Pair of Linear Equations in Two Variables Test

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Pair of Linear Equations in Two Variables Test - 39

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Weekly Quiz Competition

Question 1
1 / -0

Find the value of k for which the given system of equations has a unique solution.
$$(k- 3)x + 3y = k; kx + ky = 12$$

A
$$k\neq3 $$

B
$$k\neq 9$$

C
$$k\neq 6$$

D
$$k\neq 2$$

Solution

Comparing $$ \left ( k-3 \right )x +3y = k$$ with $$a_1x+b_1y+c_1=0$$,

and $$kx+ky =12$$ with $$a_2x+b_2y+c_2=0$$,

we get, $$ a_{1} = \left ( k-3 \right ) , a_{2} = k , b_{1} = 3, b_{2}= k$$

Now,
Given system of equations has unique solution
$$\Rightarrow \dfrac{a_1{}}{a_{2}} \neq \dfrac{b_{1}}{b_{2}}$$

$$ \dfrac{\left ( k-3 \right )}{k}\neq \dfrac{3}{k}$$

$$\Rightarrow k^{2}- 3k \neq 3k$$

$$\Rightarrow k^{2} \neq 6k$$

$$\Rightarrow k^{2} -6k\neq 0$$

$$\Rightarrow k(k-6) \neq 0$$

$$\Rightarrow k \neq 0$$ and $$\ k \neq 6$$

So $$ k \neq 6$$ is the correct option among given options.
Question 2
1 / -0

Find the value of $$k$$ for which the given system of equation has no solution.
$$kx + 2y - 1 = 0$$
$$5x - 3y + 2= 0$$

A
$$k=-\dfrac15$$

B
$$k=-\dfrac{10}{3}$$

C
$$k=\dfrac95$$

D
$$k=\dfrac65$$

Solution

The equations are
$$\Rightarrow kx+2y-1=0\\$$
$$\Rightarrow a_{1}=k,b_{1}=2,c_{1}=-1\\$$
$$\Rightarrow 5x-3y+2=0\\$$
$$\Rightarrow a_{2}=5,b_{2}=-3,c_{2}=2\\$$

The given system of equations has no solution
$$\Rightarrow \displaystyle \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$$

$$\Rightarrow \displaystyle \frac{k}{5}=\frac{2}{-3}\Rightarrow k=-\frac{10}{3} $$
Question 3
1 / -0

Find the value of a and b for which the given system of linear equation has an infinite number of solutions.
$$2x + 3y = 7$$ and $$(a - b) x + (a + b)y = 3a + b - 2$$

A
$$a=2, b=3$$

B
$$a=5, b=1$$

C
$$a=-5, b=3$$

D
$$a=-1, b=2$$

Solution

Comparing $$2x+3y =7$$ with $$a_1x+b_1y+c_1=0$$,
we get,
$$a_1=2, b_1=3,c_1=-7$$

Comparing $$\left ( a-b \right )x + \left ( a +b \right )y = 3a + b - 2$$ with $$a_2x+b_2y+c_2=0$$,
we get,
$$a_2=\left ( a-b \right ), b_2=\left ( a +b \right ), c_2=-(3a + b - 2)$$

Given system of equation have infinite solutions:
$$\Rightarrow \dfrac{a_1{}}{a_{2}} = \dfrac{b_{1}}{b_{2}} = \dfrac{c_{1}}{c_{2}}$$

$$\Rightarrow \dfrac{2}{a-b}= \dfrac{3}{a+b}= \dfrac{7}{3a + b -2}$$

From $$\dfrac{a_1{}}{a_{2}} = \dfrac{b_{1}}{b_{2}}$$

$$\Rightarrow \dfrac{2}{a-b}= \dfrac{3}{a+b}$$

$$\Rightarrow 2a + 2b = 3a-3b$$

$$\Rightarrow a-5b = 0 ......... (I)$$

Now from $$\dfrac{b_{1}}{b_{2}} = \dfrac{c_{1}}{c_{2}}$$

$$\Rightarrow \dfrac{3}{a+b}= \dfrac{7}{3a + b -2}$$

$$\Rightarrow9a + 3b -6 = 7a + 7b$$

$$\Rightarrow 9a-7a+3b-7b =6$$

$$\Rightarrow 2a-4b = 6 ...... (II)$$

Now, for solving eq $$I$$ and eq $$II$$

$$a-5b = 0 $$

$$2a-4b = 6$$

Multiplying eq $$I$$ by $$2$$

$$2a - 10b =0$$

$$2a - 4b =6$$

on subtraction we get

$$\Rightarrow -6b = -6$$

$$\Rightarrow b = 1$$

Now, substituting $$b$$ value in eq $$I$$

$$\Rightarrow 2a-10\left ( 1 \right ) = 0$$

$$2a - 10 = 0$$

$$ 2a = 10$$

$$ a = \dfrac{10}{2}$$

$$ a = 5$$

$$\Rightarrow a = 5 , b = 1$$
Question 4
1 / -0

Determine whether the following system of linear equations have no solution, infinitely many solution or unique solutions.

$$x + 2y = 3, 2x + 4y = 15$$

A
No solution

B
Infinitely many solution

C
Unique solution

D
Cannot be determined

Solution

The equations are
$$\Rightarrow x+2y-3=0$$
$$\Rightarrow a_{1}=1,b_{1}=2,c_{1}=-3$$
$$\Rightarrow 2x+4y-15=0$$
$$\Rightarrow a_{2}=2,b_{2}=4,c_{2}=-15$$
$$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{1}{2}$$
$$\Rightarrow \frac{b_{1}}{b_{2}}=\frac{2}{4}=\frac{1}{2}$$
$$\Rightarrow \frac{a_{1}}{b_{1}}=\frac{a_{2}}{b_{2}} \neq\dfrac{c_{1}}{c_{2}}$$
Hence, the given system of equations has no solution
Question 5
1 / -0

Find the value of $$k$$ for which the given system of equations has infinite number of solutions.
$$5x + 2y = 2k$$ and $$2(k+ 1)x + ky = (3k+ 4)$$

A
4

B
7

C
3

D
6

Solution

Comparing $$ 5x+2y=2k$$ with $$a_1x+b_1y+c_1=0$$, we get,
$$a_1=5, b_1=2,c_1=-2k$$

Comparing $$2(k+1)x+ky=3k+4$$ with $$a_2x+b_2y+c_2=0$$, we get,
$$a_2=2(k+1), b_2=k, c_2=-(3k+4)$$

Now,
Given: the equations has infinite no of solutions.
$$\therefore \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$$

$$\dfrac{5}{2(k+1)}=\dfrac{2}{k}=\dfrac{-2k}{-(3k+4)}$$

When,

$$\dfrac{5}{2(k+1)}=\dfrac{2}{k}$$

$$\implies 5k=4(k+1)$$
$$\implies 5k-4k=4$$
$$\implies k=4$$

and

$$\dfrac{2}{k}=\dfrac{-2k}{-(3k+4)}$$

$$\implies 2(3k+4)=2k^2$$
$$\implies 3k+4=k^2$$
$$\implies k^2-3k-4=0$$
$$\implies k^2-4k+k-4=0$$
$$\implies k(k-4)+(k-4)=0$$
$$\implies (k-4)(k+1)=0$$

Either, $$k=4\space or\space k=-1$$
$$k=4$$ is the value that satisfies both the conditions.

So, $$Option \ A \ is\ correct$$
Question 6
1 / -0

Find the value of k for which the given system of equations has no solution.$$kx + 3y = k - 3; 12x + ky = k$$

A
$$k=-5$$

B
$$k=-2$$

C
$$k=-3$$

D
None of these

Solution

A system of equations, $$a_1x+b_1y+c_1=0$$ and $$a_2x+b_2y+c_2=0$$ has no solution for the following condition:
$$\displaystyle \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$$

The equations in the given question are:
$$kx+3y-k+3=0$$ and $$a_{1}=k,b_{1}=3,c_{1}=3$$
$$12x+ky-k=0$$ and $$a_{2}=12,b_{2}=k,c_{2}=-k$$
Substituting the above values in the condition mentioned above for no solution:
$$\dfrac{a_{1}}{b_{1}}=\dfrac{a_{2}}{b_{2}}$$

$$\dfrac{k}{12}=\dfrac{3}{k}$$

$$k^{2}=36$$
$$\Rightarrow k=6$$
Question 7
1 / -0

Determine by drawing graphs whether the following pair of equations has a unique solution or not: $$2x - 3y = 6, 4x - 6y = 9$$. If yes, find the solution also

A
No

B
Yes, $$x=3$$, $$y=-2$$

C
Ambiguous

D
Data insufficient
Question 8
1 / -0

Find the value of $$k$$ for which the given system of equations has a unique solution:
$$x - ky = 2$$;
$$3x + 2y = - 5$$

A
$$k\neq -\dfrac23$$

B
$$k\neq \dfrac12$$

C
$$k\neq \dfrac32$$

D
$$k\neq -\dfrac52$$

Solution

The equations are
$$\ x-ky-2=0 $$
$$\Rightarrow a_{1}=1,b_{1}=-k,c_{1}=-2$$
$$\ 3x+2y+5=0$$
$$\Rightarrow a_{2}=3,b_{2}=2,c_{2}=5$$
The given system of equations has a unique solution
$$\Rightarrow \dfrac{a_{1}}{a_{2}}\neq \dfrac{b_{1}}{b_{2}}$$
$$\Rightarrow \dfrac{1}{3}\neq -\dfrac{k}{2}$$
$$\Rightarrow$$$$k\neq-\dfrac{2}{3}$$
Question 9
1 / -0

Solve:

$$\frac {4}{9}x+\frac {1}{3}y=1, 5x+2y=13$$

A
$$x=3, y=-5$$

B
$$x=3, y=-6$$

C
$$x=3, y=-1$$

D
$$x=1, y=0$$

Solution

Given system of equations are:
$$\frac{4}{9}x + \frac{1}{3}y = 1$$ ... (1)
$$5x+2y = 13$$ ... (2)
From equation 1 will get:
$$ 4x + 3y = 9$$
Now subtracting eq1 and eq2
$$ 4x +3y = 9$$
$$ 5x +2y = 13$$
First multiply eq1 from 2 and eq2 from 2 will get:
$$8x + 6y = 18$$
$$15x+6y = 39$$
now will subtract the equation
$$ \Rightarrow 8x +6y = 18$$
$$ -15x -6y = -39$$
$$\Rightarrow -7x = -21$$
$$ \Rightarrow x = 3$$
now substituting x = 3 in eq 2
$$\Rightarrow 15\left ( 3 \right )+ 6y = 39$$
$$\Rightarrow 45 + 6y = 39$$
$$\Rightarrow 6y = 39-45$$
$$\Rightarrow 6y = -6$$
$$\Rightarrow y = -1$$
$$ \Rightarrow x = 3 , y = -1$$
Question 10
1 / -0

Determine how many solution exist for the following pair of equations $$8x +5y = 9, 16x +10y = 27$$

A
Both equations form coincident line hence they have no solutions.

B
Both equations form parallel lines hence they have no solutions.

C
Both equations form intersecting lines hence they have no solutions.

D
None of these

Solution

The following pair of equations can be written as

$$8x+5y-9=0 \ ; 16x+10y-27=0;$$

Here $$\displaystyle \frac{a_1}{a_2}=\frac{8}{16}=\frac{1}{2}$$

$$\dfrac{b_1}{b_2}=\dfrac{5}{10}=\dfrac{1}{2}$$

$$\dfrac{c_1}{c_2}=\dfrac{-9}{-27}=\dfrac{1}{3}$$,

We can see $$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\neq\dfrac{c_1}{c_2}$$

Hence, the given pair of the linear equation is inconsistent and has no solution.

For a graphical solution,

$$8x+5y=9$$

$$y=\dfrac{9-8x}{5}$$

$$16x+10y=27$$

$$y=\dfrac{27-16x}{10}$$

Now, we can make the table and make the graph as shown in the solution.

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Re-Attempt Weekly Quiz Competition

Pair of Linear Equations in Two Variables Test - 39

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Pair of Linear Equations in Two Variables Test - 39

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SHARING IS CARING

Question 1

$$k\neq3 $$

$$k\neq 9$$

$$k\neq 6$$

$$k\neq 2$$

Solution

Question 2

$$k=-\dfrac15$$

$$k=-\dfrac{10}{3}$$

$$k=\dfrac95$$

$$k=\dfrac65$$

Solution

Question 3

$$a=2, b=3$$

$$a=5, b=1$$

$$a=-5, b=3$$

$$a=-1, b=2$$

Solution

Question 4

No solution

Infinitely many solution

Unique solution

Cannot be determined

Solution

Question 5

4

7

3

6

Solution

Question 6

$$k=-5$$

$$k=-2$$

$$k=-3$$

None of these

Solution

Question 7

No

Yes, $$x=3$$, $$y=-2$$

Ambiguous

Data insufficient

Question 8

$$k\neq -\dfrac23$$

$$k\neq \dfrac12$$

$$k\neq \dfrac32$$

$$k\neq -\dfrac52$$

Solution

Question 9

$$x=3, y=-5$$

$$x=3, y=-6$$

$$x=3, y=-1$$

$$x=1, y=0$$

Solution

Question 10

Both equations form coincident line hence they have no solutions.

Both equations form parallel lines hence they have no solutions.

Both equations form intersecting lines hence they have no solutions.

None of these

Solution

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