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

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

#### Question No - 20

In the below figure $AB\left \| CD\left \| EF\: \: and \: \: \: GH\left \| KL$. 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?

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

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

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

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

## Is x=4, y = 0, the solution of y - 4 = 0 ?

No $(\because 0-4\neq 0)$

## If the linear equation has solutions (-5, 5) , (0,0) (5-5) then the equation is ..............

$x+y = 0$

## If (0,2) is a solution of the linear equation 2x+3y = k, then find the value f k.

$\because (0,2)$ is the solution of given equation

$\therefore$it satisfies the equation

$\therefore 2(0)+3(2)= k$

$k =6$

(1,-1)

## x = 3, y = 2 is a solution of the linear equation ............

$2x+3y = 12$

x-y = 3

## If  $x = 1$ , $y = -1$ is a solution of the equation  $px-2y=10,$   the value of p is..................

$p=8$

## Any solution of a linear equation $2x+0y+9=0$  in two variables is .........

$\left ( -\frac{9}{2}, m \right )$

## Find three solutions of the linear equation $7x-5y = 35$  in two variables

So, when $y = \frac{7x-35}{5}$

 x 5 0 10 y 0 -7 7

## Check  which of the following is (are)  the solution (s) of the equation  3y - 2x = 1.   (i) (4,3)             (ii) $2(\sqrt{2}, 3\sqrt{2})$

Put x = 4 and  y = 3

then $3y-2x=3(3)-2(4)=1$

So, (4,3)  is the solution of the equation

Again put $x = 2\sqrt{2}$

and $y = 3\sqrt{2}$ ,  then

$3y-2x =3(3\sqrt{2})-2(2\sqrt{2})$

$=5\sqrt{2}\neq 1$

So, $(2\sqrt{2},3\sqrt{2})$ is not a solution of the given equation

## If x = 2 and  y = 1 is the solution of the linear equation $2x+3y+k = 0$ find the value of k.

$2x+3y+k =0$

If x = 2 and y = 1 is the solution of the linear equation $2x+3y+k =0$ then

$2(2)+3(1)+k =0$

$\rightarrow k =-7.$

## Find the value of k so that x= -1 and y = -1 is a solution of the linear equation $9kx+12ky =63$

Substituting $x = -1$ and y = -1 in $9kx+12ky =63.$

We get,

$\Rightarrow 9k (-1)+12k(-1)=63$

$\Rightarrow -9k-12k=63$

$\Rightarrow -21k =63\Rightarrow k =-3$

## Write the equation $y\sqrt{3}=8x+\sqrt{3}$ in the form  $ax+by+c = 0.$check weather  (0,-1) and $(\sqrt{3},9)$ are the solution of this equation

$y =\sqrt{3}=8x+\sqrt{3}$

$\Rightarrow 8x-y \sqrt{3}+\sqrt{3}=0$

Putting x = 0,  y = -1

$\Rightarrow \sqrt{3}+\sqrt{3}\neq 0.$

$\therefore (0,-1)$  is not the solution of the given equation

Putting $x =\sqrt{3} , y=9$

$\Rightarrow 8\sqrt{3} -9\sqrt{3}+\sqrt{3}=0$

Which is correct

$\therefore (\sqrt{3},9)$ is a solution   of the given equation

## Given the equation 2x+y= 7, (i)  What is the value of x, when the value of y is 3 (ii) What is the value of y, when the value of x is 4? (iii)Find the one more solution for the above equation ?

(i) when y =3 then

$2x+3 = 7$

$\Rightarrow 2x = 4$

$\Rightarrow x = 2$

(ii) When x = 4, then

$2(4)+y = 7$

$\Rightarrow y=7-8=-1$

(iii) When x = 1, then

$2+y = 7 \Rightarrow y = 5$

$\therefore$ One more solution is (1,5)

## For what value of k, the linear equation 2x+ky =8 has x = 2 and y = 1  as its solution  if x=4, then find the value of y.

The linear equation is 2x+ky = 8

At  x = 2, y = 1

$2(2)+k(1)=8$

$\Rightarrow 4+k=8$

$\therefore k =4$

If x = 4 then

$\Rightarrow 2(4)+4y=8$

$\Rightarrow 8+4y=8$

$\Rightarrow 4y =0$

$\therefore y = 0$

## For what value of p; x =2,y=3 is a solution of (p+1)x-(2p+3) y - 1 = 0 and write the equation

Given equation is

$(p+1)x-(2p+3)y-1=0....(i)$

If x = 2, y = 3  is  the solution of the equation (i) then

$(p+1)2-(2p+3)3-1=0$

$\Rightarrow 2p+2-6p-9-1=0$

$\Rightarrow -4p-8=0$

$\Rightarrow p =-2$

Put the equation (i) then

$-x+y-1=0$

$x-y+1 =0$  is the required equation

## If the point (3,4) lie on the graph of the linear equation 3y = kx+7, then find the value  of k. Also find two more solutions of the equation

On putting (3,4) in the equation of the line

3y = kx +7

$\Rightarrow 3(4) = k(3)+7$

$\Rightarrow 12= 3k+7$

$\Rightarrow 3k = 5$

$\therefore k = \frac{5}{3}$

The equation becomes

9y=5x+21

Two more solutions are (-6,-1) and (12,9)

## Find k in each case if x = 2, y = 1 is a solution o f the equation :  (i) 3x+2y = k,            (ii) 2x-ky= 6 (iii)$\frac{x}{4}+\frac{y}{3} = 5k$

Given,

3x+2y = k

Put x = 2, y = 1 then

$3(2)+2(1)= k \rightarrow k = 8$

(ii) $2x-ky = 6$

Put x = 2, y = 1 then

2(2)-k (1) = 6

$\Rightarrow 4-k= 6\Rightarrow k = 4-6=-2$

(iii) $\frac{x}{4}+\frac{y}{3} = 5k$

Put x = 2, y = 1  then

$\frac{2}{4}+\frac{1}{3} = 5k$

$5k =\frac{10}{12} = \frac{5}{6}$

$\Rightarrow k =\frac{1}{6}$

## Find three different solutions for the equation 3x+2y =1.

Given  $3x+2y = 1$

$\Rightarrow \frac{1-3x}{2}= y$

(i) Put x = 1, then   $y =\frac{1-3(1)}{2}=\frac{-2}{2}=-1$

(ii) Put x = 3 then  $y =\frac{1-3(3)}{2}=\frac{1-9}{2}=-4$

(iii) Put x = 5, then $y =\frac{1-3(5)}{2}=\frac{1-15}{2}=-7$

Hence three different solutions for the equations 3x+2y = 1

 x 1 3 5 y -1 -4 -7

## Find two different solutions of the equation 4x+3y = 12 from its graph .

Given equation

4x+3y = 12

Put x= 0, then 0+3y = 12

$\Rightarrow y =4$

Hence point on y-axis is (0,4)

Now put y = 0, then

4x+0 = 12

x = 3

Hence the point on x-axis is (3, 0)

## Solve the equation $\frac{x}{3}+2=2x-3$ and represent the solution on the   cartesian plane.

$\frac{x}{3} +2 = 2x-3$

$\Rightarrow \frac{x}{3} -2x=-3-2$

$\Rightarrow x -6x=-5 \times 3$

$\Rightarrow -5x=-15$

$\Rightarrow x = 3$

Put x = 3 on the Cartesian plane

On Cartesian plane, x = 3 is a parallel to y-axis

## Write three solutions of the equation 3x = y +3 Draw its graph and find the points where the graph intersects the axes.

$3x=y+3.$three solutions are $x = 1, y = 0: x = 2, y = 3 and x = 0 , y = -3$

From graph it is clear that line meets x- axis  at (1,0) and y- axis at (0,-3)

## Solve 5x-2 = 3x-8 and represent the solution  (i) On a  number line  (ii) In the cartesian plane

$5x-2 = 3x-8$

$\Rightarrow 2x = -6$

$\Rightarrow x = -3$

(i) Point P (-3,0)  represent the solution x = -3 on the number line

(ii)Line AB represent the solution in the cartesian plane.

## Draw the graph of the linear equation x+y =7. Verify from the graph that (8,-1) is a solution of the equation x+ y = 7

x+ 7  = 7

y = 7-x

 X 5 7 4 Y 2 0 3

From graph it is clear that (8,-1) lies on the line AB.

Hence (8,-1) is a solution of the given equation.

## 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)

$\Rightarrow$y = 10+4x -4

$\Rightarrow y =4x +6$

Table of solutions

 X 0 1 -1 Y 6 10 2

## Solve the equation 3x+4 = 5x+8  and represent the solution on (i) the number line (ii) the cartesian plane. What do you get as the representation of the solution on the cartesian plane ? In cartesian planes, how many solutions this equation has?

$3x+4 = 5x+8$

$\Rightarrow 2x = -4$

$\Rightarrow x = -2$

(i) On the number line the point P (-2,0)

represent the solution

(ii) On the Cartesian plane x = -2 is a line parallel to the y-axis at a distance of 2 units to left of it.

It has infinite solutions.

## Find whether the pair of linear equations  y = 0 and y = -5 has no solution, unique solution or infinitely many solutions

The pair of equation y = 0 and  y = -5 has no  solution

## For what the value of 'k' the system of equation kx + 3 y = 1, 12x + ky = 2 has  no solution

The condition for no solution

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}\Rightarrow \frac{k}{12}=\frac{3}{k}\neq \frac{1}{2}$

When $\frac{k}{12}=\frac{3}{k}$, we  get k2 = 36

$i.e., k = \pm 6$

$k \neq 6$

So  k = -6                     ($\because$ For k = 6   $\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$   )

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

Given equations are :

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

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

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

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

I           II         III

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

$\alpha = 4$

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  $\beta = 8$

## Determine graphically whether the following pair of linear equations has; 3x- y = 7  2x + 5y +1= 0 has : (i) a unique solution  (ii) Infinitely many solutions or   (iii) No solution

$3x - y = 7$

$\Rightarrow 3x - y-7 = 0$

$a_{1} = 3, b_{1} = -1, c_{1} = -7$

$2x + 5y +1 = 0$

$a_{2} = 2, b_{2} = 5 , c_{2} = 1$

$\frac{a_{1}}{a_{2}} = \frac{3}{2}$

$\frac{b_{1}}{b_{2}} = \frac{-1}{5}$

$\frac{3}{2} \neq \frac{-1}{5}$

thus   $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$

Hence, given pair of equations has a unique solution

We have

3x- y - 7 = 0

$y = 3x-7$

 x 0 2 3 y -7 -1 2

and  2x +5y +1 = 0

$y = \frac{-1-2x}{5}$

 x -1/2 2 -3 Y 0 -1 1

x = 2 and y = -1 is the solution

## 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.

$\Rightarrow$2x+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

$\triangle ABC$is the region between the two lines represented by the above equations and the x-axis.

## Is  x = 4, y = 0 the solution of y - 4 = 0

No

$(\because 0-4\neq 0)$

x+y = 0

## If (0,2) is a solution of the linear equattion 2x+3y = k then find the value of k.

(0,2) is the solution of given equation

$\therefore$ It satisfies the equation

$\therefore 2(0)+3(2)= k$

$\therefore k = 6$

## ABCD is a rhombus. Show that the diagonal AC bisects $\angle$A as well as $\angle$C and diagonal BD bisects $\angle$B as well as $\angle$D.

Given ABCD is a rhombus,

So, AB = BC

To prove :

$\angle BAC = \angle DAC$ and $\angle DCA = \angle BCA$

Solution : In  $\Delta ADC$ and ABC

CD =  CB (Given)

AC is common

So, $\Delta ADC \cong \Delta ABC$

So, $\angle BAC = \angle DAC$   (c.p.c.t)

$\angle DCA = \angle BCA$        (c.p.c.t)

## If am = bl, then find whether the pair of linear equation ax + by =  c  and lx+ my = n  has no solution, unique solution or infinitely many solutions.

Since, am = bl

$\Rightarrow \frac{a}{l} = \frac{b}{m} \neq \frac{c}{n}$

So ax +by  = c and lx + my = n  has no solution.

## If  $ad \neq bc$  then finds whether the pairs of linear equations ax+by = p and cx+dy = q has no solution, unique solution or infinitely many solutions.

$ad \neq bc \Rightarrow \frac{a}{c} \neq \frac{b}{d}$

Hence the pair of given linear equations has unique solution.

## For what value of p does the pair of linear equations given below has  unique solution ? 4x +py +8 = 0  and  2x+2y + 2 = 0.

Given equation are :

4x + py + 8 = 0

2x + 2y + 2 = 0

The condition of unique solution, $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$

Hence,  $\frac{4}{2} \neq \frac{p}{2}\: \: or \: \frac{2}{1} \: \: \neq \: \: \frac{p}{2}$

$p \neq 4$

## For what value of k, the pair of linear equation kx-4y = 3, 6x-12y =9  has an infinite number of solutions?

Pair of linear equation   kx-4y -3 = 0

and                                6x-12y - 9 = 0

Condition for infinite solutions :

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

$\frac{ k}{6} = \frac{-4}{-12}= \frac{3}{9}$

$k =2$

## For what value of k, 2x+3y = 4  and  (k+2) x+6y = 3k+2 will have infinitely many solutions?

For equation,  2x+3y - 4 = 0

$a_{1}=2, b_{1} = 3, c_{1}= -4$

For equation, $(k+2) x+6y-(3k+2)=0$

$a_{2}= k+2, b_{2} = 6, c_{2} =-(3k+2)$

For infinite solutions

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

$\frac{2}{k+2}=\frac{3}{6}=\frac{4}{3k+2}\Rightarrow 12= 3k +6$

$6 = 3k \Rightarrow k = 2$

## 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.

## Determine the values of m and n so that the following system of linear equations have infinite number of solutions : (2m-1) x +3y - 5 = 0  3x + (n-1) y -2 = 0

(2m-1)x+3y-5 = 0

On comparing with the equation

$a_{1}x+b_{1}y+c_{1} = 0.$

$a_{1} = 2m-1, b_{1} = 3 , c_{1} = -5$

$3x+ (n-1)y-2 = 0$

On comparing with the equation

$a_{2}x+b_{2}y+c_{2} = 0.$

$a_{2} = 3, b_{2} = (n-1) , c_{2} = -2$

For a pair of linear equations to have an infinite number of solutions

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

$\Rightarrow \frac{ 2m-1}{3} = \frac{3}{n-1} = \frac{5}{2}$

$\Rightarrow 2 (2m-1) = 15 \, \, and\, \, 5(n-1) = 6$

$\Rightarrow m = \frac{17}{4}, n = \frac{11}{5}$

## Represent the following pair of linear equations graphically and hence comment on the condition of consistency of this pair.  x-5y = 6, 2x -10 y = 12.

Given x-5y = 6 $\Rightarrow$ 2x-10y = 12.

Since of the lines are coincident, so the system of linear equations is consistent with infinite solutions.

## If x= $x =- \frac{1}{2}$, is a solution of the quadratic equation,$3x2+2kx-3=0$ find the value of k.

Putting  $x =- \frac{1}{2}$ in  $3x^{2}+2k x-3 = 0$

$3\left (- \frac{1}{2} \right )^{2} +2k \left ( -\frac{1}{2} \right ) - 3 = 0$

$\rightarrow \frac{3}{4} -k-3 = 0$

$\rightarrow k= \frac{3}{4} - 3$

$\rightarrow k = \frac{3-12}{4}$

$\rightarrow k =\frac{-9}{4}$

## 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.

## For what value of P will be the following system of equations have no solution? $(2p-1) x+(p-1) y = 2p + 1; y+3x-1 = 0.$

$(2p-1) x+(p-1) y = 2p + 1; y+3x-1 = 0.$

$a_{1} = 2p-1, b _{1} =p-1 \, \, and \, \, c_{1} = - (2p+1)$

and for  3x + y - 1 = 0

$a_{2} = 3, b_{2} = 1 \, \, and \, \, c_{2} = -1$

The condition for no solution is

$\frac{a_{1} }{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

$\Rightarrow \frac{2p-1}{3} = \frac{p-1}{1} \neq \frac{2p+1}{1}$

By  $\frac{2p-1}{3}=\frac{p-1}{1}$

$\Rightarrow 3p-3=2p-1$

$\Rightarrow 3p-2p= 3-1$

$\therefore p = 2$

from $\frac{p-1}{1}\neq 2p+1$

We have $p-1 \neq 2p+1$

$\Rightarrow -1-1 \neq 2p -p$

$\therefore p \neq - 2$

from $\frac{2p-1}{3}\neq \frac{2p+1}{1}$

$\Rightarrow 2 p - 1 \neq 6p+3$

$4p\neq -4$

$\therefore p \neq -1$

## The solution of    $\frac{3}{7} + x = \frac{17}{7} \: \: is ...................$ a. x = 14       b. x = 2      c. x = 10       d. x = 4

b. x= 2

$\frac{3}{7}+x =\: \frac{17}{7}$

$x = \frac{17}{7}-\frac{3}{7}=\frac{14}{7}=2$

## If the equation x2 + kx + k = 0 has only one solution, find the possible values of k.

If there is only one solution the discriminant = 0

b2 - 4ac = 0

k2 - 4 x 1 x k = 0

k2 = 4k

k = 4

Possible value of k is 4.

## 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.

## Write $p(x)=x^2 -9x+20$ as a product of two first degree polynomial. Write also the solutions of the equation p(x)=0

$p(x)=x^2 -9x+20$

If $x^2 -9x+20=x^2 +(a+b)x+ab,$ then

a+b =-9   ........ (1)

ab=20

$(a-b)^2 =(a+b)^2 -4ab=81-80=1$

$a-b=\sqrt{1}=1$ ....... (2)

(1) +(2), 2a=-8, a=-4

From (1), -4+b=-9, b=-9+4=-5

$\therefore x^2 -9x+20=(x-4)(x-5)$

Solution of $x^2 -9x+20=0$ are 4 and 5