MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 3 - MCQExams.com
CBSE
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:
Report Question
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
Report Question
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
Report Question
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?
Report Question
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 $$2$$
$$2y-4x=0$$
......................................................(ii
i
)
Subtracting equation
(ii
i
)
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
Report Question
0%
$$0$$
0%
$$4$$
0%
$$2$$
0%
$$6$$
Explanation
Firstly mak
e 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 $$y$$:
$$4x+\dfrac{6}{y}=15 $$ ; $$3x-\dfrac{4}{y}=7$$
Report Question
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.
Report Question
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
Report Question
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
Report Question
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 $$y=-3$$
If $$(3)^{x + y} = 81$$ and $$(81)^{x - y} = 3$$, then the values of $$x$$ and $$y$$ are
Report Question
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 $$y = \frac{15}{8}$$
If $$\left (x+y,1 \right )$$ $$=$$ $$\left (3,y-x \right )$$, then $$x$$ $$=$$
, $$y$$ $$=$$
Report Question
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:
$$3x\, +\, y\, =\, 10;\quad 9x\, +\, py\, =\,23$$
Report Question
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 t
he 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
$$ \dfrac { 3 }{ 9 } \neq \dfrac { 1 }{ p } \\ \Rightarrow p\neq 3\quad
$$
If $$\displaystyle \frac{3}{x}- \frac{2}{y} =5$$ and $$\displaystyle \frac{4}{x} - \frac{5}{y} = 2$$, then $$\displaystyle \frac{1}{x} - \frac{1}{y} =?$$
Where $$ (x, y \neq 0)$$
Report Question
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
Report Question
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
Report Question
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.
If $$2^{2x - y} = 32$$ and $$2^{x + y} = 16$$ then $$x^{2} + y^{2}$$ is equal to
Report Question
0%
$$9$$
0%
$$10$$
0%
$$11$$
0%
$$13$$
Explanation
$$ Given\quad { 2 }^{ 2x-y }=32\\ \Rightarrow { 2 }^{ 2x-y }={ 2 }^{ 5 }\\ \Rightarrow 2x-y=5----(1)\\ Again\quad { 2 }^{ x+y }=16\\ \Rightarrow { 2 }^{ x+y }={ 2 }^{ 4 }\\ \Rightarrow x+y=4----(2)\\ Adding\quad (1)\quad and\quad (2)\quad we\quad get\\ 3x=9\\ \Rightarrow x=3\\ Substituting\quad x=3\quad in\quad (2)\quad we\quad get\\ y=1.\\ \therefore \quad { x }^{ 2 }+{ y }^{ 2 }={ 3 }^{ 2 }+{ 1 }^{ 2 }=10\quad (Ans) $$
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
Report Question
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.
$$\displaystyle \frac{x-2y}{3} =\, 1;\quad 2x\, -\, 4y\, =\, \displaystyle \frac{9}{2}$$
Report Question
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.
(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.
Without actually solving the simultaneous equations given below, decide whether the system has unique solution, no solution or infinitely many solutions.
$$8y= x - 10; 2x = 3y + 7$$
Report Question
0%
Unique solution
0%
infinitely many solutions.
0%
no solution
0%
cannot be determined
Explanation
As we know t
he 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:
$$8x\, -\, py\, +\, 7\, =\, 0;\quad 4x\, -\, 2y\, +\, 3\, =\, 0$$
Report Question
0%
All values of $$p$$ except $$4$$
0%
$$p = 7$$
0%
$$p = 6$$
0%
All values of $$p$$ except $$5$$
Explanation
As we know t
he 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?
Report Question
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:
$$ 4x\, +\, y\, =\, 7;\quad 16x\, +\, ky\, =\, 28$$
Report Question
0%
$$k\, =\, 2$$
0%
$$k\, =\, 6$$
0%
$$k\, =\, 3$$
0%
$$k\, =\, 4$$
Explanation
As we know t
he 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:
$$4y = kx- 10; 3x = 2y + 5$$
Report Question
0%
$$k\, =\, 2$$
0%
$$k\, =\, 6$$
0%
$$k\, =\, 8$$
0%
$$k\, =\, 4$$
Explanation
As we know t
he 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.
$$3x+5y=16; 4x-y=6$$
Report Question
0%
No Solution
0%
Infinitely many solutions
0%
Unique solution
0%
Cannot be determined
Explanation
As we know t
he 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:
$$kx\, -\, y\, +\, 3\, -\, k\, =\, 0;\quad 4x\, -\, ky\, +\, k\, =\, 0$$
Report Question
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.
$$\displaystyle \dfrac{x}{2} + \displaystyle \dfrac{y}{3} = 4; \displaystyle \dfrac{x}{4} +\displaystyle \frac{y}{6} = 2$$
Report Question
0%
no solution
0%
Infinite solutions
0%
unique solution
0%
Cannot be determined
Explanation
As we know t
he 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.
$$3y=2-x; 3x=6-9y$$
Report Question
0%
Infinite solutions
0%
unique solution
0%
no solution
0%
Cannot be determined
Explanation
As we know t
he 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.
$$ x - 2y - 4=0 $$
$$ 2x + y= 3 $$
Report Question
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:
$$\displaystyle \frac{1}{x}\, +\, \frac{1}{y}\, =\, 8;\quad \frac{4}{x}\, -\, \frac{2}{y}\, =\, 2$$
Report Question
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 $$ b = 5$$
So $$ x = \dfrac { 1 }{ 3 }$$ and $$y= \dfrac { 1 }{ 5 }$$
Solve the following equations by substitution method.
$$3a-2b=-10; \, \, 2a+3b=2$$
Report Question
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 $$b = 2 $$
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
0
Answered
0
Not Answered
0
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 10 Maths Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page
