Pair of Linear Equations in Two Variables Test

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Pair of Linear Equations in Two Variables Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

If $$2^{2x - y} = 32$$ and $$2^{x + y} = 16$$ then $$x^{2} + y^{2}$$ is equal to

A
$$9$$

B
$$10$$

C
$$11$$

D
$$13$$

Solution

$$ Given\quad { 2 }^{ 2x-y }=32\\ \Rightarrow { 2 }^{ 2x-y }={ 2 }^{ 5 }\\ \Rightarrow 2x-y=5----(1)\\ Again\quad { 2 }^{ x+y }=16\\ \Rightarrow { 2 }^{ x+y }={ 2 }^{ 4 }\\ \Rightarrow x+y=4----(2)\\ Adding\quad (1)\quad and\quad (2)\quad we\quad get\\ 3x=9\\ \Rightarrow x=3\\ Substituting\quad x=3\quad in\quad (2)\quad we\quad get\\ y=1.\\ \therefore \quad { x }^{ 2 }+{ y }^{ 2 }={ 3 }^{ 2 }+{ 1 }^{ 2 }=10\quad (Ans) $$
Question 2
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The system of equations $$3x - 4y = 12$$ and $$6x - 8y = 48$$ has

A
$$2$$ solution

B
$$1$$ solution

C
Infinite number of solutions

D
no solution

Solution

$$3x-4y=12$$ $$\Rightarrow a_1=3,\,b_1=-4,\,c_1=12$$
$$6x-8y=48$$ $$\Rightarrow a_2=6,\,b_2=-8,\,c_2=48$$

$$\dfrac{a_1}{a_2}=\dfrac36=\dfrac12$$
$$\dfrac{b_1}{b_2}=\dfrac{-4}{-8}=\dfrac{1}{2}$$
$$\dfrac{c_1}{c_2}=\dfrac{12}{48}=\dfrac{1}{4}$$

$$\therefore\, \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\ne\dfrac{c_1}{c_2}$$
The given system of linear equations are inconsistent
Therefore, the given system of equations has no solution.
Question 3
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If $$6$$ kg of sugar and $$5$$ kg of tea together cost Rs. $$209$$ and $$4$$ kg of sugar and $$3$$ kg of tea together cost Rs. $$131$$, then the cost of $$1$$ kg sugar and $$1$$ kg tea are respectively

A
Rs. $$11$$ and Rs. $$25$$

B
Rs. $$12$$ and Rs. $$20$$

C
Rs. $$14$$ and Rs. $$20$$

D
Rs. $$14$$ and Rs. $$25$$

Solution

Let the price of sugar be $$x$$ and tea be $$y$$.
Then,
$$6x + 5y = 209....(1)$$
$$4x + 3y = 131....(2)$$

Multiply equation $$(1)$$ by $$4$$ and equation $$(2)$$ by $$6$$ and then subtract:
$$(6x + 5y = 209 )4$$
$$(4x + 3y = 131)6$$
$$.............................................$$
$$24x + 20y - 24x - 18y = 209\times4 - 131\times 6$$
$$.............................................$$
$$2y = 836-786$$
$$ 2y = 50$$
$$y = 25$$
Therefore, $$x = 14$$
Price of sugar is $$Rs.14$$ and price of tea is $$Rs.25$$ .
Question 4
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Without actually solving the simultaneous equations given below, decide whether simultaneous equations have unique solution, no solution or infinitely many solutions.
$$\displaystyle \frac{x-2y}{3} =\, 1;\quad 2x\, -\, 4y\, =\, \displaystyle \frac{9}{2}$$

A
No solution

B
Infinitely many solutions

C
Unique solutions

D
Data insufficient

Solution

The given equations are
$$ \dfrac { x-2y }{ 3 } =1$$ and $$ 2x-4y=\dfrac { 9 }{ 2 } $$.

The equations can be written as,
$$ x-2y-3=0$$ ..........(i)
and $$4x-8y-9=0$$ ........(ii)

We know that, for two linear equations
$$a_{1}x+b_{1}=c_{1}$$ and $$a_{2}x+b_{2}y=c_{2}$$:
(a) If $$\dfrac{a_{1}}{a_{2}}\ne\dfrac{b_{1}}{b_{2}}$$, the system has exactly one solution.

(b) If $$\dfrac{a_{1}}{a_{2}}=\dfrac{b_{1}}{b_{2}}=\dfrac{c_{1}}{c_{2}}$$, the system has infinitely many solutions.

(c) If $$\dfrac{a_{1}}{a_{2}}=\dfrac{b_{1}}{b_{2}}\neq \dfrac{c_{1}}{c_{2}}$$, the system has no solution.

Here $$ { a }_{ 1 }=1, { a }_{ 2 }=4, b_{ 1 }=-2, { b }_{ 2 }=-8, { c }_{ 1 }=-3$$ and $$ { c }_{ 2 }=-9$$

$$ \therefore \ \displaystyle \frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { 1 }{ 4 } , \frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { -2 }{ -8 } =\frac { 1 }{ 4 } , \frac { { c }_{ 1 } }{ { c }_{ 2 } } =\frac { -3 }{ -9 } =\frac { 1 }{ 3 } $$

$$ \Rightarrow \dfrac { { a }_{ 1 } }{ { a }_{ 2 } } =\dfrac { { b }_{ 1 } }{ { b }_{ 2 } } \neq \dfrac { { c }_{ 1 } }{ { c }_{ 2 } } $$

Hence, the system of equations is inconsistent and has no solution.
Question 5
1 / -0

Without actually solving the simultaneous equations given below, decide whether the system has unique solution, no solution or infinitely many solutions.
$$8y= x - 10; 2x = 3y + 7$$

A
Unique solution

B
infinitely many solutions.

C
no solution

D
cannot be determined

Solution

As we know 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 -
$$\displaystyle \frac{a_1}{a_2} \neq \frac{b_1}{b_2} $$
Here the given system of linear equation is
$$
x\quad -\quad 8y\quad =\quad 10;\\ 2x\quad -3y\quad =7\\$$
we have $$\dfrac { 1 }{ 2 } \neq \dfrac { -8 }{ -3 } ;\\$$
$$\therefore$$ Unique Solution
Question 6
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Solve the following equations using Graphical method:
$$4x=y-5; y=2x+1$$

Then $$(x,y)$$ is equal to

A
$$(-4, -2)$$

B
$$(6, -2)$$

C
$$(0, -4)$$

D
$$(-2, -3)$$

Solution

Writing both the equation as intercept form,
$$\dfrac { x }{ a } +\dfrac { y }{ b } =1$$
$$4x=y-5$$
$$-4x+y=5$$
$$\Rightarrow \dfrac { x }{ -\dfrac { 5 }{ 4 } } +\dfrac { y }{ 5 } =1$$
$$x$$ axis intercept $$=$$ $$-\dfrac { 5 }{ 4 } $$
$$y$$ axis intercept $$=5$$
Now $$y=2x+1$$
$$-2x+y=1$$
$$\Rightarrow \dfrac { x }{ -\dfrac { 1 }{ 2 } } +\dfrac { y }{ 1 } =1$$
$$x$$ axis intercept $$=$$ $$-\dfrac { 1 }{ 2 } $$
$$y$$ axis intercept $$=1$$
Drawing both the graphs, we get the following graphs
$$-2x+y=1\ \implies$$ black line
$$-4x+y=5\ \implies$$ brown line.
From the point of contact of both the lines, we get the value of $$x$$ and $$y$$
$$x=-2$$ and $$y=-3$$
Answer is $$ (-2,-3)$$.
Question 7
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Find the value of $$k$$ for which the given simultaneous equations have infinitely many solutions:
$$ 4x\, +\, y\, =\, 7;\quad 16x\, +\, ky\, =\, 28$$

A
$$k\, =\, 2$$

B
$$k\, =\, 6$$

C
$$k\, =\, 3$$

D
$$k\, =\, 4$$

Solution

As we know the condition for the pair of equations $$a_1x+b_1y+c_1=0$$ and $$a_2x+b_2y+c_2 = 0$$ to have infinite solutions is
$$\displaystyle \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} $$

$$\therefore$$ For infinite number of solutions,
$$
\\ \\ \dfrac { 4 }{ 16 } =\dfrac { 1 }{ k } =\dfrac { -7 }{ -28 } ;\quad \quad\\\therefore \quad k\quad =\quad 4
$$
Question 8
1 / -0

Find the value of $$k$$ for which the given simultaneous equations have infinitely many solutions:
$$4y = kx- 10; 3x = 2y + 5$$

A
$$k\, =\, 2$$

B
$$k\, =\, 6$$

C
$$k\, =\, 8$$

D
$$k\, =\, 4$$

Solution

As we know the condition for the pair of equations $$a_1x+b_1y+c_1=0$$ and $$a_2x+b_2y+c_2 = 0$$ to have infinite solutions is
$$\displaystyle \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} $$

$$
kx - 4y = 10; \\ 3x - 2y = 5\\ \\$$
So for infinite no of solutions,
$$\dfrac { k }{ 3 } =\dfrac { -4 }{ -2 } =\dfrac { -10 }{ -5 }$$
$$\therefore k =6$$
Question 9
1 / -0

Find the value of $$k$$ for which the given simultaneous equations have infinitely many solutions:
$$kx\, -\, y\, +\, 3\, -\, k\, =\, 0;\quad 4x\, -\, ky\, +\, k\, =\, 0$$

A
$$k\, =\, 3$$

B
$$k\, =\, 4$$

C
$$k\, =\, 2$$

D
$$k\, =\, 1$$

Solution

Compare the equations $$kx-y+3-k=0$$ and $$4x-ky+k=0$$ with the general equations $$a_1x+b_1y+c_1=0$$ and $$a_2x+b_2y+c_2=0$$ respectively.

Here, $$\cfrac { { a }_{ 1 } }{ { a }_{ 2 } } =\cfrac { k }{ 4 } ,\cfrac { { b }_{ 1 } }{ { b }_{ 2 } } =\dfrac { -1 }{ -k } =\dfrac { 1 }{ k }$$ and $$\cfrac { { c }_{ 1 } }{ { c }_{ 2 } } =\cfrac { 3-k }{ k }$$.

For a pair of linear equations to have infinitely many solutions : $$\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { { c }_{ 1 } }{ { c }_{ 2 } }$$

So, we have $$\cfrac { k }{ 4 } =\cfrac { 1 }{ k } =\cfrac { 3-k }{ k }$$

We first solve $$\cfrac { k }{ 4 } =\cfrac { 1 }{ k }$$

which gives $$k^2 = 4$$, i.e., $$k = ± 2$$.

Also, $$\dfrac { 1 }{ k } =\cfrac { 3-k }{ k }$$

gives $$k = 3k-k^2$$, i.e., $$ k^2-2k=0$$, which means $$k = 0$$ or $$k = 2$$.

Therefore, the value of $$k$$, that satisfies both the conditions, is $$k = 2$$.

Hence,for $$k = 2$$, the pair of linear equations has infinitely many solutions.
Question 10
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Without actually solving the simultaneous equations given below, decide whether the system has unique solution, no solution or infinitely many solutions.
$$\displaystyle \dfrac{x}{2} + \displaystyle \dfrac{y}{3} = 4; \displaystyle \dfrac{x}{4} +\displaystyle \frac{y}{6} = 2$$

A
no solution

B
Infinite solutions

C
unique solution

D
Cannot be determined

Solution

As we know the condition for the pair of equations $$a_1x+b_1y+c_1=0$$ and $$a_2x+b_2y+c_2 = 0$$ to have infinite solutions is
$$\displaystyle \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} $$
Here the gine system of linear equations is
$$\displaystyle \dfrac{x}{2} + \displaystyle \dfrac{y}{3} -4=0; \displaystyle \dfrac{x}{4} +\displaystyle \frac{y}{6} - 2=0$$
We have $$\dfrac{4}{2}=\dfrac{6}{3}=\dfrac{-4}{-2}$$
Hence, the system has infinite no. of solutions.

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Re-Attempt Weekly Quiz Competition

Pair of Linear Equations in Two Variables Test - 23

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Pair of Linear Equations in Two Variables Test - 23

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SHARING IS CARING

Question 1

$$9$$

$$10$$

$$11$$

$$13$$

Solution

Question 2

$$2$$ solution

$$1$$ solution

Infinite number of solutions

no solution

Solution

Question 3

Rs. $$11$$ and Rs. $$25$$

Rs. $$12$$ and Rs. $$20$$

Rs. $$14$$ and Rs. $$20$$

Rs. $$14$$ and Rs. $$25$$

Solution

Question 4

No solution

Infinitely many solutions

Unique solutions

Data insufficient

Solution

Question 5

Unique solution

infinitely many solutions.

no solution

cannot be determined

Solution

Question 6

$$(-4, -2)$$

$$(6, -2)$$

$$(0, -4)$$

$$(-2, -3)$$

Solution

Question 7

$$k\, =\, 2$$

$$k\, =\, 6$$

$$k\, =\, 3$$

$$k\, =\, 4$$

Solution

Question 8

$$k\, =\, 2$$

$$k\, =\, 6$$

$$k\, =\, 8$$

$$k\, =\, 4$$

Solution

Question 9

$$k\, =\, 3$$

$$k\, =\, 4$$

$$k\, =\, 2$$

$$k\, =\, 1$$

Solution

Question 10

no solution

Infinite solutions

unique solution

Cannot be determined

Solution

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