MCQ Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 12

Solve graphically:

$5x-6y+30=0$ ;

$5x+4y-20=0$ . Also, find the vertices of the triangle formed by the above two lines and

$x$ -axis.

0%
$(3,5),(4,2),(6,9)$
0%
$(0,5),(-6,0),(4,0)$
0%
$(0,2),(7,2),(0,4)$
0%
$(2,3),(1,7),(3,9)$

Explanation

For

$5x - 6y + 30 = 0$ :

x	0	-6	6
y	5	0	10

For

$x - y - 1 = 0$ :

x	0	4	-4
y	5	0	10

As the graph shows, vertices of triangle are

$(0, 5), (-6, 0) \& (4, 0)$ .

In a zoo there are some pigeons and some rabbits. If the total number of heads is

$300$ and the total number of legs is

$750$ , then how many pigeons are there?

0%
$150$
0%
$175$
0%
$200$
0%
$225$

Explanation

Let, the total number of pigeons be

$p$ and the total number of rabbits be

$r$

Both the pigeons and rabbits has one head each

so,

$p+r=300$ ...........................................(1)

Each pigeon has two legs and each rabbit has four legs

so,

$2p+4r=750$ ................................................(2)

Multiply by (1) by

$4$

Then

$4p+ 4r=1200$ ................................................(3)

By subtraction (2) from (3)

$(4p+ 4r)-(2p+4r)=1200-750$

$\Rightarrow 4p+ 4r-2p-4r=450$

$\Rightarrow 2p=450$

$\Rightarrow p=225$

Hence, the number of pigeons in the zoo is

$225$ .

Using substitution method find the value of x and y:

$2x - 3y = 2$ and

$5x - 6y = 2$

0%
-4 and -1
0%
-2 and 10
0%
-2 and -11
0%
-2 and -2

Explanation

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

$5x- 6y = 2$ ------ (2)
From equation 1:

$2x- 2 = 3y$

$y =\dfrac{2x-2}{3}$
Substitute the value of y in equation 2:

$5x- 6y = 2$

$5x- 6(\dfrac{2x-2}{3}) = 2$

$5x- 2(2x- 2) = 2$

$5x- 4x + 4 = 2$

$x = 2- 4$

$x = -2$
Now, Substitute

$x = -2$ in equation 1:

$5x- 6y = 2$

$5(-2)- 6y = 2$

$-10 -2 = 6y$

$6y = -12$

$y = \dfrac{-12}{6} = -2$
Therefore the solution is:

$x = -2$ and

$y = -2$

The total salary of

$15$ men and

$8$ women is

$Rs.\ 3050$ . The difference of salaries of

$5$ women and

$3$ men is

$Rs.\ 50$ . Find the sum of the salaries of

$3$ men and

$3$ women

0%
$Rs.\ 900$
0%
$Rs.\ 850$
0%
$Rs.\ 750$
0%
$Rs.\ 1000$

Explanation

Let the salary of

$1$ men be

$x$ and the salary of

$1$ woman be

$y.$

As per the question,

$15x + 8y =3050\quad\quad...(1)$

$3x-5y=-50\quad\quad...(2)$

Multiply equation

$(2)$ by

$5$ ,

$15x -25y = -250\quad\quad...(3)$

Subtract equation

$(3)$ from equation

$(1),$

$\left( {15x + 8y} \right) - \left( {15x - 25y} \right) = 3050 - \left( { - 250} \right)$

$\Rightarrow 15x + 8y - 15x + 25y = 3050 + 250$

$\Rightarrow 33y = 3300$

$\Rightarrow y= 100$

Substitute the value of

$y$ in equation

$(2),$

$3x-500 =-50$

$\Rightarrow 3x= 450$

$\Rightarrow x= 150$

Now, total salary of

$3$ men and

$3$ women will be,

$3x + 3y= 3 \times 150 + 3 \times 100$

$=450 + 300$

$=Rs.\ 750$

A students walks from his house at less 4 km per hour and reaches his school late by 5 minutes. If his speed has been increase by 5 km per hour then he would have reached 10 minutes early. The distance of the school from his house is

0%
$\displaystyle \frac{3}{5}km$
0%
$5$ km
0%
$6$ km
0%
$4$ km

Explanation

Let the actual time be 't'
and distance be x
Now according to first condition, we have

$\dfrac{x}{4}=t+\dfrac{5}{60}$
And according to Second condition, we have

$\dfrac{x}{5}=t-\dfrac{10}{60}$

$60x=4t+20$

$60x=5t-50$
Subtracting equation 1 from 2, we have

$0=t-70$

$\therefore t=70$
Putting the value of t we get

$x=5km$

Solve for

$x$ and

$y$ , if

$2x+ 3y= 8$ and

$x+2y=5$ .

0%
$x=1, y=2$
0%
$x=-1, y=-2$
0%
$x=-2, y=2$
0%
$x=-1, y=2$

Explanation

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

$x+2y=5$ .......(2)
From (1), we get

$x=\frac {8-3y}{2}$ .......... (3)
From (2), we get

$x=5-2y$ .......... (4)

$\therefore \frac {8x-3y}{2}=5-2y$
or

$8x -3y= 10 -4y$ or

$3y= 10 -4y$ or

$y= 2$
Using

$y = 2$ in (4), we get

$x = 5 -4 = 1$ . Therefore,

$x=1,y=2$

The sum of the digits of a two-digit number is

$5$ . The digit obtained by increasing the digit in tens' place by unity is one-eighth of the number. Then the number is

0%
less than $30$
0%
lies between $30$ and $40$
0%
more than $37$
0%
lies between $40$ and $50$

Explanation

$x+y=5$
and

$x+1=\dfrac {1}{8}(10x+y)$ .........(1)
or

$8x+8=10x+y$
or

$2x+y=8$ .........(2)
Subtract equation (1) from equation (2)

$2x+y=8$

$+ x+y=5$

$\underline {(-) (-) (-)}$

$x=3$

$\therefore y=2$
Number is

$32$ .

10 years ago the age of the father was 5 times that of the son. 20 years hence the age of the father will be twice that of the son. The present age of the father (in years) is

0%
$40$
0%
$45$
0%
$60$
0%
$70$

Explanation

Let the present age of father be

$x$ years. and that of son be

$y$ years
10 years ago the age of the father was 5 times that of the son.

$\therefore$

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

$\therefore$

$x-5y=-40$ ...(1)
20 years hence, the age of the father will be twice that of the son.

$\therefore$

$x+20=2(y+20)$

$\therefore$

$x-2y=20$ ...(2)
Subtracting (1) and (2) we get ,

$y=20$ and

$x=60$

$\therefore$ The present age of father is

$60$ years

The graphs of

$2x+3y-6=0, 4x-3y-6=0, x=2$ and

$\displaystyle y=\dfrac{2}{3}$ intersect in

0%
four points
0%
one point
0%
in no points
0%
infinite number of points

Explanation

Solving

$2x + 3y - 6 = 0$ and

$4x - 3y - 6 = 0$ we get

$x = 2$ and

$\displaystyle y=\dfrac{2}{3}$
Hence from all the four given equations the
value of

$x$ is

$2$ and

$y$ is

$\displaystyle \frac{2}{3}$
Therefore the graphs of all the four lines intersect
in one point
1292219_375307_ans_17c9bbe6198c4cf78621191bb0273b82.jpg

1292219_375307_ans_17c9bbe6198c4cf78621191bb0273b82.jpg

What values of x and y satisfies the equations

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

$\displaystyle \frac{x}{3}-\frac{y}{12}=4\frac{3}{4}$

0%
$(6,15)$
0%
$(18,15)$
0%
$(15,18)$
0%
$(12,30)$

Explanation

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

Let us consider the equation

$\frac { x }{ 3 } -\frac { y }{ 12 } =4\frac { 3 }{ 4 }$ or

$4x-y=57$ and write it as

$y=4x-57...........(1)$

Substitute the value of

$y$ in Equation

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

$15x+6y=360$ . We get

$15x+6(4x-57)=360$

i.e.

$15x+24x-342=360$

i.e.

$39x=702$

i.e.

$x=18$

Therefore,

$x=18$

Substituting this value of

$x$ in equation 1, we get

$y=72-57$

i.e.

$y=15$

Hence, the solution is

$x = 18, y =15$ .

The ratio of the ages of A and B ten years before was

$3 : 5$ . The ratio of their present ages is

$2 : 3$ . Their ages are respectively

0%
$30,50$
0%
$20,30$
0%
$40,60$
0%
$16,24$

Explanation

Let the present ages of

$A$ and

$B$ are

$x$ and

$y$ years respectively

$\displaystyle \therefore \dfrac{x-10}{y-10}=\dfrac{3}{5}$ .....(i)
or

$5x -50=3y-30$ ......(ii)
and

$\displaystyle \dfrac{x}{y}=\dfrac{2}{3}$ or

$3x=2y$ or

$y=\dfrac{3x}{2}$
putting the value of y in equation (i) we get

$5x-50=$

$\displaystyle \dfrac{9x}{2}-30$

$10x-100=9x-60$
or

$x=40$

$\displaystyle \therefore y=\dfrac{3\times 40}{2}=60$

Solve the equations using graphical method:

$x + y = 1$ and

$2x -3y = 7$

0%
(2, -1)
0%
(-2, -1)
0%
(-2, 1)
0%
(2, 1)

Explanation

$x + y = 1$ ------- (I)

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

Find the value of x and y using substitution method:

$5x - 6y = 2$ and

$6x - 5y = 9$

0%
$4$ and $-3$
0%
$-3$ and $-4$
0%
$-4$ and $-3$
0%
$4$ and $3$

Explanation

$5x -6y = 2$ -------

$(1)$

$6x - 5y = 9$ -------

$(2)$

From equation

$(1)$ :

$\Rightarrow 6y = 5x- 2$

$\Rightarrow y =\dfrac{5x-2}{6}$

Substitute the value of

$y$ in equation

$(2)$ :

$\Rightarrow 6x - 5y = 9$

$\Rightarrow6x- 5(\dfrac{5x-2}{6}) = 9$

$\Rightarrow36x -25x + 10 = 54$

$\Rightarrow11x = 54- 10$

$\Rightarrow x = \dfrac{44}{11} = 4$

Now, Substitute

$x = 4$ in equation

$(1)$

$\Rightarrow5x - 6y = 2$

$\Rightarrow5(4)-6y = 2$

$\Rightarrow 20 - 2 = 6y$

$\Rightarrow18 = 6y$

$\Rightarrow y =\dfrac{18}{6}=3$

$\therefore$ the solution is

$x = 4$ and

$y = 3$

Find the value of

$x$ and

$y$ using substitution method:

$x- y = 2$ and

$2x - y = 9$

0%
$7$ and $7$
0%
$7$ and $5$
0%
$-7$ and $-5$
0%
$5$ and $7$

Explanation

$x- y = 2$ -------

$(1)$

$2x - y = 9$ ------

$(2)$
From equation

$(1)$ :

$y = x - 2$
Substitute the value of

$y$ in equation

$(2)$ :

$2x - y = 9$

$2x -x + 2 = 9$

$x = 9 -2$

$x = 7$
Now, Substitute

$x = 7$ in equation

$(1)$ :

$x- y = 2$

$7- y = 2$

$y = 7 -2$

$y = 5$
Therefore the solution is:

$x = 7$ and

$y = 5$

Solve equations using substitution method:

$x-y = 3$ and

$x + y = 0$

0%
$\dfrac{3}{2}$ and $1$
0%
$\dfrac{-3}{2}$ and $-1$
0%
$\dfrac{3}{2}$ and $-\dfrac{3}{2}$
0%
$0$ and $\dfrac{3}{2}$

Explanation

$x- y = 3\dots (1)$

$x + y = 0\dots (2)$
From equation

$(1)$ :

$y = x - 3$
Substitute the value of

$y$ in equation

$(2)$ :

$x + y = 0$

$x + x- 3 = 0$

$2x = 3$

$x=\dfrac{3}{2}$

Now, Substitute

$x = \dfrac{3}{2}$ in equation

$(1)$ :

$x- y = 3$

$\dfrac{3}{2}- y = 3$

$y = \dfrac{3}{2}- 3$

$y = -\dfrac{3}{2}$
Therefore the solution is:

$x =\dfrac{3}{2}$ and

$y = -\dfrac{3}{2}$

Solve equations using substitution method:

$2x + y = 5$ and

$2x- 3y = 3$

0%
$\frac{1}{4}$ and $\frac{-1}{2}$
0%
$\frac{1}{4}$ and $\frac{1}{2}$
0%
$\frac{-9}{4}$ and $\frac{-1}{2}$
0%
$\frac{9}{4}$ and $\frac{1}{2}$

Explanation

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

$2x- 3y = 3$ ------ (2)
From equation 1 :

$y = 5 -2x$
Substitute the value of y in equation 2 :

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

$2x -15 + 6x = 3$

$8x = 3 + 15$

$x = \dfrac{18}{8} = \dfrac{9}{4}$
Now, Substitute

$x = \dfrac{9}{4}$ in equation 1 :

$2(\dfrac{9}{4}) + y = 5$

$18 + 4y = 20$

$4y = 20 -18$

$y = \dfrac{2}{4}$

$y = \dfrac{1}{2}$
Therefore the solution is:

$x = \dfrac{9}{4}$ and

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

Solve equations using substitution method:

$2x -y = 3$ and

$4x + y = 3$

0%
$-1$ and $1$
0%
$-2$ and $1$
0%
$1$ and $-1$
0%
$-1$ and $-1$

Explanation

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

$4x + y = 3$ ------ (2)
From equation 1 :

$y = 2x - 3$
Substitute the value of

$y$ in equation 2 :

$4x + y = 3$

$4x + 2x -3 = 3$

$6x = 3 + 3$

$x = \dfrac{6}{6} = 1$
Now, Substitute

$x = 1$ in equation 1 :

$2x -y = 3$

$2 (1)- y = 3$

$2- 3 = y$

$y = -1$
Therefore the solution is:

$x = 1$ and

$y = -1$

$2$ tables and

$3$ chairs cost

$Rs. 4500$ and

$3$ tables and

$2$ chairs cost

$Rs. 5000$ , then how much does a table cost?

0%
Rs. $1100$
0%
Rs. $1200$
0%
Rs. $1150$
0%
Rs. $1300$

Explanation

Let the cost of table and chair is

$x$ and

$y$ .

Then,

$2x+3y=4500$ ....(1)

and

$3x+2y=5000$ ...(2)

Add eq(1) and eq(2)

$5x+5y=9500$

$x+y = 1900$ ....(3)

Subtract eq(1) from eq(2)

$x-y= 500$ ....(4)

Solving eq(3) and eq(4) we get

$2x=2400$

$x =1200$

Father says to son "I am

$5$ times as old as you were when i was as old as you are". If sum of their ages is

$48$ , what is the age of the father?

0%
$20$
0%
$9$
0%
$45$
0%
$40$

Explanation

Let the present age of father and son is x and y

Then as per question

$x+y=48$ ........................(1)

Given

$x=5y$ .......................................................(2)

substituting equation 2 in equation 1 we get

$5y+y=48$

$6y=48$

$y=8$ .........................................................................(3)

Putting value of y in eqn 2 we get

$x=5(8)=40$

We get

$x=40$

Then age of father is

$40$ years

Graph the equation by determining the missing values needed to plot the ordered pairs.

$y+x=4;$

$\left( 1, y_1\right) ,\left( 4, y_2 \right) ,\left( 3, y_3\right)$

Explanation

We have given line,

$y+x=4\Rightarrow y=4-x$
Now, at

$x=1, y=4-1=3$
At

$x=4, y=4-4=0$ and at

$x=3, y=4-3=1$
Hence, given points on the line are

$(1,3),(4,0),(3,1)$ .
Clearly these points lying on the graph of line in option B.
Hence option B is correct choice.

Use the method of substitution to solve the equations

$x+2y=-4$ and

$4x+5y=2$

0%
$\dfrac{-16}{3}$ and $\dfrac{-2}{3}$
0%
$\dfrac{-16}{3}$ and $\dfrac{2}{3}$
0%
$8$ and $-6$
0%
$\dfrac{16}{3}$ and $\dfrac{2}{3}$

Explanation

$x+2y=-4$

$\implies x=-4-2y$

Substituting

$x=-4-2y$ in

$4x+5y=2$

$\implies 4(-4-2y)+5y=2$

$\implies -16-8y+5y=2$

$\implies -16-3y=2$

$\implies -16-2=3y$

$\implies y=\dfrac{-18}{3}=-6$

Substituting

$y=-6$ in

$x=-4-2y$ , we get,

$x=-4-2(-6)$

$=-4+12=8$

$\therefore x=8, y=-6$

Solve the equation using substitution method:

$3x-5y =11$ and

$+10y+6x=7$

0%
$\left(\frac{29}{12}, \frac{3}{4}\right)$
0%
$\left(\frac{3}{4}, \frac{29}{12}\right)$
0%
$\left(\frac{-29}{12}, \frac{3}{4}\right)$
0%
$\left(\frac{29}{12}, \frac{-3}{4}\right)$

Explanation

$3x-5y =11$ ---------- (1)

$+10y+6x=7$ ---------(2)
From equation (1)

$3x-5y=11$

$\frac{3x-11}{5}=y$
Substitute the value of

$y$ in equation (2)

$+10\left(\frac{3x-11}{5}\right)+6x = 7$

$+6x-22+6x = 7$

$12x=7+22$

$x=\frac{29}{12}$
Substitute the value of

$x$ in equation (1)

$3x-5y = 11$

$3\left(\frac{29}{12}\right) - 5y = 11$

$29-20y = 44$

$29-44=20y$

$\frac{-153}{20} = y$

$y=\frac{-3}{4}$
therefore solution is

$x=\frac{29}{12}$ and

$y=\frac{-3}{4}$

Using substitution method find the value of

$x$ and

$y:$

$x + 4y = -4$ and

$2x- 3y = 2$

0%
$\dfrac{-4}{11}$ and $\dfrac{-10}{11}$
0%
$\dfrac{-2}{11}$ and $\dfrac{10}{11}$
0%
$\dfrac{-2}{11}$ and $\dfrac{-10}{11}$
0%
$\dfrac{2}{11}$ and $\dfrac{-2}{11}$

Explanation

$x + 4y = -4$ -------

$(1)$

$2x- 3y = 2$ ------

$(2)$
From equation

$1:$

$4y = -4 -x$

$y =\dfrac{-4-x}{4}$
Substitute the value of

$y$ in equation

$2:$

$2x- 3y = 2$

$2x- 3\left(\dfrac{-4-x}{4}\right) = 2$

$8x + 12 + 3x = 8$

$11x = 8-12$

$11x = -4$

$x = \dfrac{-4}{11}$
Now, Substitute

$x =\dfrac{-4}{11}$ in equation

$1:$

$x + 4y = -4$

$\dfrac{-4}{11} + 4y = -4$

$-4 + 44y = -44$

$44y = -44 + 4$

$44y = -40$

$y = \dfrac{-40}{44} \\= \dfrac{-10}{11}$
Therefore the solution is:

$x = \dfrac{-4}{11}$ and

$y = \dfrac{-10}{11}$

Using substitution method find the value of x and y:

$3x+5y$

$= 1$ and

$4x + 5y$

$= 2$

0%
$-1$ and $\dfrac{-2}{5}$
0%
$-1$ and $\dfrac{2}{5}$
0%
$1$ and $\dfrac{-2}{5}$
0%
$1$ and $\dfrac{2}{5}$

Explanation

$3x+5y=1$

$\implies 3x=1-5y$

$\implies x=\dfrac{1-5y}{3}$

Now, Substituting

$x=\dfrac{1-5y}{3}$ in

$4x+5y=2$

$\implies 4(\dfrac{1-5y}{3})+5y=2$

$\implies \dfrac{4-20y+15y}{3}=2$

$\implies 4-5y=6$

$\implies -2=5y$

$\implies y=\dfrac{-2}{5}$

Substituting

$y=\dfrac{-2}{5}$ in

$x=\dfrac{1-5y}{3}$

$x=\dfrac{1-5(\dfrac{-2}{5})}{3}$

$x=\dfrac{1+2}{3}=1$

$\therefore x=1, y=\dfrac{-2}{5}$

Using substitution method find the value of x and y:

$4x + 9y = 5$ and

$-5x + 3y = 8$

0%
$-1$ and $\dfrac{-2}{5}$
0%
$-1$ and $\dfrac{2}{5}$
0%
$-1$ and $1$
0%
$1$ and $\dfrac{2}{5}$

Explanation

$4x+9y=5$

$\implies 4x=5-9y$

$\implies x=\dfrac{5-9y}{4}$

Substituting

$x=\dfrac{5-9y}{4}$ in

$-5x+3y=8$ , we get,

$-5(\dfrac{5-9y}{4})+3y=8$

$\implies \dfrac{-25+45y+12y}{4}=8$

$\implies -25+45y+12y=8\times 4$

$\implies -25+57y=32$

$\implies 57y=57$

$\implies y=1$

Substituting

$y=1$ in

$x=\dfrac{5-9y}{4}$

$x=\dfrac{5-9(1)}{4}$

$\implies x=\dfrac{-4}{4}$

$\implies x=-1$

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

Solve the equation using substitution method:

$3x+y+1=0$ and

$3x-4y=10$

0%
(0.4, 2.2)
0%
(0.4, -2.2)
0%
(0.4, 1.2)
0%
(-0.4, 2.2)

Explanation

$3x+y+1=0$ ------- (1)

$3x-4y=10$ --------(2)
From equation (1)

$3x+y=-1$

$y=-1-3x$
Substitute the value of

$y$ in equation (2)

$3x-4(-1-3x)=10$

$3x+4+12x=10$

$15x=10-4$

$x=\dfrac{6}{15}$ or

$\dfrac{2}{5}\simeq 0.4$
The substitute the value of x in equation (1)

$3x+y=-1$

$3(\dfrac{2}{5})+y=-1$

$6+5y=-5$

$5y=-5-6$

$5y=-11$

$y=\dfrac{-11}{5}$ or -2.2
therefore solution is

$x= 0.4$ and

$y=-2.2$

Using substitution method find the value of x and y:

$x - 9y = 18$ and

$-x + 3y = -15$

0%
$-13.5$ and $-3.5$
0%
$-13.5$ and $2.5$
0%
$13.5$ and $-2.5$
0%
$13.5$ and $-0.5$

Explanation

$x-9y=18$

$\implies x=18+9y$

Substituting

$x=18+9y$ in

$-x+3y=-15$ , we get,

$-(18+9y)+3y=-15$

$\implies -18-9y+3y=-15$

$\implies -6y=-15+18$

$\implies -6y=3$

$\implies y=-0.5$

Substituting

$y=-0.5$ in

$x=18+9y$

$x=18+9(-0.5)$

$\implies x=18-4.5$

$\implies x=13.5$

$\therefore x=13.5, y=-0.5$

Solve the equations using the method of substitution method :

$3x+4y=-43$ and

$-2x+3y=11$

0%
7 and 5
0%
5 and -7
0%
-5 and 7
0%
5 and 7

Explanation

$3x+4y=-43$ ------- (1)

$-2x+3y=11$ ------ (2)
From equation 1:

$3x+4y=-43$

$4y=43-3x$

$y=\frac{43-3x}{4}$
Substitute the value of

$y$ in equation 2:

$-2x+3(\frac{43-3x}{4})=11$

$-8x + 129-9x= 44$

$-17x = 44-129$

$x =\frac{-85}{-17}=5$
The substitute the value of

$x$ in equation 1:

$3x+4y=43$

$3(5)+4y=43$

$15+4y=43$

$4y=43-15$

$4y=28$

$y=\frac{28}{4}$

$y=7$
Therefore the solution is:

$x = 5$ and

$y=7$

Find the value of a and b using substitution method:

$a + b = 3$ and

$-3a + 2b = 1$

0%
1 and 2
0%
2 and 1
0%
-2 and -1
0%
2 and -1

Explanation

$a + b = 3$ ------- (1)

$-3a + 2b = 1$ ------ (2)
From equation 1:

$a = 3 - b$
Substitute the value of a in equation 2 :

$-3a + 2b = 1$

$-3(3- b) + 2b = 1$

$-9 + 3b + 2b = 1$

$5b = 10$

$b = \dfrac{10}{5} = 2$
Now, Substitute

$b = 2$ in equation 1:

$a + b = 3$

$a + 2 = 3$

$a = 3- 2$

$a = 1$
Therefore the solution is:

$a = 1$ and

$b = 2$

Solve the equations using graphical method:

$2x- y = 2$ and

$-4x- y = -4$

0%
(1, -1)
0%
(1, 0)
0%
(2, -1)
0%
(0, 1)

Explanation

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

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

$y = 2x - 2$ ------ (3)
Assume the value of x = 1, 2 and put those values in equation (3)
If

$x = 1, y = 2(1)- 2 = 0$
If

$x = 2, y = 2(2) - 2=2$
Now plotting (1, 0), (2, 2) and joining them, we get a straight line.
From equation (2),

$-4x- y = -4$

$y = -4x + 4$ ------ (4)
Assume the value of

$x = 1, 2$ and put those values in equation (3)
If

$x = 1, y =-4(1)+4= 0$
If

$x = 2, y = -4(1)+4 = -2$
Plotting (1, 0), (2, -2) and joining them, we get another straight line.
These lines intersect at the point (1, 0) and therefore the solution of the equation is

$x = 1, y = 0$

CBSE Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 12 - MCQExams.com

0:0:1

Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 12 - MCQExams.com

0:0:1

Practice Class 10 Maths Quiz Questions and Answers

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