How many linear equations in x and y can have a solution as (x = 1, y = 3)?

about 2 years ago 0 Answer 291 views

Show graphically that this system equation 2 X + 3 y is equal to 10, 4x+ 6yis equal to 12. has no solution

about 7 months ago 0 Answer 62 views

Question No - 20

 

In the below figure ABleft | CDleft | EF: : and : : : GHleft | KL. Find angle HKL ?

In this question from lines and angles exercise.in the solution it was said that:-

-----------------------------X----------------------------X------------------------X------------X-----

Now, alternate angles are equal

angle CHG=angle HGN=60^0

angle HGN=angle KNF=60^0           [ corresponding angles]

Hence, angle KNG=180^{o}-60^{o}=120^{o}

Rightarrow angle GNK=angle AKL=120^0      [ corresponding angles]

angle AKH=angle KHD=25^0           [ alternative angles]

Therefore, angle HKL=angle AKH+angle AKL=25+120=145^0

-------------X-------------------------X--------------------------X------------------------------X-----------------------X----------

In the above said solution where did the 60 degree come from?

 

about 4 months ago 0 Answer 0 views

In figure a-b=80 and POQ is a straight line,then find a and b

about 4 months ago 1 Answer 58 views

Starting from Q6 i am not able to find solutions of it

about 2 months ago 1 Answer 6 views

I'm not getting solutions of surface area and volumes starting from q6 . And there no teach me how icon also ! What to do ?

about 2 months ago 1 Answer 58 views

I'm not getting solutions of surface area and volumes starting from q6 . And there no teach me how icon also ! What to do ?

about 2 months ago 0 Answer 0 views

The auto rikshaw fare in a city is charged RS . 10 for first kilometer and @Rs.4 per kilometer for subsequent distance covered. Write the linear equation to express the above statment. Draw the graph of the linear  equation. 

 

Total distance covered =  x km. 

Total fare = y km. 

Fare for the first kilometer 

Subsequent distance  = (x-1) km

Fare for the subsequent distance = Rs. 4(x-1)

According to the question, 

 y = 10+4 (x-1)

Rightarrowy = 10+4x -4 

Rightarrow y =4x +6

Table of solutions

X 0 1 -1
Y 6 10 2

 

Find the values of alpha and beta for which the following pair of linear equations has the infinite number of solutions :  2x +3y = 7;  2alpha x+ (alpha +beta ) y = 28

Given equations are : 

2x +3y = 7 and 2alpha x + (alpha +beta ) y = 28

We know that the condition for a pair of linear equations to be consistent and having infinite number of solution is 

frac{a_{1}}{a_{2}} = frac{b_{1}}{b_{2}}= frac{c_{1}}{c_{2}}

frac{2}{2alpha }= frac{3}{alpha +beta } = frac{7}{28}

I           II         III

From I and III,    frac{2}{2alpha } = frac{7}{28} 

              alpha = 4

From II and III,  frac{3}{alpha +beta } =frac{7}{8}

                   Rightarrow alpha +beta = 12

                         Rightarrow beta = 12-alpha 

                         Rightarrow beta = 12-4

                         therefore beta = 8

Hence alpha = 4 ,and  beta = 8

 

 

The cost of 2 kg of apples and 1 kg of grapes on a day was found to be  Rs. 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs. 300. Represent the situation algebraically and geometrically.

 

Let the cost of 1 kg of apples be Rs. 'x' and cost of 1 kg of grapes  be Rs 'y'.Then the algebraic representation is given by the following equations : 

2x + y  = 160

         y = 160 -2x

4x+ 2y = 300

         y = 150 -2x

To find the equivalent geometric representation 

We find two points on the line representing each equation i,e, we find two solutions of each equation.

Rightarrow2x+y = 160

         y = 160-2x

X 50 45
Y 60 70

4x + 2 y = 300

          y  = 150-2x

X 50 40
Y 50 70

 

 

Hence geometric representation is shown above which is a pair of parallel lines.

Solve the following pair of linear equations graphically:

x +3y = 6, 2x -3y  = 12

Also shade the region bounded by the line  2 x -3y =12  and both the co-ordinate axes. 

 

x + 3y = 6 Rightarrow y = frac{6-x}{3}: : : : cdot cdot cdot cdot cdot cdot cdot cdot cdot cdot cdot cdot cdot cdot left ( 1 right )        

X 3 6 0
Y 1 0 2

2x-3y = 12: : : cdot cdot cdot cdot cdot cdot cdot cdot cdot cdot cdot cdot left ( 2 right )

y = frac{2x-12}{3}

X 0 6 3
Y -4 0 -2

Plotting the  above points and drawing a line joining them, we get the graphs of the equations x + 3y = 6 and  2x- 3y = 12.

Clearly, the two lines intersect at point B (6,0) 

 Hence x = 6 and  y  = 0 is the solution of the system. 

Again triangle AOB  is the region bounded by the line 2x-3y = 12 and both co-ordinate axes.

Solve the following pair of equations graphically :

2x + 3 y = 12, x -y -1 = 0.

Shade the region between the two lines represented  by the above equations and the x-axis.

2x +3y = 12

 y = frac{12-2x}{3}

X 0 6 3
Y 4 0 2

x -y = 1
y = x -1

X 0 1 3
Y -1 0 2

plotting the above points and drawing a line joining them, we get the graph of equations 2x + 3y = 12 and  x-y -1 = 0.

Clearly, the two lines intersect at point (3,2) Hence x = 3 and  y = 2 is the required solution 

triangle ABCis the region between the two lines represented by the above equations and the x-axis.

Half of the perimeter of a rectangular garden whose length is 4 m more than its width, is 36 m. Find the dimension of garden.

 

Let the length of the garden be x m and its width be y m. 

Then, the perimeter of rectangular garden 

= 2 (Length + width )  = 2 (x+y)

Therefore, Half perimeter  = (x+y)

But its given as 36 m. 

therefore (x+y) = 36

Also, x = y + 5

i.e., x-y = 4

For finding the solution of eqs.(i) and (ii) graphically, 

we form the following table : 

For x+y = 36 

 

 

 

 

Draw the graph by joining the points (20,16) and (24,12) and points (10,6) and (16,12). The two lines intersect at a point (20,16) as shown in the graph.

 

The coach of a cricket team nuys 3 bats and 6 balls for Rs.39,00. Later, she buys another bat and  2 more balls of the same kind for Rs. 1,300. Represent this situation algebraically and geometrically 

Let the cost of one bat be x. and one ball be y. Then, the algebraic representation is given by the following equations:

3x+6y = 3,9000

and  x+2y = 1,300

To obtain the equivalent geometric representation we find two points on the line representing each equation ie., we find two solutions of each equation. these solutions are given below in the table:

For 3x+6y = 3,900

y = frac{1,300-x}{2}

x 0 1,300
y 650 0

for  x+2y = 1,300

y = frac{1,300-x}{2}

x 500 100
y 400 600

We plot the points A (0, 650) and B (1,300,0) to obtain the geometric representation of 3x+6y = 3,900 and C (500,400) and D(100,600) to obtain the geometric representaion of x+2y = 1,300. We observe these lines are coincident.

Aftab tell his daughter, 7 years ago, I Was seven times as old as you were then. Also, 3 years from now, I shall be three times as old as you will be represented this situation algebraically and graphically .

 

Let the present age of father be x years and the age of daughter be y years.

7 years ago father's age = (x-7) years 

7 years ago daughter's age = (y -7)years 

According to the question 

(x-7) = 7 ( y-7)

x-7y = -42

After 3 years father's age = (x+3) years

After 3 years  daughter's age = (y +3)years 

According to the condition,

x+3 = 3 (y+3)

x- 3y = 6

To fid the equivent geometric representation we find some points on the line represnting each question, these solutions are given below in the table

From eq : 

x-7y = -42

x 0 7 14 12
y =frac{x+42}{7} 6 7 8 12
Points A B C G

 

 

x 6 12 18 42
y =frac{x-6}{3} 0 2 4 12
Points  D E F G

Plot the points A(0,6), B (7,7) and C(14,8) and join them to get a straight line ABC. Similarly, plot the points D (6,0), E (12,2) and F (18,4) and join them to get a straight line DEF. These two lines intersect at point G (42,12). Thus we conclude that the present ages of afthaf and his daughter are 42 years and 12 years respectively.

a. Find the centre of the circle with the line joining te points (3,-1), (13,-9) as diameter. Find the equation of this circle.

b. Show taht there is no point on the circle whose x and y coordiantes are equal.

a. Centre of the circle = midpoint of the line joining the points (3,-1) and (13,-9)=left ( frac{3+13}{2},frac{-1+-9}{2} right )=(8,-5)

Radius of teh circle = distance between the point (3,-1) and (8,-5)=sqrt{(3-8)^2 +(-1--5)^2 }=sqrt{5^2 +4^2 }=sqrt{25+16}=sqrt{41}

Eqaution of the circle with centre (8,-5) and radius sqrt{41} is (x-8)^2 +(y--5)^2 =left ( sqrt{41} right )^2

(x-8)^2 (y+5)^2 =41

b. Ket x=y

(x-8)^2 +(x+5)^2 =41

x^2 -16x+64+x^2 +10x+25-41=0

2x^2 -6x+48=0,x^2 -3x+24=0

sqrt{b^2 -4ac}=sqrt{9-96}=sqrt{-87}

Since negative numbers has no squareroot, there is no solution to the equation x^2 -3x+24=0.

So there cannot be any point on the circle where x and y coordinates are equal.