: a1 = 1, a2 = -2, b1 = -3, b2 = 6, c1 = -2, and c2 = 5.

The formulas divide to give us a1/a2 = - 1/2,

b1/b2 = - 3/6 = - 1/2,

and c1/c2 = 2/5.

a1/a2 = b1/b2 c1/c2, etc.

They are parallel lines as a result.

As a result, the following formulas' representation of both linear routes always crosses them. The two run parallel to one another.

Explanation:

A1/A2 b1/B2 is a requirement for the set of devices to possess a specific answer.

Let's look at both of-variable linear problems a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0.

The sole solution to both of those solutions is x = - 1 and y = 3.

Therefore, it must meet the equations above.

The equation is: a1(-1) + b1(3) + c1 = 0 - a1 + 3b1 + c1 = 0 ------------------ (1)

"a2(- 1) + b2(3) + c2 = 0" means "a2 + 3b2 + c2 = 0" (2)

The formulas (1) and (2) can be satisfied with any number of a1, b1, c1, and a2, b2, c2.

There are hence an endless number of alternative linear equation couples.

Explanation:

The linear formulas from the aforementioned query are as follows:

2x + y = 23---------------(i)

4x - y = 19----------------(ii)

6x = 42 x = 7 when (i) and (ii) are added together.

By replacing the x value in (i), we obtain 2(7) + y = 23,

14 + y = 23, 

and y = 23 - 14

y = 9.

5y - 2x = 5 9 - 2 - 7 

= 45 - 14

= 31

y/x - 2 = 9/7 -2 = -5/7 is the result of substituting the values of x and y in 5y - 2x and y/ x - 2.

5y - 2x = 31 y/x - 2 = -5/7.

Consequently, 31 and -5/7 are the figures for (5y - 2x) and y/x - 2 correspondingly.

Explanation:

The opposite corners of a circle are equivalent by definition.

The lengths are equivalent, hence CD = AB and x+3y = 13.(i) The breadth is equal, hence AD = BC and 3x+y=7.(ii)

We obtain 9x+3y=21x+3y=13 by multiplying Eq (ii) by 3 and then removing Eq (i).                    –––––––––––––8x =8 x = 1

When we enter x=1 in equation (i), we obtain 3y=12 y=4.

As a result, 1 and 4 are the necessary numbers for x and y, respectively.

Explanation:

The answers to the aforementioned query are x + y = 3.3----------(1) and 0.6/3x - 2y = -1.

Cross-multiplying, we get 0.6 = -3x + 2y.

Multiplying (1) by 2 and combining (2) results in 3x - 2y = -0.6-------(2). 

5x = 6

x = 6/5

x = 1.2.

When the value of x in (1) is substituted, 1.2 + y = 3.3 

y = 3.3 - 1.2

y = 2.1,

where x = 1.2 

y = 2.1.

In light of this, x = 1.2 and y = 2.1.

Explanation:

The answer to the previous query is x/10 + y/5 - 1 = 0-----------(1).

x/8 + y/6 = 15-----------------(2)

With the equation rearranged, we obtain x + 2y = 10-------------(3)

3x + 4y = 360--------------(4)

Let's resolve these both of-variable linear problems.

When (3) is multiplied by two and deducted by (4), the result is (3x + 4y) - (2x + 4y) = 360 - 20 x = 340.

In (3), replace x with the number 340 + 2y = 10 2y = 10 - 340 = -330 y = -165.

The formula is y = x + 5.

When x and y are substituted in the formula above, -165 = (340) + 5 340 = -170 = -1/2. x = 340. 

y = -165.

λ = -1/2.

The answers to the two formulas are thus x = 340, y = -165, and the necessary value of is -1/2.

Explanation:

The formula for the given query is 3x + y + 4 = 0--------------(1).

6x - 2y + 4 = 0----------------(2)

When compared to the basic shape of the linear line, axe + by + c = 0, we obtain the following: a1/a2 = 1/2;

b1/b2 = 1/-2; 

c1/c2 = 1. 

a1/a2 b1/b2.

As a result, the supplied set of linear problems crosses at one location resulting in one answer.

The linear equation pairs that are supplied are hence consistent.

Let's use an image to show this.

We have y = -4 3x + y + 4 = 0.

6x - 2y + 4 = 0

2y = 6x + 4

 y = 3x + 2

We obtain the straight edge AB by displaying the locations B(0, -4) and A(-2, 2) on a graph.

We obtain a horizontal line PQ by displaying the locations of Q (0, 2) and P (1, 5).

At point C (-1, -1), the lines AB and PQ converge.

The aforementioned combinations thus appear equivalent.

Explanation:

Chart for the path 2x + y = 4 and 2x - y = 4.

Worksheet for liner y = 2x - 4 and x = 2x.

The two lines are shown graphically as follows:

The y-axis and lines in this form the letters ABC.

As a result, A (0, 4), B (2, 0), and C (0, - 4) are the vertices of an ABC.

Amount of AOB = 2 times the area of ABC

The area of ABC is 2 1/2 4 2 sq units or 8 square units.

The shape's area is 8 square units as a result.

Explanation:

We can deduce the linear equations from the previous problem as x + y - 2 = 0----------(1).

2x - y - 1 = 0---------(2)

When compared to the generic form of the straight line axe + by + c = 0, we have al = 1, bl = 1, cl = -2,

a2 = 2, b2 = 1,

c2 = 1, al/a2 = 1/2,

bl/b2 = 1,

and cl/c2 = 2.

Al/a2 bl/b2.

In this, the two lines meet at a single place.

Thus, there is only one solution to the combination of equations.

These equations are coherent as a result.

Now, y = 2 - x and x + y = 2.

Because y = 2x - 1 and 2x - y - 1 = 0.

The intersection of the lines is at E(1,1).

Thus, the connection of the lines formed by the formulas for x + y = 2 and 2x - y = 1, or E(1, 1), can contain endless lines, such as y = x, 2x + y = 3, x + 2y = 3, and so on.

Therefore, the points where the linear equations x + y = 2 and 2x - y = 1 intersect can be reached by infinite lines.

Explanation:

Since f(x) = 2x3 + ax2 + 2bx + 1 and f(-1) = 0, (x + 1) is a factor of that equation.

F(-a) = 0 if (x + a) is a factor of f(x) = ax2 + bx + c.

2(-1)3 + a(-1)2 + 2b (-1) + 1 = 0

After some additional math, -2 + a - 2b + 1 = 0

a - 2b - 1 = 0-----------(1)

2a - 3b = 4

3b = 2a - 4 b = (2a-4/3).

The result of replacing the value of b in (1) is a - 2(2a-4/3)- 1 = 0 3a - 2(2a - 4) - 3 = 0

3a - 4a + 8 - 3 = 0 

-a + 5 = 0

 a = 5.

When the value of an is substituted in (1), we have 5 - 2b - 1 = 0 2b = 4 b = 2, where a = 5

As a result, a and b have values of 5 and 2, respectively.

Explanation:

The shape of the triangle's angles is equal to x, y, and 40°.

Because the total number of angles is 180°, x + y + 40° = 180°, x + y = 140°.------------(1)

x-y = 30°--------------(2)

The result of adding (1) and (2) is 2x = 170° x = 85°.

When x = 85° is substituted in (1), we obtain 85° + y = 140° and y = 55°.

x = 85° y = 55°.

As a result, x and y have values of 85° and 55°, respectively.

Explanation:

Assume Salim to be x years old.

Let's assume that his daughter is y years old.

Salim was three times his daughter's age two years ago based on the first condition: x - 2 = 3(y - 2)

x - 2 = 3y - 6 

x - 3y = -4-----------(1) Salim will be four years older than twice her age six years from now, according to the second condition: x + 6 = 2(y + 6) + 4

x + 6 = 2y + 12 + 4

x -2y = 16 - 6

x - 2y = 10-------(2) Right now, solve equations (1) and (2).

y = 14 when (1) is subtracted from (2).

In (2), change x to y, where x - 2 x 14 = 10

x = 10 + 28

x = 38.

So, we have x = 38 and y = 14.

Salim and his daughter are so 38 and 14 years old, respectively.

Explanation:

Assume that the father is x years old and that the combined ages of the two children are y years.

The answer is x=2y. (1) After 20 years, 

x+20=y+20+20

x+20=y+40 

x=y+20.(2) By combining (1) and (2), y equals 20.

When y = 20 is substituted in solution (1), we obtain x = 40.

Consequently, the father is 40 years old.

Explanation:

Considering that, the ratio equals 4: 5 if 8 is deducted from each of the integers.

(5x - 8) : (6x - 8) = 4 : 5 (5x - 8) / (6x - 8) = 4 / 5

5(5x - 8) = 4(6x - 8)

25x - 40 = 24x - 32

25x - 24x = 40 - 32 x = 8

initial number: 5x = 5 (8) = 40

next figure = 6x = 6 (8) = 48

Therefore, the figures are 40 and 48, and follow the 5:6 ratio.

Explanation:

Let's say there are x and y students in halls A and B, accordingly.

Consequently, x - 10 = y + 10 x - y = 20 and (x + 20) = 2 (y - 20) x - 2y = - 60 are the results of the provided condition.

When we put y = 80 in Eq. (i), we get (x - y) - (x - 2y) = 20 + 60 

x - y - x + 2y x = 100 and y = 80, meaning that there are 100 students in hall A and 80 students in hall B.

Explanation:

The set fee for one of the two days is x.

Y represents the supplementary fee for each.

Latika paid $22 for a book held for six days under the first requirement. (x + 4y = 22)(1) In accordance with the second requirement, Anand paid 16 for a book maintained for four days. Solve equations (1) and (2) now.

The result of deducting (2) from (1) is 2y = 6y = 3.

When the value of y is replaced in (2), we get the following: x + 2 x 3 = 16

x = 16 - 6 = 10 

x = 10. 

x = 10. y = 3.

As a result, the fixed fee is equal to 10, and the fee for each additional day is equal to 3.

Explanation:

Assume that there are x accurate responses.

Consider that there are (120 - x) incorrect responses.

In accordance with the stated circumstance, x 1 - (120 - x) 1/2 = 90.

Now calculate x.

x - 6 + x/2 = 90

3x/2 = 150 x = 150 × 2/3 x = 300/3 x = 100.

Jayanti successfully responded to 100 questions as a result.

Explanation:

Depending on the issue, either 7x + y = 170 or y = 170 - 7x or 6x + 10 + x + y = 180. One more time, 5x + 3y - 10 = 180 or 5x + 3y = 190 or 5x + 3(170-7x) = 190 or 5x + 510 - 21x = 190 or -16x = -320 or x = 20 So, y = 170 - 7x = 170 - 140 = 30 Therefore, the sum of all angles is 130°, 100°, 50°, and 80°.

Explanation:

As a result of the above query, we have the equation 2x + y = 6.

2x - y + 2 = 0.

Equation 2x + y = 6 table.

Equation 2x - y + 2 = 0 table

The visual intersection of the two equations is at point E(1, 4), where x = 1 and y = 4.

Let A1 stand in for the ACE area, and let A2 stand in for the BDE area.

A1 = Area of ACE = 1/AC/PE.

ACE's region is 1/2 x 4 x 4.

The area of ACE equals 8.

A2 = Area of BDE = BD x QE / half

Area of BDE = 1/2 x 4 x 1, 

which equals 2.

The necessary ratio is therefore A1: A2 = 8: 2 = 4: 1.

Explanation:

The answer to the previous query is y = x, 

3y = x,

and x + y = 8.

Graph for the line y = x,

Graph for the line x = 3y

Chart for the line with x+y=8

We obtain a chart of the provided lines by displaying the points.

At Q and D, accordingly, the lines y = x and 3y = x cross the line x + y = 8.

These lines together comprise the OQD.

Therefore, O(0, 0), Q(4, 4), and D(6, 2) are the vertices of the OQD.

Explanation:

The answer to the previous query is 2x - y - 4 = 0, 

x = 3, and x = 5.

The graph of 2x - y - 4 = 0 is obtained by displaying the locations (0, -4) and (2, 0) and connecting them.

Create the lines 3x and 5x.

According to the chart, the quadrilateral created by the provided lines and x-axis is ABCD.

The aspect of the hexagon ABCD is equal to half of the sum of both sides and the parallel line distance.

The area of the ABCD hexagon is equal to 1/2 x AB x (AD + BC).

ABCD's quadrilateral area is equal to 1/2 x 2 x (2 + 6). [OA = 5-3 = 2, AD = 2, and BC = 6]

ABCD's sided polygon's area is 8 square units.

As a result, the rectangle formed by the lines and the x-axis has an area of 8 square units.

Explanation:

Let's say a pen costs Rs. x and a pencil box costs Rs. y.

Following the above requirement, x + y = 25 when 4x + 4y = 100.

Additionally, 3x = y + 15 3x - y = 15

Eqs. (i) and (ii) are added together to give us 4x = 40 x = 10.

When x = 10 is used in Eq. (i), we obtain

Consequently, a pen box costs Rs. 10 and a pencil box costs Rs. 15, respectively.

Explanation:

The formula for the aforementioned query is,

3x-y = 3 --------------(1)

2x - 3y = 2------------(2)

x + 2y = 8-------------(3)

Let's assume that lines (1), (2), and (3), which stand for AB, BC, and CA, respectively, are the sides of an ABC.

By resolving lines (1) and (2), point B where they connect can be found.

With (1) multiplied by 3 and (2) subtracted, the result is (9x - 3y) - (2x - 3y) = 9 - 2

7x = 7

x = 1.

When the value of x is replaced in (1), we get 3 x 1 - y = 3

y = 0.

B's point or vertex's coordinates are (1, 0).

Finding the intersection point C of lines (2) and (3) requires solving.

By dividing (3) by 2, then multiplying the result by 2, we arrive at (2x + 4y) - (2x - 3y) = 16 - 2

7y = 14 

y = 2.

Replace with the

We obtain x + 2 x 2 = 8 x = 8 - 4 x = 4 as the value of y in (3).

The point of vertex C's coordinates is (4,2).

The intersection point A will be obtained by solving lines (3) and (1).

By calculating (1) by 2, then adding (3), we get at (6x - 2y) + (x + 2y) = 6 + 8 7x = 14 x = 2, which is equal to 6.

When the value of x is replaced in (1), we obtain 3 x 2 - y = 3

 y = 6 - 3

 y = 3.

Corner or vertex A's coordinates are (2, 3).

A(2, 3), B(1, 0), and C(4, 2) are the vertices of the ABC created by the provided lines.

Explanation:

Make the rickshaw's speed x km/h.

Let the bus's speed be y km/h.

2 kilometres travelled in a rickshaw in t1 = 2/x hours.

Time required to cover the rest of the route by bus (14 – 2) = 12 kilometres, or 12 hours.

t1 + t2 = 1/2 under the first condition, and 2/x + 12/y = 1/2 under the second condition.(1)

The time required for a 4-kilometre rickshaw ride, t3, equals 4/x hours.

The rest of the distance will be travelled in (14 - 4) hours by bus and will be 10 kilometres.

The second condition is that t3 + t4 = 1/2 + 9/60 = 1/2 + 3/20.

4/x + 10/y = 13/20-----------------------------(2)

Think about 1/x equals u and 1/y = v.

2u + 12v = 1/2 is the result of rearranging the equation. ------------------(3)

4u + 10ν = 13/20----------------------Solve the linear equations in (3) and multiply it by (2) in (4).

When we subtract 4, we get 1-13/20 (4u + 24v - 4u + 10v).

14v = 7/20

2v = 1/20 v = 1/20.

When the value of v in (3) is changed, we get 2u + 12(1/40) = 1/2.

2u = 1/2 - 3/10

2u = 5-3/10

2u = 2/10 u = 1/10.

1/x = = 1/10 x = 10.

1/y = 1/40 y = 40. 

x = 10.

 y = 40.

As a result, the bus travels at 40 km/h and the rickshaw at 10 km/h, respectively.

Explanation:

Assume that the stream is moving at v km/h.

Provided,

In still water, a human can row at a speed of 5 km/h.

The speed of a rower coming upstream is (5 + v) km/h.

A person can row upstream at a speed of (5 - ) km/h.

The distance traveled 40 km downstream in t1 is 40/5+v hours.

The time needed to travel 40 kilometers upstream, t2 = 40/5-v hours.

The provided condition states that t2 = t1 x 3. 40/5-v = 40/5+v x 3. 1/5-v = 3/5+v

Solve this linear equation, please.

2.5km/h is equal to 5+ v = 15 - 3v = 4v = 10 v = 10/4 v.

Consequently, the stream is moving at a 2.5 km/h rate.

Explanation:

Let's assume that the motorboat travels at a speed of u km/hr in still water.

Let's assume that the stream is moving at v km/h.

motorboat's downstream speed is equal to (u + ) km/h.

Boat's upstream speed is equal to (u - ) km/h.

30 km upstream journey time is given by t1 = 30/u-v hours.

Traveling 28 km downstream takes t2 = 28/u+v hours.

The first condition states that t1 + t2 = 7 hours.

30/u-v + 28/u+v = 7------------------------------------------(1)

T3 = 21/u-v hours for a 21 km upstream journey.

T4 = 21/u+v hours for a downstream journey of 21 kilometers.

The second condition is that t4 + t3 equals 5 hours.

21/u+v + 21/u+v = 5------------------------------------------(2) Now imagine that x = 1/u+v and y = 1/u-v

Rearranging (1) and (2), we obtain 30x + 28y = 7---------------------------(3)

21x + 21y = 5---------------------------------------------------(4)

x + y = 5/21.

Conquer the equations (3) and (4) in linear form

Equation (4) is multiplied by 28 before being subtracted from equation (3) to yield (30x - 28y). - (28x + 28y) = 7 - 140/21

2x = 7 - 20/3

2x = 1/3 x = 1/6.

When the value of x is replaced in (4), we obtain 1/6 + y = 5/21 y = 5/21 = 1/6 = 10 -7/42 = 3/42 y = 1/14. x = 1/u+v = 1/6 u + v = 6---------------------(5).

y = 1/u-v = 1/14 u-v = 14----------------------------(6)

(5) and (6) added together give us 2u = 20 u = 10.

The result of replacing the value of u in equation (5) is 10 + v = 6 = -4.

As a result, the stream moves at a speed of 4 km/h while the motorboat travels at 10 km/h in still water.

Explanation:

Both of-digit figures are thought of as 10x + y.

Case 1: By multiplying the total by 8 and taking away 5, the result is 10x + y.

By applying the distributive property, 8x + 8y - 5 = 10x + y 2x-7y = -5(1)

Case 2: By dividing the digits' difference by 16 and adding 3, 16 x (x - y) + 3 becomes 10x+ y.

Using the distributive property, 6x - 17y = -3------------ 16x - 16y + 3 = 10x + y2. Resolve equations (1) and (2) in linear form.

By dividing by two and multiplying by three, (1) becomes (6x - 17y) - (6x - 21y) = - 3 - (- 15) 

4y=12

y = 3.

Using the value of y in (1) as a substitute, 2x - 7 x 3 = -5

2x = 21 - 5 = 16 x = 8.

Number with two digits = 10x + y = 10 x 8 + 3 = 80

Explanation:

Assume that luxury travel costs x.

Assume that the price of half first class is x/2.

The cost of each reservation ticket is y.

Case I: The cost of one reservation excellent ticket from station A to station B is 2530 x plus 2530 y.-------------------------(1)

Case II: The cost of one reserved first-class ticket from station A to station B is equal to 3810 x + y + x/2 + y = 3810.

3x/2 + 2y = 3810 3x + 4y = 7620-----------------------(2)

The linear formulas (1) and (2) must be solved. By multiplying (1) by 4 and then removing it from (2), we receive the following result: (3x + 4y) - (4x + 4y) = 7620 - 10120 -x = -2500 x = 2500.

Changing the value of x in (1) results in the following: 2500 + y = 2530 y = 2530 - 2500 y = 30.

Thus, we have x = 2500 and y = 30.

Explanation:

Think of x as the saree's purchase price.

Think of y as the sweater's list price.

Case I: A saree sold for 8% profit plus a jumper sold for 10% off equaled $108.

It can be expressed mathematically as (100 + 8)% of x and (100 - 10)% of y = 1008

Thus, we have 108% of x plus 90% of y = 1008

1.08x + 0.9 y = 1008--------------(1)

Case II: A 10% profit on the saree and an 8% discount on the jumper equals $1028.

You can express it mathematically as (100 + 10)% of x + (100 - 8)% of y = 1028.

Therefore, we have 110% of x + 92% of y = 1028.

1.1x + 0.92y = 1028----------------(2) Find the solution to the linear equations (1) and (2) by substituting the value of y from (1) in (2) to obtain 1.1 x 0.92(1008-1.08x/0.9) 1.1 xThe formula reads as follows: 0.9x + 927.36 - 0.9936x = 1028 x 0.9 0.99x - 0.9936x = 925.2 - 927.36 -0.0036x =-2.16 x = 2.16/0.0036 x = 600.

With the value of x substituted in (1), we have 1.08 x 600+ 0.9y=1008

648 + 0.9y=1008

0.9y=1008 - 648.

0.9y = 360

 y = 360/0.9 

y = 400.

So, x = 600 and y = 400.

As a result, the sweater's list price (price before discount) is $400 and the saree's cost price is 600.

Explanation:

Take into account the number of investments in plan A as x.

Take into account y as the amount invested in plan B.

Principal times rate times time/100 equals simple interest.

Case I: 8% annual interest on Scheme A + 9% annual interest on Scheme B = 1860 x x 8% x 1/100 + y x x 9% x 1/100 = 1860. 8x + 9y = 186000--------------------(1)

Case II: 9% annual interest on plan A plus 8% annual interest on scheme B is 1860 + 20 x x 9 x 1/100 + y x 8 x 1/100 = 1860 9x/100 + 8y/100 = 1880 9x + 8y= 188000.-----------------------(2) Conquer the linear equations (1) and (1) multiplied by 9 and by 8 respectively.

The result is (72x + 81y) - (72x + 64y) = 9 x 186000 - 8 x 188000 17y = 1000[(9 x 186) - (8 x 188)] = 1000(1674 - 1504) = 1000 x 170 17y = 170000 y =10000

The result of replacing the value of y in (1) is 8x + 9 x 10000 = 186000.

8x = 186000 - 90000

8x = 96000 x = 12000.

She thus put $12,000 and $10,000, respectively, into two schemes A and B.

Explanation:

Let's assume that there are x bananas in lot A.

Let's assume that there are y bananas in lot B.

Case I: The first lot cost $2 for every three bananas, plus the second lot cost $1 for each banana, for a total of $400.

2/3x + y = 400

2x + 3y = 1200---------------(1)

Case #2

Amount received x + 4/5 y = 460 5x + 4y = 2300 First lot cost at a rate of 1 per banana + Second lot cost at a rate of 4 for 5 bananas----------------(2) Compute the solutions to the linear equations (1) and (2). By multiplying (1) by 4 and (2) by 3, respectively, and then removing the results, we obtain (8x + 12y) - (15x + 12y) = 4800, 6900, 7x, and 2100 

x = 300.

In (1), change the value of x.2 x 300 plus 3 y equals 1200.

600 + 3y = 1200

3y = 1200 - 600

3y = 600 y = 200.

Total bananas are equal to the sum of the bananas in lots A and B.

= x + y = total number of bananas

300 + 200 = total quantity of bananas

There are 500 bananas in total.

He had 500 bananas as a result.