Pair of Linear Equations in Two Variables Test

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Pair of Linear Equations in Two Variables Test - 18

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Question 1
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A system of linear equations is given as follows :
$$a_1x+b_1y+c_1=0$$
$$a_2x+b_2y+c_2=0$$
Both the lines are parallel only if,

A
$$\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}$$

B
$$\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}\neq \frac{c_1}{c_2}$$

C
$$\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$$

D
None of these

Solution

$$ Let\quad us\quad take\quad two\quad equations\quad as,\quad$$
$$ \\ { a }_{ 1 }x+{ b }_{ 1 }y+{ c }_{ 1 }=0........(i)\quad and\\ { a }_{ 2 }x+{ b }_{ 2 }y+{ c }_{ 2 }=0.......(ii).$$

$$\\ If\quad the\quad the\quad lines\quad are\quad parallel\quad then\quad the\quad slopes\quad of\quad the\quad two\quad lines\quad \\ should\quad be\quad equal\quad and\quad the\quad intercepts\quad of\quad the\quad two\quad lines\quad \\ should\quad NOT\quad be\quad equal.$$

$$\\ Writing\quad the\quad equations\quad of\quad lines\quad in\quad slope-intercept\quad form\\ we\quad have,\quad$$

$$ (i)\quad as\quad y=\dfrac { -{ a }_{ 1 } }{ b_{ 1 } } x+\left( \dfrac { -{ c }_{ 1 } }{ { b }_{ 1 } } \right) \quad and\quad \\ (ii)\quad as\quad y=\dfrac { -{ a }_{ 2 } }{ b_{ 2 } } x+\left( \dfrac { -{ c }_{ 2 } }{ { b }_{ 2 } } \right) $$

$$\\ Now\quad if\quad the\quad lines\quad are\quad parallel\quad then\quad the\quad slopes\quad are\quad equal.$$

$$\\ i.e\quad \dfrac { -{ a }_{ 1 } }{ b_{ 1 } } =\dfrac { -{ a }_{ 2 } }{ b_{ 2 } } \Longrightarrow \dfrac { { a }_{ 1 } }{ { a }_{ 2 } } =\dfrac { b_{ 1 } }{ b_{ 2 } } ........(iii)$$

$$\\ Also\quad the\quad intercepts\quad are\quad NOT\quad equal.$$

$$\\ i.e\quad \left( \dfrac { -{ c }_{ 1 } }{ { b }_{ 1 } } \right) \neq \left( \dfrac { -{ c }_{ 2 } }{ { b }_{ 2 } } \right) \Longrightarrow \dfrac { { c }_{ 1 } }{ { c }_{ 2 } } \neq \dfrac { { b }_{ 1 } }{ { b }_{ 2 } } ......(iv)$$

$$\\ Combining\quad (iii)\quad \& \quad (iv)\quad we\quad have\quad \dfrac { { a }_{ 1 } }{ { a }_{ 2 } } =\dfrac { b_{ 1 } }{ b_{ 2 } } \neq \dfrac { { c }_{ 1 } }{ { c }_{ 2 }}$$

$$\\ Ans-\quad Option\quad B.\\ \\ \\ $$
Question 2
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STATEMENT - 1 : If the graphs of the two equations are parallel lines, there exists no solution.
STATEMENT - 2 : The system is called an inconsistent system.

A
Statement - 1 is True, Statement - 2 is True, Statement - 2 is a correct explanation for Statement - 1

B
Statement - 1 is True, Statement - 2 is True : Statement 2 is NOT a correct explanation for Statement - 1

C
Statement - 1 is True, Statement - 2 is False

D
Statement - 1 is False, Statement - 2 is True

Solution

If the graphs of the two equations are parallel lines, then there will be no solution, and the system is called an inconsistent system because the two lines will never intersect each other.
Question 3
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The condition for the pair of equations $$a_1x+b_1y+c_1=0$$ and $$a_2x+b_2y+c_2 = 0$$ to have a unique solution is -

A
$$a_1b_2-a_2b_1=0$$

B
$$a_1b_2-a_2b_1 \neq 0$$

C
$$a_1b_1 - a_2 b_2 =0$$

D
$$a_1b_1 - a_2b_2 \neq 0$$

Solution

For unique solution
$$\displaystyle \frac{a_1}{a_2} \neq \frac{b_1}{b_2} $$
$$\Rightarrow a_1b_2 - a_2b_1 \neq 0$$
Question 4
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The solution set of $$x + y -1=0$$ and $$3x + 3y -2=0$$ is

A
(1,0)

B
(0, 1)

C
An empty set

D
(1, 1)

Solution

$$x + y -1=0\Rightarrow a_1=1,\,b_1=1,\,c_1=-1$$

$$3x + 3y -2=0\Rightarrow a_2=3,\,b_2=3,\,c_2=-2$$

$$\displaystyle \dfrac{1}{3} = \dfrac{1}{3} \neq \dfrac{-1}{-2}$$

$$ \displaystyle \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$$,

Hence no solution.

Therefore, solution set for $$x + y -1=0$$ and $$3x + 3y -2=0$$ is an empty set.
Question 5
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Find the solution set of the system of equations: $$\displaystyle \frac{4}{x}+5y=7\:and\:\frac{3}{x}+4y=5.$$

A
$$\displaystyle \left ( \frac{1}{3},-1 \right )$$

B
$$\displaystyle \left ( \frac{1}{3},-3 \right )$$

C
$$\displaystyle \left ( \frac{1}{3},-2 \right )$$

D
$$\displaystyle \left ( \frac{1}{3},1 \right )$$

Solution

We put $$\dfrac { 1 }{ x } =p$$
Then the system of given equations becomes
$$4p+5y=7............(i)$$ &
$$3p+4y=5........(ii)$$
$$(i)\times 3\Rightarrow 12p+15y=21........(iii)\quad \& \\ (ii)\times 4\Rightarrow 12p+16y=20.........(iv).\\ (iii)-(iv)\Rightarrow -y=1\\ \Rightarrow y=-1.\quad $$
Putting $$y=-1$$ in $$(i)$$,
$$4p+5\times (-1)=7\\ \Rightarrow p=\dfrac { 12 }{ 4 } =3.\\ \therefore x=\dfrac { 1 }{ 3 } .\\ \therefore \left( x,y \right) =\left( \dfrac { 1 }{ 3 }, -1 \right) .\quad $$
Hence, the answer is $$\left(\dfrac{1}{3},-1\right)$$
Question 6
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The system of linear equation $$ax+by=6$$, $$cx+dy=8$$ has no solution if:

A
$$ad-bc>0$$

B
$$ad-bc<0$$

C
$$ad+bc=0$$

D
$$ad-bc=0$$

Solution

The given equations are $$ax+by = 6$$ and $$cx+dy = 8$$.

Here, $$a_1=a, b_1=b, c_1=-6\,\,and\,\, a_2=c, b_2=d, c_2=-8$$

They have no solution if,

$$\dfrac { a_1 }{ a_2 } =\cfrac { b_1 }{ b_2 } \neq \cfrac { c_1 }{ c_2 }$$.

Hence, $$\dfrac { a }{ c } = \cfrac { b }{ d }$$

$$\Rightarrow ad - bc = 0$$

So, option D is correct.
Question 7
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The cost of 9 chairs and 3 tables is Rs. 306, while the cost of 6 chairs and 3 tables is Rs. 246. Then the cost of 6 chairs and 1 table is

A
Rs. 161

B
Rs. 162

C
Rs. 169

D
Rs. 175

Solution

$$9x+3y=306$$ ---- (1)
$$6x+3y=246$$ -----(2)
Subtracting eq(2) from eq(1)
$$3x=60$$
$$x=20$$
Substituting $$x=20$$ in eq(1)
$$9(20)+3y = 306$$
$$180+3y=306$$
$$3y=126$$
$$y=42$$
Hence, $$6x+y =162$$.
Question 8
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Check whether the pair of equations $$x + 3y = 6$$, and $$2x - 3y = 12$$ is consistent. If so, solve graphically

A
Yes; $$x=6, y=0$$

B
No; $$x=6, y=0$$

C
Ambiguous

D
Data insufficient

Solution

The given pair can be written as
$$x+3y-6=0$$ and $$ 2x-3y-12=0$$
Here$$a_1=1,b1=3,c_1=-6...........eq1 and a_2=2,b_2=-3,c_2=-12.........eq2$$
$$\therefore \frac{a_1}{a_2}=\frac{1}{2},\frac{b_1}{b_2}=\frac{3}{-6}=\frac{1}{-2}$$
Hence $$\frac{a_1}{a_2}\neq\frac {b_1}{b_2}$$
Thus the given pair of equation is consistent.
To solve them graphically we have
Consider $$x+3y-6=0$$
$$x=0 \implies y =2$$
$$x=3 \implies y=1$$
Join the points by drawing line.
Consider $$ 2x-3y-12=0$$
$$x=0 \implies y =-4$$
$$x=3 \implies y=-2$$
Join the two points by drawing line.
The given pair of linear equation has unique solution $$x=6 $$ and $$y=0$$.
Question 9
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If a pair of linear equations is inconsistent. In the below graph, the lines are _________.

A
Parallel

B
Coincidence

C
Intersecting

D
Divergence

Solution

From figure, it is clear that lines are parallel as only parallel lines are inconsistent for a 2 -dimensional reference.
Question 10
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If a pair of linear equations is consistent and dependent. In the below graph, the lines are __________.

A
Parallel

B
Coincident

C
Intersecting

D
none of these

Solution

If a pair of linear equations is consistent, then the number of solutions can be either one or infinite.
Here, the lines are dependent as well and hence, have an infinite number of solutions.
$$\therefore$$ Lines having an infinite number of solutions are coincident
Answer: B

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Pair of Linear Equations in Two Variables Test - 18

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Pair of Linear Equations in Two Variables Test - 18

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Question 1

$$\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}$$

$$\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}\neq \frac{c_1}{c_2}$$

$$\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$$

None of these

Solution

Question 2

Statement - 1 is True, Statement - 2 is True, Statement - 2 is a correct explanation for Statement - 1

Statement - 1 is True, Statement - 2 is True : Statement 2 is NOT a correct explanation for Statement - 1

Statement - 1 is True, Statement - 2 is False

Statement - 1 is False, Statement - 2 is True

Solution

Question 3

$$a_1b_2-a_2b_1=0$$

$$a_1b_2-a_2b_1 \neq 0$$

$$a_1b_1 - a_2 b_2 =0$$

$$a_1b_1 - a_2b_2 \neq 0$$

Solution

Question 4

(1,0)

(0, 1)

An empty set

(1, 1)

Solution

Question 5

$$\displaystyle \left ( \frac{1}{3},-1 \right )$$

$$\displaystyle \left ( \frac{1}{3},-3 \right )$$

$$\displaystyle \left ( \frac{1}{3},-2 \right )$$

$$\displaystyle \left ( \frac{1}{3},1 \right )$$

Solution

Question 6

$$ad-bc>0$$

$$ad-bc<0$$

$$ad+bc=0$$

$$ad-bc=0$$

Solution

Question 7

Rs. 161

Rs. 162

Rs. 169

Rs. 175

Solution

Question 8

Yes; $$x=6, y=0$$

No; $$x=6, y=0$$

Ambiguous

Data insufficient

Solution

Question 9

Parallel

Coincidence

Intersecting

Divergence

Solution

Question 10

Parallel

Coincident

Intersecting

none of these

Solution

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