If $$(3)^{x + y} = 81$$ and $$(81)^{x - y} = 3$$, then the values of $$x$$ and $$y$$ are

$$\frac{17}{8}$$, $$\frac{15}{8}$$

$$\frac{17}{8}$$,$$\frac{9}{8}$$

$$\frac{17}{8}$$, $$\frac{11}{8}$$

$$\frac{11}{8}$$, $$\frac{15}{8}$$

MCQ Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 3

The graphical representation of the pair of equations

$x+2y-4=0$ and

$2x+4y-12=0$ is:

0%
intersecting lines
0%
parallel lines
0%
coincident lines
0%
all the above

Explanation

For two lines to be parallel,

$\dfrac { a_1 }{ a_2 } =\dfrac { b_1 }{ b_2 } \neq \dfrac { c_1 }{ c_2 }$ -----equation

$(1)$

where,

$a$ and

$b$ are coefficients of

$x$ and

$y$ and

$c$ is constant

Here, the lines are

$x+2y-4=0$ and

$2x+4y-12=0$

$a_1 =1 , b_1 =2 , c_1 = -4$ and

$a_2=2, b_2=4, c_2 =-12$

As we can see that,

$\dfrac { 1 }{ 2 } =\dfrac { 2 }{ 4 } \neq \dfrac { -4 }{ -12 }$

We can say that the given lines are parallel.

The pair of linear equations

$4x +6y= 9\ and\ 2x+ 3y= 6$ has

0%
No solution
0%
Many solutions
0%
Two solutions
0%
One solution.

Explanation

Given system of linear equations

$4x+6y- 9=0\Rightarrow a_1=4,\,b_1=6,\,c_1=-9$

$2x+3y-6=0\Rightarrow a_2=2,\,b_2=3,\,c_2=-6$ .

As we know a pair of linear equations is inconsistent (no solution) if

$\cfrac{{a}_{1}}{{a}_{2}} = \cfrac{{b}_{1}}{{b}_{2}} \ne \cfrac{{c}_{1}}{{c}_{2}}$

We have

$\dfrac{4}{2}=\dfrac{6}{3} \neq\dfrac{-9}{-6}$

Hence, no solution.

The ages of Hari and Harry are in the ratio

$5:7$ . If four years from now, the ratio of their ages will be

$3:4$ , then the present age of

0%
Hari is $20$ years and Harry is $28$ years.
0%
Hari is $28$ years and Harry is $20$ years.
0%
Hari is $25$ years and Harry is $35$ years.
0%
Hari is $35$ years and Harry is $25$ years.

Explanation

Suppose Hari's present age and Harry's present age are

$x$ and

$y$ years respectively.
Therefore,

$\dfrac{x}{y}=\dfrac{5}{7}$

$7x-5y=0$

$\Rightarrow x=\dfrac{5y}{7}$ ...(1)

After four years, their ages will be

$x+4$ and

$y+4$ years respectively.
Therefore,

$\dfrac{x+4}{y+4}=\dfrac{3}{4}$

$4x+16=3y+12$

$4x-3y=-4$ ...(2)

Putting the value of

$x$ from (1) in equation (2), we get

$4\left(\dfrac{5y}{7}\right)-3y=-4$

$20y-21y=-28$

$y=28$ years

Putting the above value of

$y$ in equation (1), we get

$x=\dfrac{140}{7} = 20$ years.

Hence, option A.

Suresh is half his father's Age. After

$20$ years, his father's age will be one and a half times the Suresh's age. What is his father's age now?

0%
$40$
0%
$20$
0%
$26$
0%
$30$

Explanation

Let Suresh's present age be

$x$
Let Father's present age be

$y$
Therefore applying condition no.

$1$

$x=\dfrac{y}{2}$

$y-2x=0$ ...............................................................(i)

Also applying condition no. 2

$\dfrac{3(x+20)}{2}=y+20$

$2y-3x=20$ ......................................................(ii)
Multiplying equation (i) by

$2$

$2y-4x=0$ ......................................................(iii)

Subtracting equation (iii) from equation (ii), we get

$(2y-3x-2y+4x)=20$
Therefore

$x=20$ years.
Substituting in equation (i), we get

$y=40$ years
Hence, the age of the father is

$40$ years.

$3^{x - y} = 27$ and

$3^{x + y} = 243$ , then

$x$ is equal to

0%
$0$
0%
$4$
0%
$2$
0%
$6$

Explanation

Firstly make the bases same on both the sides of equation. Apply the concept if

${ a }^{ m }={ a }^{ n }\quad then\quad m=n$ .

${ 3 }^{ x-y }=27\Rightarrow { 3 }^{ x-y }={ 3 }^{ 3 }\Rightarrow x-y=3---(1)\\ again\quad { 3 }^{ x+y }=243\Rightarrow { 3 }^{ x+y }={ 3 }^{ 5 }\Rightarrow x+y=5---(2)\\ Adding\quad (1)\quad and\quad (2)\quad we\quad get\\ 2x=8\\ or\quad x=4\quad \quad (Ans)$

Solve for

$x$ and

$y$ :

$4x+\dfrac{6}{y}=15$ ;

$3x-\dfrac{4}{y}=7$

0%
$x=1, y=2$
0%
$x=1, y=1$
0%
$x=2, y=2$
0%
none of the above

Explanation

Given equations are

$4x+\cfrac { 6 }{ y } =15$ ------equation

$(1)$

$3x-\cfrac { 4 }{ y } =7$ --------equation

$(2)$

From equation

$(1)$ and equation

$(2)$ can be written as

$4xy+6=15y$ -----equation

$(3)$

$3xy-4=7y$ ------equation

$(4)$

Multiplying equation

$(3)$ by

$3$ and equation

$(4)$ by

$4$ and subtracting

$12xy+18=45y$

$12xy-16=28y$

After solving, we get

$y=2$

From equation

$(1)$ ,

$4x+\cfrac { 6 }{ y } =15$ --------equation

$(1)$

$4x=12$

$x=3,y=2$

Choose the correct matching(s) for solving questions of the system of linear equation in two variables

	Methods	Uses/Disadvantages
$(a)$	Graphical	$(i)$ Use: When the coefficients of the variables and the solutions are integers. Disadvantage: If the solutions are not integers, they are hard to plot and read on the graph.
$(b)$	Substitution	$(ii)$ Use: When variables with coefficients that are the same or additive inverse of each other ( for example, $2x$ and $-2x$ ) are present. Disadvantage: If fractions are involved, you may have much computation.
$(c)$	Elimination	$(iii)$ Use: When one of the variables is isolated (alone) on one side of the equation Disadvantage: You may have lots of computations involving signed numbers.

0%
$(a) \rightarrow(i)$
0%
$(b) \rightarrow (iii)$
0%
$(c) \rightarrow (ii)$
0%
$(b) \rightarrow (i)$

Explanation

The correct uses and disadvantages of various methods are:

$Graphical:$

$Use -$ When the coefficient of the variables and the solutions are integers.

$Disadvantage -$ If the solutions are not integers, they are hard to read on the graph.

$Substitution:$

$Use -$ When one of the variables is isolated (alone) on one side of the equation

$Disadvantage -$ If fractions are involved, you may have much computation

$Elimination:$

$Use -$ When fractions, decimals or variables with coefficients that are the same or negative inverse of each other

$(eg: 2x\ and\ -2x)$ are present.

$Disadvantage -$ You may have lots of computations involving signed numbers.

Solve the following equations using Graphical method:

$4x=y-5; y=2x+1$

Then

$(x,y)$ is equal to

0%
$(-4, -2)$
0%
$(6, -2)$
0%
$(0, -4)$
0%
$(-2, -3)$

Explanation

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

$(-2,-3)$ .

Solution of the equations

$\cfrac{x + 3}{4} + \cfrac{2y + 9}{3} = 3$ and

$\cfrac{2x - 1}{2} - \cfrac{y + 3}{4} = 4 \cfrac{1}{2}$ is

0%
$x = - 5, y = - 3$
0%
$x = - 5, y = 3$
0%
$x = 5, y = 3$
0%
$x = 5, y = -3$

Explanation

$\dfrac{x+3}{4}+\dfrac{2y+9}{3}=3$

Multiplying both sides by

$12$ we get,

$\Rightarrow$

$3x+9+8y+36=36$

$\Rightarrow$

$3x+8y=-9$ ----- ( 1 )

$\dfrac{2x-1}{2}-\dfrac{y+3}{4}=4\dfrac{1}{2}$

$\dfrac{2x-1}{2}-\dfrac{y+3}{4}=\dfrac{9}{2}$

Multiplying both sides by

$4$ , we get

$\Rightarrow$

$4x-2-y-3=18$

$\Rightarrow$

$4x-y=23$ ---- ( 2 )

Multiplying equation ( 1 ) by

$4$ ,

$12x+32y=-36$ ----- ( 3 )

Multiplying equation ( 2 ) by

$3$ ,

$12x-3y=69$ ---- ( 4 )

Subtracting equation ( 4 ) from ( 3 ) we get,

$\Rightarrow$

$35y=-105$

$\therefore$

$y=-3$

Substituting

$y=-3$ in equation ( 1 )

$3x+8(-3)=-9$

$3x-24=-9$

$3x=15$

$\therefore$

$x=5$

$\therefore$

$x=5$ and

$y=-3$

$(3)^{x + y} = 81$ and

$(81)^{x - y} = 3$ , then the values of

$x$ and

$y$ are

0%
$\frac{17}{8}$ , $\frac{9}{8}$
0%
$\frac{17}{8}$ , $\frac{11}{8}$
0%
$\frac{17}{8}$ , $\frac{15}{8}$
0%
$\frac{11}{8}$ , $\frac{15}{8}$

Explanation

$3^{x+y} = 81$

$\Rightarrow 3^{x+y} = 3^{4}$
or

$x+y = 4$ ...(i)
Also

$(81)^{x-y} = 3$

$\Rightarrow 3^{4(x-y)} = 3^{1}$ ,

$4x-4y = 1$ ...(ii)
Solving equations (i) and (ii), we get

$x=\frac{17}{8}$ and

$y = \frac{15}{8}$

$\left (x+y,1 \right )$

$=$

$\left (3,y-x \right )$ , then

$x$

$=$ ,

$y$

$=$

0%
2, 1
0%
-1, -2
0%
1, 2
0%
2, -1

Explanation

We have,

${(x+y,1)}={(3,y-x)}$

$\implies x+y=3$ and

$y-x=1$

$\Rightarrow x+y=3$ and

$-x+y=1$

Adding both equations, we get

$2y=4$

$\Rightarrow y=2$

Put the value of

$y$ in

$x+y=3$ , we get

$x+2=3$

$\Rightarrow x=1$

So,

$\text{C}$ is the correct option.

Find the value of

$p$ for which the given simultaneous equations have unique solution:

$3x\, +\, y\, =\, 10;\quad 9x\, +\, py\, =\,23$

0%
$p = 5$
0%
All values of $p$ except $7$
0%
All values of $p$ except $3$
0%
Cannot be determined

Explanation

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

$3x+y=10;\\ 9x+py=23\\$

So for unique solution, we have

$\dfrac { 3 }{ 9 } \neq \dfrac { 1 }{ p } \\ \Rightarrow p\neq 3\quad$

$\displaystyle \frac{3}{x}- \frac{2}{y} =5$ and

$\displaystyle \frac{4}{x} - \frac{5}{y} = 2$ , then

$\displaystyle \frac{1}{x} - \frac{1}{y} =?$

Where

$(x, y \neq 0)$

0%
$1$
0%
$-1$
0%
$5$
0%
None of these

Explanation

$\displaystyle \frac{3}{x} - \frac{2}{y} = 5$ ........ eq(1)

$\displaystyle \frac{4}{x} - \frac{5}{y} = 2$ ........ eq(2)

Multiply (1) by

$-4$ and (2) by

$3$ and add them

$-\displaystyle \frac{12}{x} + \frac{8}{y} = -20$ ...........eq(3)

$\displaystyle \frac{12}{x} - \frac{15}{y}=6$ .................eq(4)
adding eq(3) and eq(4)

$\displaystyle -\frac{7}{y}= -14 \Rightarrow y = \frac{1}{2}$

Put

$y = \displaystyle \frac{1}{2}$ in eq. (1)

$\displaystyle \frac{3}{x} - 4 = 5 \Rightarrow \frac{3}{x} = 9 \Rightarrow x = \frac{1}{3}$

$\displaystyle \frac{1}{x} = 3$ and

$\displaystyle \frac{1}{y} = 2$

$\therefore$

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

Solve the following equation simultaneously using Graphical method:

$x+2y=5; y=-2x-2$

Then

$(x,y)$ is equal to

0%
$(-3, 4)$
0%
$(4, -2)$
0%
$(3, 4)$
0%
$(4, 2)$

Explanation

Given first line

$x+2y=5$
Take two points

$(5,0)$ and

$(1,2)$ and join them by drawing a line.
Another line

$y=-2x-2$
Take two points

$(0,-2)$ and

$(-1,0)$ and join them by drawing a line.
We get the graphs

$x+2y=5\ \implies$ brown line

$2x+y=2\ \implies$ blue line
the point of intersection is

$(-3,4)$

The system of equations

$3x - 4y = 12$ and

$6x - 8y = 48$ has

0%
$2$ solution
0%
$1$ solution
0%
Infinite number of solutions
0%
no solution

Explanation

$3x-4y=12$

$\Rightarrow a_1=3,\,b_1=-4,\,c_1=12$

$6x-8y=48$

$\Rightarrow a_2=6,\,b_2=-8,\,c_2=48$

$\dfrac{a_1}{a_2}=\dfrac36=\dfrac12$

$\dfrac{b_1}{b_2}=\dfrac{-4}{-8}=\dfrac{1}{2}$

$\dfrac{c_1}{c_2}=\dfrac{12}{48}=\dfrac{1}{4}$

$\therefore\, \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\ne\dfrac{c_1}{c_2}$
The given system of linear equations are inconsistent
Therefore, the given system of equations has no solution.

$2^{2x - y} = 32$ and

$2^{x + y} = 16$ then

$x^{2} + y^{2}$ is equal to

0%
$9$
0%
$10$
0%
$11$
0%
$13$

Explanation

$Given\quad { 2 }^{ 2x-y }=32\\ \Rightarrow { 2 }^{ 2x-y }={ 2 }^{ 5 }\\ \Rightarrow 2x-y=5----(1)\\ Again\quad { 2 }^{ x+y }=16\\ \Rightarrow { 2 }^{ x+y }={ 2 }^{ 4 }\\ \Rightarrow x+y=4----(2)\\ Adding\quad (1)\quad and\quad (2)\quad we\quad get\\ 3x=9\\ \Rightarrow x=3\\ Substituting\quad x=3\quad in\quad (2)\quad we\quad get\\ y=1.\\ \therefore \quad { x }^{ 2 }+{ y }^{ 2 }={ 3 }^{ 2 }+{ 1 }^{ 2 }=10\quad (Ans)$

$6$ 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

0%
Rs. $11$ and Rs. $25$
0%
Rs. $12$ and Rs. $20$
0%
Rs. $14$ and Rs. $20$
0%
Rs. $14$ and Rs. $25$

Explanation

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

$Rs.25$ .

Without actually solving the simultaneous equations given below, decide whether simultaneous equations have unique solution, no solution or infinitely many solutions.

$\displaystyle \frac{x-2y}{3} =\, 1;\quad 2x\, -\, 4y\, =\, \displaystyle \frac{9}{2}$

0%
No solution
0%
Infinitely many solutions
0%
Unique solutions
0%
Data insufficient

Explanation

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.

$\dfrac{a_{1}}{a_{2}}=\dfrac{b_{1}}{b_{2}}\neq \dfrac{c_{1}}{c_{2}}$ , 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.

Without actually solving the simultaneous equations given below, decide whether the system has unique solution, no solution or infinitely many solutions.

$8y= x - 10; 2x = 3y + 7$

0%
Unique solution
0%
infinitely many solutions.
0%
no solution
0%
cannot be determined

Explanation

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

Find the value of

$p$ for which the given simultaneous equations have unique solution:

$8x\, -\, py\, +\, 7\, =\, 0;\quad 4x\, -\, 2y\, +\, 3\, =\, 0$

0%
All values of $p$ except $4$
0%
$p = 7$
0%
$p = 6$
0%
All values of $p$ except $5$

Explanation

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 linear of equations is

$8x-py+7=0;\\ 4x-2y+3=0$

So for the unique solution

$\dfrac { 8 }{ 4 } \neq \dfrac { -p }{ -2 } \quad$

$\\ \Rightarrow p\neq 4\quad\quad$

A particular work can be completed by

$6$ men and

$6$ women in

$24$ days; whereas the same work can be completed by

$8$ men and

$12$ women in

$15$ days, according to the amount of work done , one man is equivalent to how many women?

0%
$2\dfrac{1 }{2 }$ women
0%
$5\dfrac{1 }{3 }$ women
0%
$5\dfrac{2 }{3 }$ women
0%
$\dfrac{3}{2 }$ women

Find the value of

$k$ for which the given simultaneous equations have infinitely many solutions:

$4x\, +\, y\, =\, 7;\quad 16x\, +\, ky\, =\, 28$

0%
$k\, =\, 2$
0%
$k\, =\, 6$
0%
$k\, =\, 3$
0%
$k\, =\, 4$

Explanation

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,

$\\ \\ \dfrac { 4 }{ 16 } =\dfrac { 1 }{ k } =\dfrac { -7 }{ -28 } ;\quad \quad\\\therefore \quad k\quad =\quad 4$

Find the value of

$k$ for which the given simultaneous equations have infinitely many solutions:

$4y = kx- 10; 3x = 2y + 5$

0%
$k\, =\, 2$
0%
$k\, =\, 6$
0%
$k\, =\, 8$
0%
$k\, =\, 4$

Explanation

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,

$\dfrac { k }{ 3 } =\dfrac { -4 }{ -2 } =\dfrac { -10 }{ -5 }$

$\therefore k =6$

Without actually solving the simultaneous equations given below, decide whether the system has unique solution, no solution or infinitely many solutions.

$3x+5y=16; 4x-y=6$

0%
No Solution
0%
Infinitely many solutions
0%
Unique solution
0%
Cannot be determined

Explanation

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 -

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

We have

$\$

$\dfrac{3}{4} \neq \dfrac{5}{-1}$

Hence, Unique Solution

Find the value of

$k$ for which the given simultaneous equations have infinitely many solutions:

$kx\, -\, y\, +\, 3\, -\, k\, =\, 0;\quad 4x\, -\, ky\, +\, k\, =\, 0$

0%
$k\, =\, 3$
0%
$k\, =\, 4$
0%
$k\, =\, 2$
0%
$k\, =\, 1$

Explanation

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.

Without actually solving the simultaneous equations given below, decide whether the system has unique solution, no solution or infinitely many solutions.

$\displaystyle \dfrac{x}{2} + \displaystyle \dfrac{y}{3} = 4; \displaystyle \dfrac{x}{4} +\displaystyle \frac{y}{6} = 2$

0%
no solution
0%
Infinite solutions
0%
unique solution
0%
Cannot be determined

Explanation

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.

Without actually solving the simultaneous equations given below, decide whether the system has unique solution, no solution or infinitely many solutions.

$3y=2-x; 3x=6-9y$

0%
Infinite solutions
0%
unique solution
0%
no solution
0%
Cannot be determined

Explanation

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 given system of linear equations is

$x + 3y -2=0;\\ 3x+9y -6=0\\ \dfrac { 1 }{ 3 } =\dfrac { 3 }{ 9 } =\dfrac { -2 }{ -6 } ;\\$

$\therefore$ The system has infinite number of solutions.

Solve graphically the simultaneous equations given below. Take the scale as

$1 \ \mathrm{cm} = 1$ unit on both the axes.

$x - 2y - 4=0$

$2x + y= 3$

0%
$x= 1,\:y= -1$
0%
$x= 2,\:y= -1$
0%
$x= 3,\:y= -1$
0%
$x= 7,\:y= -1$

Explanation

Plotting

$x-2y-4=0$
For

$x=0$ we get

$y=-2$
For

$y=0$ we get

$x=4$
Join the two points

$(0,-2)$ and

$(4,0)$ by drawing line.
Plotting

$2x+y=3$
For

$x=0$ we get

$y=3$
For

$y=0$ we get

$x=1.5$
Join the two points

$(0,3)$ and

$(1.5,0)$ by drawing line.
the point of intersection of both lines is

$(2,-1)$

Solve the following simultaneous equations:

$\displaystyle \frac{1}{x}\, +\, \frac{1}{y}\, =\, 8;\quad \frac{4}{x}\, -\, \frac{2}{y}\, =\, 2$

0%
$x\, =\,\displaystyle \frac{1}{3}\, ,\, y\, =\, \frac{1}{5}$
0%
$x\, =\,\displaystyle \frac{1}{2}\, ,\, y\, =\, \frac{1}{5}$
0%
$x\, =\,\displaystyle \frac{1}{3}\, ,\, y\, =\, \frac{1}{7}$
0%
$x\, =\,\displaystyle \frac{1}{2}\, ,\, y\, =\, \frac{1}{7}$

Explanation

Given equations are

$\dfrac { 1 }{ x } +\dfrac { 1 }{ y } = \ 8\\ \dfrac { 4 }{ x } -\dfrac { 2 }{ y }=\ 2$

Let

$\dfrac { 1 }{ x } =\ a,\ \dfrac { 1 }{ y } =\ b$

So,

$a+b\ =\ 8$ .......

$(1)$

$\\ 4a\ -\ 2b\ =\ 2$ ......

$(2)$

Multiply equation

$(1)$ by

$2$ , we get

$2a\ +2b\ =\ 16\$ ........

$(3)$

Adding equations

$(2)$ and

$(3)$ , we get

$6a=18$

$a=\dfrac { 18 }{ 6 }= 3$

Put

$a=3$ in

$a+b=8$

we get

$b = 5$

$x = \dfrac { 1 }{ 3 }$ and

$y= \dfrac { 1 }{ 5 }$

Solve the following equations by substitution method.

$3a-2b=-10; \, \, 2a+3b=2$

0%
$a = -2, b= 2$
0%
$a = -1, b= 2$
0%
$a= -4, b= 2$
0%
$a = -7, b= 2$

Explanation

$3a -2b =-10; \, \, 2a +3b =2$

$a = \dfrac {2b-10}{3} \Rightarrow$ Put it in second equation.

$2 \times \dfrac {2b-10}{3} +3b=2$

$4b - 20 +9b = 6$

$13b =26$

$\therefore b=2$

$a = \dfrac {2b-10}{3} = \dfrac {2(2)-10}{3} = -2$

$a = -2$ and

$b = 2$

CBSE Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 3 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 3 - MCQExams.com

0:0:1

Practice Class 10 Maths Quiz Questions and Answers

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