MCQ Questions for Class 12 Commerce Applied Mathematics Linear Equations Quiz 6

Solve the following pair of equations:

$\sqrt{2}x+3y=\sqrt{3}; \, \, \sqrt{3}x+3y=\sqrt{2}$

0%
$x=-1, \, y=\displaystyle \frac{\sqrt{3}+\sqrt{2}}{3}$
0%
$x=1, \, y=\displaystyle \frac{\sqrt{5}-\sqrt{2}}{3}$
0%
$x=1, \, y=\displaystyle \frac{\sqrt{7}-\sqrt{13}}{3}$
0%
$x=1, \, y=\displaystyle \frac{\sqrt{10}-\sqrt{5}}{3}$

Explanation

$\sqrt{2}x+3y=\sqrt{3}$ .....(i)

$\sqrt{3}x+3y=\sqrt{2}$ .....(ii)

Subtracting (ii) from (i)

$\sqrt{2}x+3y-\sqrt{3}x-3y=\sqrt{3}-\sqrt{2}$

$(\sqrt{2}-\sqrt{3})x=\sqrt{3}-\sqrt{2}$

$\Rightarrow x=-1$

On substituting

$x=-1$ in (i). we get,

$\sqrt{2}(-1)+3y=\sqrt{3}$

$3y=\sqrt{3}+\sqrt{2}$

$y=\dfrac {\sqrt{3}+\sqrt{2}}{3}$

$\therefore x=-1 , y=\dfrac {\sqrt{3}+\sqrt{2}}{3}$

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.

From the following figure, we can say:

$0.9x+0.5y=6; \, \, 0.7x+0.7y=5.6$

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

Explanation

$0.9x+0.5y=6$ .....(i)

$0.7x+0.7y=5.6$ ....(ii)
On multiplying both equations by

$10$ . we get,

$9x+5y=60$ .....(iii)

$7x+7y=56 \Rightarrow x+y =8$ ....(iv)
Multiplying equation (iv) by

$5$ . we get,

$5x+5y=40$ .....(v)
Subtracting (v) from (iii). we get,

$4x=20 \Rightarrow x=5$
Substitute

$x=5$ in equation (iii). we get,

$9(5)+5y=60$

$\Rightarrow 5y=15 \Rightarrow y=3$

$\therefore x=5 , y=3$

Solve the following pair of equations:

$6x+5y=7x+3y+1=2x+12y-2$

0%
$x=3, \, y=2$
0%
$x=1, \, y=0$
0%
$x=5, \, y=3$
0%
$x=7, \, y=1$

Explanation

Given,

$6x+5y=7x+3y+1=2x+12y-2$
By taking first two we get,

$6x+5y = 7x+3y+1$

$\Rightarrow x-2y=-1$ ....(i)
By taking last two we get,

$7x+3y+1=2x+12y-2$

$\Rightarrow 5x-9y=-3$ .....(ii)
On multiplying equation (i) by

$5$ . we get,

$5x-10y=-5$ ....(iii)
Subtracting (ii) from (iii). we get,

$5x-10y-(5x-9y)=-5-(-3)$

$\therefore -y=-2 \Rightarrow y=2$
On substituting

$y=2$ in (i). we get,

$x-2(2)=-1$

$\therefore x=3$

$\therefore x=3 , y=2$

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 following simultaneous equations:

$217x+131y= 913$

$131x+217y= 827$

0%
$x=7, y=3$
0%
$x=2, y=1$
0%
$x=3, y=2$
0%
$x=1, y=5$

Explanation

$217x+131y= 913$ ...

$(i)$

$131x+217y= 827$ ...

$(ii)$
On multiplying

$(i)$ by

$131$ and

$(ii)$ by

$217$ and subtracting, we get

$28427x+17161y=119603$

$\underline {\underset {-}28427x\underset {-}{+}47089y=\underset{-}179459}$

$-29928y=-59856$

$\therefore y=2$
On putting

$y=2$ in

$(i)$ , we get

$217x+131\times2=913\Rightarrow 217x=651\Rightarrow x=3$

$\therefore x=3,y=2$

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 given lines are

$2x\, +\, y\, =\, 6;\quad\, x\, +\, 2y\, =\, 6;\quad\, 7x\, -\, 4y\, =\, 6$

0%
Concurrent
0%
Coincident
0%
Parallel
0%
None of these

Explanation

$2x+y=6..........eq1$

$x+2y=6..........eq2$
multiply eq1 by 2 we get

$4x+2y=12........eq3$
subtracting eq2 and eq3 we get

$3x=6$

$x=2$
put x=2 in eq1 we get

$2\times 2+y=6$

$y=2$
Put x=2 and y=2 in 7x-4y=6

$L.H.S=7\times 2-4\times 2=14-8=6=R.H.S$

$\therefore$ all these lines whose equation are 2x+y=6,x+2y=6 and 7x-4y=6 passes through (2,2)

$\therefore$ They are concurrent line and their point of intersection is (2,2)

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

$x = 3, y = 2.5$ .

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),

$2x-3\times3=1$

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

Solve the following simultaneous equations:

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

$18x-7y+86= 0$

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

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 the following pairs of linear (simultaneous) equation by the method of elimination:

$x + 8y = 19$ ,

$2x + 11y = 28$

0%
$x=3$ and $y =2$
0%
$x=1$ and $y =4$
0%
$x=3$ and $y =5$
0%
$x=4$ and $y =7$

Explanation

Multiply the equation

$x+8y=19$ by

$2$ to make the coefficients of

$x$ equal. Then we get the equations:

$2x+16y=38.........(1)$

$2x+11y=28.........(2)$

Subtract Equation (2) from Equation (1) to eliminate

$x$ , because the coefficients of

$x$ are the same. So, we get

$(2x-2x)+(16y-11y)=38-28$

i.e.

$5y=10$

i.e.

$y = 2$

Substituting this value of

$y$ in the equation

$x+8y=19$ , we get

$x+16=19$

i.e.

$x=19-16=3$

Hence, the solution of the equations is

$x = 3, y = 2$ .

Solve the following pairs of linear (simultaneous) equation by the method of elimination :

$2x + 3y = 8$ ,

$2x = 2 + 3y$

0%
$x=2$ , $y=\dfrac{3}{2}$
0%
$x=3$ , $y=5$
0%
$x=\dfrac{5}{2}$ , $y=1$
0%
$x=\dfrac{3}{2}$ , $y=4$

Explanation

The equation

$2x=2+3y$ can be rewritten as

$2x-3y=2$ , therefore, the equations are:

$2x+3y=8.........(1)$

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

Subtract Equation (2) from Equation (1) to eliminate

$x$ , because the coefficients of

$x$ are the same. So, we get

$(2x-2x)+(3y+3y)=8-2$

i.e.

$6y=6$

i.e.

$y = 1$

Substituting this value of

$y$ in the equation

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

$2x=2+3$

i.e.

$2x=5$

i.e.

$x=\frac { 5 }{ 2 }$

Hence, the solution of the equations is

$x=\frac { 5 }{ 2 },y=1$ .

Solve the following pairs of linear (simultaneous) equation by the method of elimination:

$x + y = 7$ ,

$5x + 12y = 7$

0%
$x =13$ and $y=5$
0%
$x=11$ and $y=-4$
0%
$x=7$ and $y =-2$
0%
$x=5$ and $y =7$

Explanation

Multiply the equation

$x+y=7$ by

$5$ to make the coefficients of

$x$ equal. Then we get the equations:

$5x+5y=35.........(1)$

$5x+12y=7.........(2)$

Subtract Equation (2) from Equation (1) to eliminate

$x$ , because the coefficients of

$x$ are the same. So, we get

$(5x-5x)+(5y-12y)=35-7$

i.e.

$7y=-28$

i.e.

$y = -4$

Substituting this value of

$y$ in the equation

$x+y=7$ , we get

$x-4=7$

i.e.

$x=11$

Hence, the solution of the equations is

$x = 11, y = -4$ .

$1$ 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 pair of equations :

$\displaystyle \frac{3}{5}\, x\, -\, \displaystyle \frac{2}{3}\, y\, +\, 1\, =\, 0$

$\displaystyle \frac{1}{3}\, y\, +\, \displaystyle \frac{2}{5}\, x\, =\, 4$

0%
$x\, =\, 2\, ;\, y\, =\, 6$
0%
$x\, =\, 5\, ;\, y\, =\, 6$
0%
$x\, =\, 1\, ;\, y\, =\, 6$
0%
$x\, =\, 7\, ;\, y\, =\, 6$

Explanation

On simplifying

$\frac {3x}{5} - \frac {2y}{3} + 1 = 0$ , we get

$9x -10y = -15$ ---- (1)

On simplifying

$\frac {y}{3} + \frac {2x}{5} = 4$ , we get

$5y + 6x = 60$ ---- (2)

Multiplying equation

$(2)$ with

$2$ we get,

$10y + 12x = 120$ ----- equation

$(3)$
Adding equation

$(3)$ to

$(1)$ , we

get

$21x = 105 => x = 5$

Substituting

$x = 5$ in the equation

$(2)$ , we get

$5y + 6(5) = 60 => y = 6$

Solve the following pairs of linear (simultaneous) equation by the method of elimination:

$2x + 7y = 39$ ,

$3x + 5y = 31$

0%
$x = 0, y = 7$
0%
$x = 6, y = 2$
0%
$x = -1, y = 3$
0%
$x = 2, y = 5$

Explanation

Multiply the equation

$2x+7y=39$ by

$3$ and equation

$3x+5y=31$ by

$2$ to make the coefficients of

$x$ equal. Then we get the equations:

$6x+21y=117.........(1)$

$6x+10y=62.........(2)$

Subtract Equation (2) from Equation (1) to eliminate

$x$ , because the coefficients of

$x$ are the same. So, we get

$(6x-6x)+(21y-10y)=117-62$

i.e.

$11y=55$

i.e.

$y = 5$

Substituting this value of

$y$ in the equation

$2x+7y=39$ , we get

$2x+35=39$

i.e.

$2x=4$

i.e.

$x=2$

Hence, the solution of the equations is

$x = 2, y = 5$ .

Solve the following pairs of linear (simultaneous) equation by the method of elimination:

$8x + 5y = 9$ ,

$3x + 2y = 4$

0%
$x=1$ and $y =7$
0%
$x=-3$ and $y =4$
0%
$x=-2$ and $y =5$
0%
$x=0$ and $y =1$

Explanation

Multiply the equation

$8x+5y=9$ by

$3$ and equation

$3x+2y=4$ by

$8$ to make the coefficients of

$x$ equal. Then we get the equations:

$24x+15y=27.........(1)$

$24x+16y=32.........(2)$

Subtract Equation (1) from Equation (2) to eliminate

$x$ , because the coefficients of

$x$ are the same. So, we get

$(24x-24x)+(16y-15y)=32-27$

i.e.

$y =5$

Substituting this value of

$y$ in the equation

$8x+5y=9$ , we get

$8x+25=9$

i.e.

$8x=-16$

i.e.

$x=-2$

Hence, the solution of the equations is

$x = -2, y = 5$ .

Solve the following pairs of linear (simultaneous) equation by the method of elimination:

$0.2x + 0.1y = 25$ ,

$2(x - 2) - 1.6y = 116$

0%
$x= 100$ , $y= 50$
0%
$x= 80$ , $y= 30$
0%
$x= -78$ , $y= 35$
0%
$x= -65$ , $y= -75$

Explanation

Multiply the equation

$0.2x+0.1y=25$ by

$10$ and the equation

$2(x-2)-1.6y=116$ can be rewritten as

$2x-1.6y=120$ . Then we get the following equations:

$2x+y=250.........(1)$

$2x-1.6y=120.........(2)$

Subtract Equation (2) from Equation (1) to eliminate

$x$ , because the coefficients of

$x$ are the same. So, we get

$(2x-2x)+(y+1.6y)=250-120$

i.e.

$2.6y=130$

i.e.

$y =50$

Substituting this value of

$y$ in the equation

$2x-1.6y=120$ , we get

$2x-80=120$

i.e.

$2x=200$

i.e.

$x=100$

Hence, the solution of the equations is

$x = 100, y = 50$ .

Solve the following pairs of linear (simultaneous) equation by the method of elimination:

$6x = 7y +7$ ,

$7y - x = 8$

0%
$x =3$ , $y = \displaystyle \frac{11}{7}$
0%
$x=7$ , $y= \displaystyle \frac{2}{3}$
0%
$x =\displaystyle \frac{1}{3}$ , $y =5$
0%
$x= \displaystyle \frac{2}{5}$ , $y=4$

Explanation

The equation

$6x=7y+7$ can be rewritten as

$6x-7y=7$ and the equation

$7y-x=8$ can be rewritten as

$-x+7y=8$ . Then we get the following equations:

$6x-7y=7.........(1)$

$-x+7y=8.........(2)$

Add equations 1 and 2 to eliminate

$y$ , because the coefficients of

$y$ are the opposite. So, we get

$(6x-x)+(-7y+7y)=7+8$

i.e.

$5x=15$

i.e.

$x=3$

Substituting this value of

$x$ in the equation

$6x-7y=7$ , we get

$18-7y=7$

i.e.

$-7y=-11$

i.e.

$y=\frac { 11 }{ 7 }$

Hence, the solution of the equations is

$x=3,y=\frac { 11 }{ 7 }$ .

Solve the following pairs of linear (simultaneous) equation by the method of elimination:

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

$16x\, -\, 5y\, =\, 25$

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$x\, =\, \displaystyle \frac{7}{5},\, y\, =\, 10$
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$x\, =\, \displaystyle \frac{5}{2},\, y\, =\, 3$
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$x\, =\, \displaystyle \frac{1}{2},\, y\, =\, 4$
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$x\, =\, \displaystyle \frac{9}{2},\, y\, =\, 8$

Explanation

The equation

$y=4x-7$ can be rewritten as

$4x-y=7$ and multiply the equation

$4x-y=7$ by

$4$ . Then we get the following equations:

$16x-5y=25.........(1)$

$16x-4y=28.........(2)$

Subtract equation 1 from equation 2 to eliminate

$x$ , because the coefficients of

$x$ are same. So, we get

$(16x-16x)+(-4y+5y)=28-25$

i.e.

$y=3$

Substituting this value of

$y$ in the equation

$4x-y=7$ , we get

$4x-3=7$

i.e.

$4x=10$

i.e.

$x=\frac { 5 }{ 2 }$

Hence, the solution of the equations is

$x=\frac { 5 }{ 2 },y=3$ .

Solve the following pairs of linear (simultaneous) equation by the method of elimination :

$2x - 3y= 7$ ,

$5x + y =9$

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$x=2$ and $y =-1$
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$x=4$ and $y =0$
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$x=-3$ and $y =-5$
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$x=-7$ and $y =6$

Explanation

Multiply the equation

$2x-3y=7$ by

$5$ and equation

$5x+y=9$ by

$2$ to make the coefficients of

$x$ equal. Then we get the equations:

$10x-15y=35.........(1)$

$10x+2y=18.........(2)$

Subtract Equation (2) from Equation (1) to eliminate

$x$ , because the coefficients of

$x$ are the same. So, we get

$(10x-10x)+(-15y-2y)=35-18$

i.e.

$-17y =17$

i.e.

$y = -1$

Substituting this value of

$y$ in the equation

$2x-3y=7$ , we get

$2x+3=7$

i.e.

$2x=4$

i.e.

$x=2$

Hence, the solution of the equations is

$x = 2, y = -1$ .

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

$x = 3, y = 7$ .

Solve the following pair of equations :

$x\, -\, y\, =\, 0.9$

$\displaystyle \frac{11}{2\, (x\, +\, y)}\, =\, 1$

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$x\, =\, 1.2\, ;\, y\, =\, 2.3$
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$x\, =\, 7.2\, ;\, y\, =\, 2.3$
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$x\, =\, 3.2\, ;\, y\, =\, 2.3$
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$x\, =\, 4.2\, ;\, y\, =\, 2.3$

Explanation

Given, equations are $x - y = 0.9$ ---- (1)

And $\dfrac {11}{2(x+y)} = 1 => 2x + 2y = 11$ ---- (2)
Multiplying equation $(1)$ with $2$ we get, $2x - 2y = 1.8$ ----- equation $(3)$

Adding

equations $1$ and $2$ , we get $4x = 12.8 => x = 3.2$

Substituting

$x = 3.2$ in the equation $(1)$ , we get $3.2 - y = 0.9 => y = 2.3$

Solve the following pair of equations :

$7x + 6y = 71$

$5x - 8y = -23$

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$x = 2; y = 7$
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$x = 5; y = 6$
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$x = 4; y = 9$
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$x = 1; y = 8$

Explanation

Given equations are

$7x + 6y = 71$ --- (1)

and

$5x - 8y = -23$ --- (2)

Multiplying equation $(1)$ with $4$ we get, $28x + 24y = 284$ ----- equation $(3)$

Multiplying equation $(2)$ with $3$ we get, $15x - 24y = - 69$ ----- equation $(4)$

Adding equations $(4)$ and $(3)$ , we get $43x = 215 => x = 5$

Substituting $x = 5$ in the equation $(2)$ , we get

$5(5) - 8y = -23 => y = 6$

CBSE Questions for Class 12 Commerce Applied Mathematics Linear Equations Quiz 6 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Applied Mathematics Linear Equations Quiz 6 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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