MCQ Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 6

Solve the following simultaneous equation.

$12x+17y=53; \, \, 17x+12y=63$

0%
$x = 3,\, y=1$
0%
$x = 2,\, y=1$
0%
$x = 1,\, y=1$
0%
$x = 9,\, y=1$

Explanation

$12x+17y=53$ ...(i)

$17x+12y=63$ ...(ii)
On adding (i) and (ii) , we get

$29x+29y=116$

$\Rightarrow x+y=4$ ....(iii)
On subtracting (i) from (ii) , we get

$5x-5y=10$

$\Rightarrow x-y=2$ ....(iv)
On adding (iii) and (iv) , we get

$2x=6$

$\Rightarrow x=3$
By putting

$x=3$ in (iii), we get

$3+y=4$

$\Rightarrow y=1$

Solve the following equations by substitution method.

$2x+y=-2;\, \, 3x-y=7$

0%
$x = 1, y= -4$
0%
$x = 1, y= -1$
0%
$x = 1, y= -3$
0%
$x = 1, y= -5$

Explanation

$2x+y =-2; \, \, 3x-y =7$

$y = -2x -2$

Put it in second equation.

$3x -(-2x-2) =7$

$5x +2 = 7$

$x =1$

$y = -2x-2 = -2 (1)-2 = -4$

$x =1$ and

$y = -4$

$\displaystyle \frac{x+y-8}{2} = \frac{x+2y-14}{3}=\frac{3x-y}{4}$ , then the values of

$x$ and

$y$ is

0%
$x=1, \, y=3$
0%
$x=5, \, y=2$
0%
$x=3, \, y=3$
0%
$x=2, \, y=6$

Explanation

Given,

$\displaystyle \frac{x+y-8}{2} = \frac{x+2y-14}{3}=\frac{3x-y}{4}$

By taking first two. we get,

$\displaystyle \frac{x+y-8}{2} = \frac{x+2y-14}{3}$

$\Rightarrow 3x+3y-24=2x+4y-28$

$\Rightarrow x-y=-4$ ......(i)

By taking last two. we get,

$\displaystyle \frac{x+2y-14}{3}=\frac{3x-y}{4}$

$\Rightarrow 4x+8y-56=9x-3y$

$\Rightarrow 5x-11y=-56$ .....(ii)

On multiplying equation (i) by

$5$ . we get,

$5x-5y=-20$ ....(iii)

On subtracting (ii) from (iii). we get,

$5x-5y-(5x-11y)=-20-(-56)$

$\therefore 6y=36$

$\therefore y=6$

Substitute

$y=6$ in (i). we get,

$x-6=-4$

$\therefore x=2$

$\therefore x=2 , y=6$

Solve the following simultaneous equations by the method of equating coefficients.

$3x-4y=7;\, \, 5x+2y=3$ .

0%
$x = 1, y= -1$
0%
$x = 2, y= 1$
0%
$x = 5, y= 2$
0%
$x = 4, y= -3$

Explanation

$3x-4y=7$ ...(i)

$5x+2y=3$ ...(ii)
On multiplying (i) by 5 and (ii) by 3, we get

$15x-20y=35$

$\underline {\underset {-}15x\underset {-}{+}6y=\underset{-}9}$

$-26y=26$

$\therefore y=-1$
On putting

$y=-1$ in (i), we get

$3x-4(-1)=7$

$x=1$

$\therefore x=1,y=-1$

From the following figure, we can say:

$\displaystyle \frac{x}{3}+\frac{y}{4}=4; \, \, \frac{5x}{6}-\frac{y}{8}=4$

0%
$x=1, \, y=8$
0%
$x=7, \, y=8$
0%
$x=6, \, y=8$
0%
$x=3, \, y=8$

Explanation

Given ,

$\displaystyle \frac{x}{3}+\frac{y}{4}=4; \, \, \frac{5x}{6}-\frac{y}{8}=4$
Rewriting the given equations, we get

$4x+3y=48$ ...(i)

$20x-3y=96$ ...(ii)
On adding (i) and (ii), we get

$24x=144$

$\Rightarrow x=6$
On putting

$x=6$ in (i), we get

$4(6)+3y=48$

$\Rightarrow y=8$

$\therefore x=6,y=8$

Solve the following pair of equations.

$25x-24y=197; \, \, 24x-25y=195$

0%
$x=1,\, y=-3$
0%
$x=9,\, y=-3$
0%
$x=5,\, y=-3$
0%
$x=3,\, y=-3$

Explanation

$25x-24y=197$ ...(i)

$24x-25y=195$ ...(ii)
On adding (i) and (ii) , we get

$49x-49y=392$

$\Rightarrow x-y=8$ ....(iii)
On subtracting (ii) from (i) , we get

$x+y=2$ ....(iv)
On adding (iii) and (iv) , we get

$2x=10$

$\Rightarrow x=5$
By putting

$x=5$ in (iv), we get

$5+y=2$

$\Rightarrow y=-3$

The ratio of the present ages of mother and son is

$12: 5$ . The mother's age at the time of the birth of the son was

$21$ years. Find their present ages.

0%
mother age $=$ $36$ years, son age $=$ $15$ years
0%
mother age $=$ $56$ years, son age $=$ $5$ years
0%
mother age $=$ $46$ years, son age $=$ $25$ years
0%
mother age $=$ $47$ years, son age $=$ $26$ years

Explanation

Let mother age be

$M$ and that of son age be

$S$

Therefore according to question,

$M = S+21$ ...(1)
and

$5M =12S$ ....(2)

Substitute equation (1) in equation (2), we get

$5\left(S+21\right) = 12S$

$\Rightarrow 5S+105 = 12S$

$\Rightarrow 7S= 105$

$\Rightarrow S= 15$

Using equation (1), we get

$M= 15+21 = 36$

Mother age is

$36$ years and Son age is

$15$ years.

Solve graphically:

$\displaystyle 2x-3y=7$ and

$\displaystyle 5x+y=9$

0%
$\displaystyle x=2,y=-6$
0%
$\displaystyle x=8,y=-9$
0%
$\displaystyle x=0,y=-3$
0%
$\displaystyle x=2,y=-1$

Explanation

In equation

$2x-3y=7$
Let

$y$ =

$1$
then

$x$ will be

$5$
The one point is

$(5,1)$
Taking

$x=-1$ we get

$y=-3$
The other point is

$(-1,-3)$
Join both the points by drawing line.
In equation

$5x+y=9$
Let

$y$ =

$-1$
then

$x$ will be

$2$
The one point is

$(2,-1)$
Taking

$x=-1$ we get

$y=14$
The other point is

$(-1,14)$
Join both the points by drawing line.
Now the point of intersection of both lines is

$(2,-1)$

A two-digit number is 3 more than six times the sum of its digits. If 18 is added to the number obtained by interchanged by interchanging the digits, we get the original number. Find the number.

0%
25
0%
45
0%
75
0%
None of these

Explanation

Let the tens and the units digits in the number be x and y, respectively.
So, the number may be written as

$10x+y$ .
According to the given condition.

$10x+y=6(x+y)+3 \Rightarrow 4x-5y=3$ .....(i)
If we interchange the digits of Original number then we get new number. i.e

$10y+x$
According to the given condition

$10y+x+18=10x+y$

$\Rightarrow 9x-9y = 18 \Rightarrow x-y=2$ .....(ii)
Now multiplying equation (ii) by

$4$ . we get,

$4x-4y=8$ ....(iii)
On subtracting (iii) from (i). we get,

$4x-5y-(4x-4y)=3-8$

$\therefore y=5$
On substituting

$y=5$ in (ii). we get,

$x-5=2 \Rightarrow x=7$

$\therefore$ number is

$10x+y = 10(7)+5=75$

Point

$A$ and

$B$ are

$70\ km$ apart on a highway. A car starts from

$A$ and another car starts from

$B$ at the same time. If they travel in the same direction, they meet in

$7$ hours, but if they travel towards each other they meet in one hour. What are their speeds?

0%
$30\ km/hr$ and $40\ km/hr$
0%
$36\ km/hr$ and $40\ km/hr$
0%
$19\ km/hr$ and $20\ km/hr$
0%
$40\ km/hr$ and $50\ km/hr$

Explanation

Let car speed from point

$A=x\, km/hr$ .
and car speed from point

$B=y\, km/hr$ .

Distance = speed

$\times$ time

$\therefore$ Distance from

$A = x \times 7 = 7x$
Distance from

$B = y \times 7 = 7y$
If they travel in the same direction, they meet in

$7$ hours. Point

$A$ and

$B$ are

$70 km$ apart.

$\Rightarrow 7x-7y=70$

$\Rightarrow x-y=10$ ...

$(i)$

if they travel towards each other they meet in one hour. Point

$A$ and

$B$ are

$70 km$ apart.

$x+y=70$ ...

$(ii)$

Adding

$(i)$ and

$(ii)$ . we get

$2x=80$

$\therefore x=40$

substituting

$x=40$ in

$(i)$ . we get

$40-y=10$

$\therefore y=30$

$\therefore x=40 km/hr$ and

$y=30 km/hr$ .

Solve the following simultaneous equation :

$ax\, +\, by\, =\, 5$ and

$\, bx\, +\, ay\, =\, 3,$ where

$a$ and

$b$ are constants.

0%
$\displaystyle x\, =\, \frac{5a\, -\, 3b}{a^2\, +\, b^2}, \quad\, y\, =\, \frac{3a\, -\, 5b}{a^2\, +\, b^2}$
0%
$\displaystyle x\, =\, \frac{5a\, -\, 3b}{a^2\, -\, b^2}, \quad\, y\, =\, \frac{3a\, -\, 5b}{a^2\, -\, b^2}$
0%
$\displaystyle x\, =\, \frac{3a\, -\, 5b}{a^2\, -\, b^2}, \quad\, y\, =\, \frac{5a\, -\, 5b}{a^2\, -\, b^2}$
0%
$\displaystyle x\, =\, \frac{3a\, -\, 5b}{a^2\, +\, b^2}, \quad\, y\, =\, \frac{5a\, -\, 5b}{a^2\, +\, b^2}$

Explanation

$ax + by = 5$ ...(i)

$bx + ay = 3$ ...(ii)
Multiply (i) by

$a$ and (ii) by

$b$ and subtract

$a^2x + aby - b^2x - aby = 5a - 3b$

$(a^2 - b^2) x = 5a - 3b$

$x = \dfrac{5a - 3b}{a^2 - b^2}$
Similarly,
Multiply (i) by

$b$ and (ii) by

$a$ and subtract

$abx + b^2y - abx - a^2y = 5b - 3a$

$(b^2 - a^2)y = 5b - 3a$

$y = \dfrac{5b - 3a}{b^2 - a^2}$

The area of a rectangle gets reduced by

$9$ sq. units, if its length is reduced by

$5$ units and the breadth, is increased by

$3$ units. If we increase the length by

$3$ units and breadth by

$2$ units, the area is increased by

$67$ sq. units. Find the length and breadth of the rectangle.

0%
length $= 12$ , breadth $=6$
0%
length $= 20$ , breadth $=15$
0%
length $= 17$ , breadth $=9$
0%
length $= 21$ , breadth $=8$

Explanation

Let the length and breadth of the rectangle be

$x$ and

$y$ .

$\therefore$ Area of rectangle

$=xy$

According to given condition:

$(x-5)(y+3)=xy-9$

$\therefore xy+3x-5y-15=xy-9 \Rightarrow 3x-5y=6$ ....

$(i)$

and

$(x+3)(y+2)=xy+67$

$\therefore xy+2x+3y+6=xy+67 \Rightarrow 2x+3y=61$ .....

$(ii)$

On multiplying equation

$(i)$ by

$2$ , we get,

$6x-10y=12$ ....

$(iii)$

On multiplying equation

$(ii)$ by

$3$ , we get,

$6x+9y=183$ .....

$(iv)$

On subtracting

$(iv)$ from

$(iii)$ , we get,

$-19y=-171 \Rightarrow y=9$

by putting

$y=9$ in

$(i)$ . we get,

$3x-5(9)=6 \Rightarrow x=17$

$\therefore$ length

$=17$ and breadth

$=9$

A man gets Rs. 100 per day if he works, but he is fined by Rs. 10 per day if he is absent. In the whole month of April he received Rs. 1900 only. How many days did he work ?

0%
20 days
0%
22 days
0%
29 days
0%
23 days

Explanation

Let the man works for

$x$ days in April and he is absent for

$y$ days.
Since, April has total 30 days.

$\therefore x+y=30$ .........(1)
According to question:

$100x-10y=1900 \Rightarrow 10x-y=190$ ....(2)
Adding (1) and (2). we get,

$11x=220$

$\therefore x=20$
Hence the man worked for

$20$ days in April.

Solvw the following pair of linear equations:

$2x\, +\, 5y\, =\, 13$ and

$4x\, -\, 9y\, =7$

0%
$(1,4)$
0%
$(2,-3)$
0%
$(-2,3)$
0%
$(4,1)$

Explanation

$2x+5y=13$ ......

$(1)$

$4x-9y=7$ ....

$(2)$

Multiplying equation

$(1)$ by

$4$ . we get,

$8x+20y=52$ ....(3)

Now, Multiplying equation

$(2)$ by

$2$ . we get,

$8x-18y=14$ ......

$(4)$

On subtracting

$(4)$ from

$(3)$ . we get,

$38y=38\\ \Rightarrow y=1$

On substitute the value of

$y=1$ in

$(1)$ . we get,

$2x+5(1)=13 \\\Rightarrow 2x=8 \\\Rightarrow x=4$

Thus solution of given equations is

$(4,1)$

Option D is correct.

$12x\, +\, 13y\, =\, 29\, and\, 13x\, +\, 12y\, =\, 21,$ find

$x\, +\, y$ .

0%
$2$
0%
$7$
0%
$4$
0%
$11$

Explanation

$12x\, +\, 13y\, =\, 29\, and\, 13x\, +\, 12y\, =\, 21,$
Add the two equations:

$25x + 25y = 50 \\x + y = 2$

In the following figure,

$ABCD$ is a parallelogram. Find the values of

$x$ .

0%
$3$
0%
$5$
0%
$7$
0%
$9$

Explanation

Given:

$ABCD$ is a

$\parallel gm$ , therefore

$AB=CD$ and

$AD=BC$ [Opposite sides of

$\parallel gm$ are equal]

$4x=6y+2 \Rightarrow 4x-6y=2$

$2x-3y=1 \rightarrow (1)$

$4y=3x-3 \Rightarrow 3x-4y=3 \rightarrow (2)$

Multiply (1) by 3 and (2) by 2 and subtracting we get,

$6x-9y=3$

$6x-8y = 6$

------------------------

$0-y = -3$

$y=3$

From (1),

$2x-3\times3=1$

$\Rightarrow x= \dfrac{9+1}{2}=5$

$1$ is added to each of the two certain numbers, their ratio is

$1:2$ ; and if

$5$ is subtracted from each of the two numbers, their ratio becomes

$5:11$ . Find the numbers.

0%
$35$ and $70$
0%
$35$ and $71$
0%
$35$ and $72$
0%
$35$ and $73$

Explanation

Let the numbers be

$x$ and

$y$

Given, if

$1$ is added to the numbers, then ratio

$= \dfrac {x + 1}{y + 1} = \dfrac {1}{2}$

$=> 2x + 2 = y + 1$

$=> 2x - y = -1$ ...(1)

Also, if

$5$ is subtracted from the numbers, then ratio

$= \dfrac {x - 5}{y - 5} = \dfrac {5}{11}$

$=> 11x - 55 = 5y - 25$

$=> 11x - 5y = 30$ ...(2)

Multiplying equation $(1)$ with $5$ we get, $10x - 5y = -5$ ----- equation $(3)$

Subtracting equation $(3)$ from $(1)$ , we get $x = 35$

Substituting $x = 35$ in the equation $(1)$ , we get $70 - y = -1 => y = 71$

Hence, the numbers are $35$ and $71$

Solve the following simultaneous equations:

$8x+13y-29= 0$ ,

$12x-7y-17= 0$

0%
$x=3;\, y=2$
0%
$x=0;\, y=5$
0%
$x=2;\, y=1$
0%
$x=-1;\, y=3$

Explanation

We pick either of the equations and write one variable in terms of the other.

Let us consider the equation

$8x+13y=29$ and write it as

$x=\frac { 29-13y }{ 8 }.........(1)$

Substitute the value of

$x$ in Equation

$12x-7y=17$ . We get

$12\left( \frac { 29-13y }{ 8 } \right) -7y=17\\ \Rightarrow 87-39y-14y=34\\ \Rightarrow 87-53y=34\\ \Rightarrow -53y=-53\\ \Rightarrow y=1$

Therefore,

$y = 1$

Substituting this value of

$y$ in Equation (1), we get

$x=\frac { 29-13 }{ 8 } =\frac { 16 }{ 8 } =2$

Hence, the solution is

$x = 2, y = 1$ .

Solve the set of equations:

$3\left ( 2u+v \right )= 7uv$ and

$3\left ( u+3v \right )= 11uv$

0%
$u= 1; v= \displaystyle \frac{3}{2}$
0%
$u= 2; v= \displaystyle \frac{1}{2}$
0%
$u= 2; v= \displaystyle \frac{5}{4}$
0%
$u= 5; v= \displaystyle \frac{7}{4}$

Explanation

The equation

$3(2u+v)=7uv$ can be rewritten as

$6u+3v=7uv$ and the equation

$3(u+3v)=11uv$ can be rewritten as

$3u+9v=11uv$ . Then we get the equations:

$6u+3v=7uv.........(1)$

$3u+9v=11uv.........(2)$

Multiply equation 2 by

$2$ :

$6u+18v=22uv.........(3)$

Subtract Equation 1 from equation 3 to eliminate

$u$ , because the coefficients of

$u$ are the same. So, we get

$(6u-6u)+(18v-3v)=22uv-7uv$

i.e.

$-15v=-15uv$

i.e.

$u=1$

Substituting this value of

$u$ in equation 1, we get

$6+3v=7v$

i.e.

$4v=6$

i.e.

$v=\frac { 3 }{ 2 }$

Hence, the solution of the equations is

$u=1,v=\frac { 3 }{ 2 }$ .

A triangle is formed by the straight lines:

$\displaystyle x + 2y - 3 = 0, 3x - 2y + 7 = 0$ and

$\displaystyle y + 1 = 0$ .

Find graphically, the co-ordinates of the vertices of the triangle.

0%
$(-1,2), (5,-1), (-3,-1)$
0%
$(-1,2), (7,4), (-3,-1)$
0%
$(7,-4), (5,-1), (-3,-1)$
0%
$(-12,5), (5,-1), (-3,-1)$

Explanation

Plotting the equation

$x+2y=0$ ,

$3x-2y+7=0$ and

$y+1=0.$ in the graph we get,the vertices of the triangle are

$(-1,2),(5,-1)$ and

$(-3,-1)$

Solve :

$\displaystyle \frac{9}{x}\, -\, \displaystyle \frac{4}{y}\, =\, 8$ and

$\displaystyle \frac{13}{x}\, +\, \displaystyle \frac{7}{y}\, =\, 101$

0%
$x\, =\, \displaystyle \frac{1}{7}\; \, y\, =\, \displaystyle \frac{1}{6}$
0%
$x\, =\, \displaystyle \frac{3}{4}\; \, y\, =\, \displaystyle \frac{1}{6}$
0%
$x\, =\, \displaystyle \frac{1}{4}\; \, y\, =\, \displaystyle \frac{1}{7}$
0%
$x\, =\, \displaystyle \frac{2}{7}\; \, y\, =\, \displaystyle \frac{1}{7}$

Explanation

Let

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

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

$\displaystyle \frac{9}{x}\, -\, \displaystyle \frac{4}{y}\, =\, 8$ ......(i)

$\Rightarrow 9a-4b=8$ .....(ii)

$\displaystyle \frac{13}{x}\, +\, \displaystyle \frac{7}{y}\, =\, 101$ ......(iii)

$\Rightarrow 13a+7b=101$ .....(iv)
Multipliying equation (ii) by

$7$ and (iv) by

$4$ , we get

$63a-28b=56$

$52a+28b=404$

$115a =460$

$\Rightarrow a =\dfrac{460}{115}$

$\Rightarrow a =4$
Putting the value in (ii), we get

$9a-4b=8$

$\Rightarrow 9\times 4-4b=8$

$\Rightarrow -4b=8-36$

$\Rightarrow -4b=-28$

$\Rightarrow b=7$
Now, resubstituting the value, we get

$a=\dfrac{1}{x}=4$

$\Rightarrow x=\dfrac{1}{4}$
Again,

$b=\dfrac{1}{y}=7$

$\Rightarrow y=\dfrac{1}{7}$

Solve the following simultaneous equations :

$3x-5y+1= 0$ ,

$2x-y+3= 0$

0%
$x= 2;\, y= -1$
0%
$x= 7;\, y= -9$
0%
$x= -1;\, y= -4$
0%
$x= -2;\, y= -1$

Explanation

We pick either of the equations and write one variable in terms of the other.

Let us consider the equation

$2x-y+3=0$ and write it as

$y=3+2x...........(1)$

Substitute the value of

$y$ in Equation

$3x-5y+1=0$ or

$3x-5y=-1$ . We get

$3x-5(3+2x)=-1$

i.e.

$3x-15-10x=-1$

i.e.

$-7x=14$

i.e.

$x=-2$

Therefore,

$x=-2$

Substituting this value of

$x$ in the equation

$2x-y+3=0$ , we get

$-4-y+3=0$

i.e.

$-y-1=0$

i.e.

$y=-1$

Hence, the solution is

$x = -2, y = -1$ .

Solve the following simultaneous equations:

$3x+2y= 14$ ,

$-x+4y= 7$

0%
$x= 0;\, y= -4$
0%
$x= 3;\, y= 2.5$
0%
$x= 3.5;\, y= -2$
0%
$x= 4.5;\, y= 3$

Explanation

We pick either of the equations and write one variable in terms of the other.

Let us consider the equation

$-x+4y=7$ and write it as

$x = 4y-7...........(1)$

Substitute the value of

$x$ in Equation

$3x+2y=14$ . We get

$3(4y-7)+2y=14$

i.e.

$12y-21+2y=14$

i.e.

$14y=35$

i.e.

$2y = 5$

Therefore,

$y=\frac { 5 }{ 2 }=2.5$

Substituting this value of

$y$ in Equation (1), we get

$x = 10-7=3$

Hence, the solution is

$x = 3, y = 2.5$ .

Solve :

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

$\displaystyle \frac{9}{x+y}-\displaystyle \frac{4}{x-y}= 1$

0%
$x= \displaystyle \frac{2}{5}; y= \displaystyle \frac{7}{2}$
0%
$x= \displaystyle \frac{5}{2}; y= \displaystyle \frac{1}{2}$
0%
$x= \displaystyle \frac{6}{7}; y= \displaystyle \frac{4}{3}$
0%
$x= \displaystyle \frac{7}{3}; y= \displaystyle \frac{1}{2}$

Explanation

Let

$\dfrac1{x+y}=X$ and

$\dfrac1{x-y}=Y$

$3X+2Y=2$ ...(i)

$9X-4Y=1$ ...(ii)

On multiplying (i) by 3 and (ii) by 1 and subtracting, we get

$9X+6Y=6$

$\underline {\underset {-}9X\underset {+}{-}4Y=\underset{-}1}$

$10Y=5$

$\therefore Y=\dfrac12$

On putting

$Y=\dfrac12$ in (i), we get

$3X+2\times\dfrac12=2\Rightarrow X=\dfrac13$

$\therefore \dfrac1{x+y}=\dfrac13$ and

$\dfrac1{x-y}=\dfrac12$

$\Rightarrow x+y=3$ ...(iii)

$x-y=2$ ...(iv)

On adding (iii) and (iv), we get

$2x=5\Rightarrow x=\dfrac52$

On putting

$x=\dfrac52$ in (iii), we get

$\dfrac52+y=3\Rightarrow y=3-\dfrac52=\dfrac12$

$\therefore x=\dfrac52, y=\dfrac12$

Solve:

$4x+\displaystyle \frac{6}{y}= 15$ and

$6x-\displaystyle \frac{8}{y}= 14$ . Hence find the value of

$k$ , if

$y= kx-2$ .

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$x= 4, y= 1$ and $k= \displaystyle \frac{2}{3}$
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$x= 5, y= 3$ and $k= \displaystyle \frac{3}{5}$
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$x= 3, y= 2$ and $k= \displaystyle \frac{4}{3}$
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$x= 1, y= 2$ and $k= \displaystyle \frac{9}{2}$

Explanation

The equation

$4x +\frac { 6 }{ y } =15$ can be solved as:

$4x+\frac { 6 }{ y } =15\\ \Rightarrow \frac { 4xy+6 }{ y } =15\\ \Rightarrow 4xy+6=15y\quad .........(1)$

The equation

$6x -\frac { 8 }{ y } =14$ can be solved as:

$6x-\frac { 8 }{ y } =14\\ \Rightarrow \frac { 6xy-8 }{ y } =14\\ \Rightarrow 6xy-8=14y\quad .........(2)$

Multiply the equation 1 by

$3$ and equation 2 by

$2$ . Then we get the equations:

$12xy+18=45y.........(3)$

$12xy-16=28y.........(4)$

Subtract equation 4 from equation 3 to eliminate

$xy$ , because the coefficients of

$xy$ are same. So, we get

$(12xy-12xy)+(18+16)=45y-28y$

i.e.

$17y=34$

i.e.

$y=2$

Substituting this value of

$y$ in the equation 1, we get

$8x+6=30 \\ \Rightarrow 8x=24 \\ \Rightarrow x=3$

Therefore, the solution of the equations is

$x=3,y=2$ .

Now Substitute the values of

$x$ and

$y$ in the equation

$y=kx-2$ as follows:

$2=3k-2$

$3k=4$

$\Rightarrow k=\frac { 4 }{ 3 }$

Hence,

$k=\frac { 4 }{ 3 }$ .

Solve:

$3\left ( 2x+y \right )= 7xy$ and

$3\left ( x+3y \right )= 11xy$ ;

where,

$x\neq 0, y\neq 0$

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$x= 3; y= \displaystyle \frac{1}{2}$
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$x= 1; y= \displaystyle \dfrac{3}{2}$
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$x= 4; y= 2$
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$x= 3; y= 4$

Explanation

Dividing both the equations by xy, we get:

$\dfrac3x+\dfrac6y=7;\\ \dfrac3y+\dfrac9x=11$

Multiplying the 2nd equation by -2 and Adding

$\dfrac{ -15 }{x}= -15;\\ y= \dfrac32 \quad and \quad x=1$

Solve the following simultaneous equations:

$12x+15y+18= 0$ ,

$18x-7y+86= 0$

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$x=-4;\, y=2$
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$x=1;\, y=0$
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$x=-3;\, y=5$
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$x=4;\, y=1$

Explanation

We pick either of the equations and write one variable in terms of the other.

Let us consider the equation

$12x+15y+18=0$ and write it as

$x=\frac { -18-15y }{ 12 }..........(1)$

Substitute the value of

$x$ in Equation

$18x-7y=-86$ . We get

$18\left( \frac { -18-15y }{ 12 } \right) -7y=-86\\ \Rightarrow -54-45y-14y=-172\\ \Rightarrow -59y=-118\\ \Rightarrow y=2$

Therefore,

$y = 2$

Substituting this value of

$y$ in Equation (1), we get

$x=\frac { -18-30 }{ 12 } =-\frac { 48 }{ 12 } =-4$

Hence, the solution is

$x = -4, y = 2$ .

Solve graphically, the following pairs of equations:

$\displaystyle \frac{x + 1}{4} = \frac{2}{3}(1 - 2y)$

$\displaystyle \frac{2 + 5y}{3} = \frac{x}{7} - 2$

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$x = 3, y = -1$
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$x = 7, y = -1$
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$x = 5, y = -1$
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$x = 4, y = -1$

Explanation

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

$x+1=\dfrac{8}{3}(1-2y)$

$\implies x=\dfrac{8(1-2y)}{3}-1$

$\implies x=\dfrac{8(1-2y)-3}{3}$

$\implies x=\dfrac{8-16y-3}{3}$

$\implies x=\dfrac{5-16y}{3}$

Let,

$y=2$ , then

$x=\dfrac{5-16(2)}{3}=-9$

Similarly, when

$y=-1, x=7$

Now,

$\dfrac{2+5y}{3}=\dfrac{x}{7}-2$

$\implies \dfrac{2+5y}{3}+2=\dfrac{x}{7}$

$\implies \dfrac{2+5y+6}{3}=\dfrac{x}{7}$

$\implies \dfrac{7(8+5y)}{3}=x$

$\implies \dfrac{56+35y}{3}=x$

Let,

$y=-1$ , then

$x=\dfrac{56+35(-1)}{3}=7$

Similarly, when

$y=-2, x=-8$

$\therefore$ the intersecting point is

$(7,-1)$

Solve the following pair of equations:

$3x\, -\, y\, =\, 23$

$\displaystyle \frac{x}{3}\, +\, \displaystyle \frac{y}{4}\, =\, 4$

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$x\, =-\, 4\, ;\, y\, =\, 1$
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$x\, =-\, 1\, ;\, y\, =\, 7$
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$x\, =\, 9\, ;\, y\, =\, 4$
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$x\, =\, 7\, ;\, y\, =\, 13$

Explanation

We pick either of the equations and write one variable in terms of the other.

Let us consider the equation

$3x-y=23$ and write it as

$y=3x-23...........(1)$

Substitute the value of

$y$ in Equation

$\frac { x }{ 3 } +\frac { y }{ 4 } =4$ or

$4x+3y=48$ . We get

$4x+3(3x-23)=48$

i.e.

$4x+9x-69=48$

i.e.

$13x=117$

i.e.

$x=9$

Therefore,

$x=-9$

Substituting this value of

$x$ in equation 1, we get

$y=27-23$

i.e.

$y=4$

Hence, the solution is

$x = 9, y =4$ .

Solve the following pair of equations:

$13 + 2y = 9x$ ,

$3y = 7x$

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$x=6$ ; $y =-7$
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$x=5$ ; $=13$
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$x=-7$ ; $y =1$
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$x=3$ ; $y =7$

Explanation

We pick either of the equations and write one variable in terms of the other.

Let us consider the equation

$3y=7x$ and write it as

$x=\frac { 3 }{ 7 }y..........(1)$

Substitute the value of

$x$ in the equation

$13+2y=9x$ or

$9x-2y=13$ . We get:

$9\left( \frac { 3 }{ 7 } y \right) -2y=13\\ \Rightarrow 27y-14y=91\\ \Rightarrow 13y=91\\ \Rightarrow y=7$

Therefore,

$y = 7$

Substituting this value of

$y$ in Equation (1), we get

$x=\frac { 3 }{ 7 } \times 7=3$

Hence, the solution is

$x = 3, y = 7$ .

CBSE Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 6 - 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 6 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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