Pair of Linear Equations in Two Variables Test

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Pair of Linear Equations in Two Variables Test - 17

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Question 1
1 / -0

If, $$\displaystyle \frac{1}{x}+\frac{1}{y}=k$$ and $$\displaystyle \frac{1}{x}-\frac{1}{y}=k$$, then the value of y .................

A
$$3$$

B
$$-4$$

C
Does not exist.

D
None of these

Solution

Let $$\displaystyle \frac{1}{x}+\frac{1}{y}=k$$ ......(1)
$$\displaystyle \frac{1}{x}-\frac{1}{y}=k$$ ......(2)
$$(1)-(2) \Rightarrow \displaystyle \frac{2}{y}=0$$
$$\therefore$$ the value of $$y$$ does not exist.
Question 2
1 / -0

If the pair of equations has no solution, then the pair of equations is :

A
consistent

B
inconsistent

C
coincident

D
none of these

Solution

If a pair of equations has no solution, then the pair of equations is said to be inconsistent.
Question 3
1 / -0

If the lines intersect at a point, then that point gives the unique solution of the two equations. In this case, the pair of equations is .................

A
Inconsistent

B
No solution

C
Consistent

D
None of the Above

Solution

If the lines intersect at a point, then that point gives the unique solution of the given linear equations, and the pair of equations are called consistent pair of equations.
Question 4
1 / -0

Which of the following pairs of equations represent inconsistent system?

A
$$3x - 2y = 8$$

$$2x + 3y = 1$$

B
$$3x - y = - 8$$

$$3x - y = 24$$

C
$$x - y = m$$

$$x + my = 1$$

D
$$5x - y = 10$$

$$10x - 6y = 20$$

Solution

For a pair of linear equations to be inconsistent, the equations must satisfy the given condition which is
$$\cfrac{{a}_{1}}{{a}_{2}} = \cfrac{{b}_{1}}{{b}_{2}} \ne \cfrac{{c}_{1}}{{c}_{2}}$$
Since, the equations $$3x - y = -8$$ and $$3x - y = 24$$ clearly satisfy the above condition $$\cfrac{3}{3} = \cfrac{-1}{-1} \ne \cfrac{-8}{24}$$, the pair of equations represents inconsistent system.
In options A, C and D, $$\cfrac{{a}_{1}}{{a}_{2}} \neq \cfrac{{b}_{1}}{{b}_{2}} $$
Question 5
1 / -0

The point of intersection of the lines $$y=3x$$ and $$x=3y$$ is :

A
(3, 0)

B
(0, 3)

C
(3, 3)

D
(0, 0)

Solution

The point of intersection of lines $$y = 3x$$(black line) and $$x = 3y$$(brown line) must satisfies both the equation.
From the graph, we can see that solution is (0,0).
Hence the point of intersection of given lines is at origin i.e. $$\left( 0, 0 \right)$$.
Question 6
1 / -0

If the lines are parallel, then the pair of equations has no solution. In this case, the pair of equations is ...................

A
inconsistent.

B
consistent.

C
Cannot be determined

D
None of above

Solution

If the lines are parallel, then the given pair of linear equations has no solution. In this case, the pair of linear equations is said to be inconsistent.
Question 7
1 / -0

Solve, graphically, the following pairs of equations:
$$x - 5 = 0$$
$$y+ 4 = 0$$

A
$$x = 1, y = -8$$

B
$$x = 5, y = -4$$

C
$$x = 4, y = -1$$

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

Solution

Given pair of equations is $$x-5=0$$ and $$y+4=0$$
Plotting these equations in the graph, we get the above image.
Clearly from the graph, we can find the point of intersection and that point is
the point of intersection is $$(5,-4)$$.
Hence, $$x=5, y=-4$$.
Question 8
1 / -0

A system of linear equations is given as follows :
$$a_1x+b_1y+c_1=0$$
$$a_2x+b_2y+c_2=0$$
Condition for two lines to have a unique solution is,

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

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

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

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

Solution

$$ Two\quad equations\quad will\quad have\quad a\quad unique\quad solution\quad if,\quad after\quad \\ solving\quad them,\quad the\quad solution\quad is\quad valid\quad i.e\quad the\quad lines\quad actually\quad \\ intersect\quad at\quad a\quad point.$$

$$\\ Let\quad there\quad be\quad equations\quad \quad { a }_{ 1 }x+{ b }_{ 1 }y+{ c }_{ 1 }=0..........(i)\quad and$$
$$\\ { a }_{ 2 }x+{ b }_{ 2 }y+{ c }_{ 2 }=0.........(ii).\quad \quad$$

$$ Multiplying\quad (i)\quad by\quad { a }_{ 2 }\quad \& \quad (ii)\quad by\quad { a }_{ 1 }\\ and\quad eliminating\quad x\quad by\quad subtraction\quad we\quad get\quad,$$

$$ \\ y=\dfrac { { a }_{ 1 }{ c }_{ 2 }-{ a }_{ 2 }{ c }_{ 1 } }{ { a }_{ 2 }{ b }_{ 1 }-{ a }_{ 1 }{ b }_{ 2 } } .$$

$$\\ Again\quad Multiplying\quad (i)\quad by\quad { b }_{ 2 }\quad \& \quad (ii)\quad by\quad { b }_{ 1 }\\ and\quad eliminating\quad y\quad by\quad subtraction\quad we\quad get\quad$$

$$ \\ x=\dfrac { { b }_{ 2 }{ c }_{ 1 }-{ b }_{ 1 }{ c }_{ 2 } }{ { a }_{ 2 }{ b }_{ 1 }-{ a }_{ 1 }{ b }_{ 2 } } .$$

$$\\ It\quad is\quad clear\quad that\quad the\quad solution\quad \left( x,y \right) =\left( \dfrac { { b }_{ 1 }{ c }_{ 2 }-{ b }_{ 2 }{ c }_{ 1 } }{ { a }_{ 2 }{ b }_{ 1 }-{ a }_{ 1 }{ b }_{ 2 } } ,\dfrac { { b }_{ 2 }{ c }_{ 1 }-{ b }_{ 1 }{ c }_{ 2 } }{ { a }_{ 2 }{ b }_{ 1 }-{ a }_{ 1 }{ b }_{ 2 } } \right) \\ will\quad be\quad valid\quad if\quad { a }_{ 2 }{ b }_{ 1 }-{ a }_{ 1 }{ b }_{ 2 }\neq 0\Longrightarrow \dfrac { { a }_{ 1 } }{ { a }_{ 2 } } \neq \dfrac { { b }_{ 1 } }{ { b }_{ 2 } } .$$

$$\\ Ans-\quad Option\quad B. $$
Question 9
1 / -0

Value of $$\alpha$$ for which equations
$$\alpha x+ 3y = \alpha -3; 12x + \alpha y = \alpha$$ has unique solution.

A
All value of $$\alpha$$ (except 4 and -4)

B
All value of $$\alpha$$ (except 6 and -6)

C
All value of $$\alpha$$ (except 2 and -2)

D
All value of $$\alpha$$ (except 3 and -3)

Solution

For a unique solution, we must have $$\displaystyle \frac{a_1}{a_2} \neq \frac{b_1}{b_2}$$
i.e., $$\displaystyle \dfrac{\alpha}{12} \neq \dfrac{3}{\alpha}$$
$$ \Rightarrow \alpha^2 \neq 36$$
$$ \Rightarrow \alpha \neq 6\ or\ -6$$
Hence, for all valus of $$\alpha$$ (except $$6$$ and $$-6$$).
The given pair of linear equations has a unique solution.
Question 10
1 / -0

Solve the following pair of linear equations using Graphical method:
$$x\, +\, y\, =\, 8; \quad x\, -\, y\, =\, 2$$.
Then $$(x, y)$$ is equal to

A
$$(5, 3)$$

B
$$(7, 6)$$

C
$$(4, 1)$$

D
$$(2, 2)$$

Solution

The given simultaneous equations are $$x+y=8 ..... (i) \text{ and } x-y=2 ..... (ii)$$
From equation $$(i)$$
$$x$$ $$0$$ $$3$$ $$5$$ $$8$$
$$y$$ $$8$$ $$5$$ $$3$$ $$0$$
From equation $$(ii)$$
$$x$$ $$0$$ $$2$$ $$4$$ $$5$$
$$y$$ $$-2$$ $$0$$ $$2$$ $$3$$
Plotting $$x+y=8$$(blue line) and $$x-y=2$$(green line) in the graph, we get the point of intersection as $$(5,3)$$.
Hence, $$(x,y)=(5,3)$$ is the correct answer.

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$$x$$	$$0$$	$$3$$	$$5$$	$$8$$
$$y$$	$$8$$	$$5$$	$$3$$	$$0$$

$$x$$	$$0$$	$$2$$	$$4$$	$$5$$
$$y$$	$$-2$$	$$0$$	$$2$$	$$3$$

Pair of Linear Equations in Two Variables Test - 17

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Pair of Linear Equations in Two Variables Test - 17

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SHARING IS CARING

Question 1

$$3$$

$$-4$$

Does not exist.

None of these

Solution

Question 2

consistent

inconsistent

coincident

none of these

Solution

Question 3

Inconsistent

No solution

Consistent

None of the Above

Solution

Question 4

$$3x - 2y = 8$$$$2x + 3y = 1$$

$$3x - y = - 8$$$$3x - y = 24$$

$$x - y = m$$$$x + my = 1$$

$$5x - y = 10$$$$10x - 6y = 20$$

Solution

Question 5

(3, 0)

(0, 3)

(3, 3)

(0, 0)

Solution

Question 6

inconsistent.

consistent.

Cannot be determined

None of above

Solution

Question 7

$$x = 1, y = -8$$

$$x = 5, y = -4$$

$$x = 4, y = -1$$

$$x = 0, y = -1$$

Solution

Question 8

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

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

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

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

Solution

Question 9

All value of $$\alpha$$ (except 4 and -4)

All value of $$\alpha$$ (except 6 and -6)

All value of $$\alpha$$ (except 2 and -2)

All value of $$\alpha$$ (except 3 and -3)

Solution

Question 10

$$(5, 3)$$

$$(7, 6)$$

$$(4, 1)$$

$$(2, 2)$$

Solution

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$$3x - 2y = 8$$

$$2x + 3y = 1$$

$$3x - y = - 8$$

$$3x - y = 24$$

$$x - y = m$$

$$x + my = 1$$

$$5x - y = 10$$

$$10x - 6y = 20$$