Pair of Linear Equations in Two Variables Test

Result

Pair of Linear Equations in Two Variables Test - 26

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Solve the given pair of equations by substitution method:
$$2a\, +\, 3b\, =\, 6$$
$$3a\, +\, 5b\, =\, 15$$

A
$$a = 3, b = 2$$

B
$$a = -15, b = 12$$

C
$$a = 8, b = 15$$

D
$$a = -7, b = 3$$

Solution

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

Let us consider the equation $$2a+3b=6$$ and write it as

$$b=\frac { 6-2a }{ 3 } .........(1)$$

Substitute the value of $$b$$ in Equation $$3a+5b=15$$. We get

$$3a+5\left( \frac { 6-2a }{ 3 } \right) =15\\ \Rightarrow 9a+5\left( 6-2a \right) =45\\ \Rightarrow 9a+30-10a=45\\ \Rightarrow -a=45-30\\ \Rightarrow a=-15$$

Therefore, $$a=-15$$

Substituting this value of $$a$$ in Equation (1), we get

$$b=\frac { 6-(-30) }{ 3 }$$
$$b=\frac { 36 }{ 3 }=12$$

Hence, the solution is $$a=-15, b=12$$.
Question 2
1 / -0

Solve the given pair of equations by substitution method:
$$x\, +\, y\, =\, 0$$
$$y\, -\, x\, =\, 6$$

A
$$(2 , 7)$$

B
$$(-3 , 3)$$

C
$$(4 , 9)$$

D
$$(-6 , 2)$$

Solution

We have, $$x+y =0\Rightarrow x = -y$$ ..........(1)
and $$y-x=6$$.......(2)
Now substitute value of x in terms of y from (1) to (2)
$$\Rightarrow y-(-y)=6$$
$$\Rightarrow y+y=6$$
$$\Rightarrow 2y=6$$
$$\Rightarrow y=\dfrac{6}{2}=3$$
$$x =-y=-(3)=-3$$
Hence solution set is $$(x,y)=(-3,3)$$
Question 3
1 / -0

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?

A
$$ 2\dfrac{1 }{2 } $$ women

B
$$ 5\dfrac{1 }{3 } $$ women

C
$$ 5\dfrac{2 }{3 } $$ women

D
$$ \dfrac{3}{2 } $$ women
Question 4
1 / -0

Solve following pair of equations by equating the coefficient method:
$$2x\, -\, y\, =\, 9$$
$$3x\, -\, 7y\, =\, 19$$

A
$$x=2$$ , $$y=1$$

B
$$x=4$$ , $$y=-1$$

C
$$x=3$$ , $$y=2$$

D
$$x=7$$ , $$y=-8$$

Solution

Multiply the equation $$2x-y=9$$ by $$7$$ to make the coefficients of $$y$$ equal. Then we get the equations:

$$14x-7y=63.........(1)$$

$$3x-7y=19.........(2)$$

Subtract Equation (2) from Equation (1) to eliminate $$y$$, because the coefficients of $$y$$ are the same. So, we get

$$(14x-3x)+(-7y+7y)=63-19$$

i.e. $$11x = 44$$

i.e. $$x = 4$$

Substituting this value of $$x$$ in (2), we get

$$12-7y=19$$

i.e. $$-7y=19-12$$

i.e. $$-7y=7$$

i.e. $$y=-1$$

Hence, the solution of the equations is $$x = 4, y = -1$$.
Question 5
1 / -0

Solve the following pair of simultaneous equations:
$$\displaystyle \frac{8}{x}\, -\, \frac{9}{y}\, =\, 1;\,\frac{10}{x}\, +\, \frac{6}{y}\, =\, 7$$

A
$$x= 2;y= 3$$

B
$$x= 7;y= 5$$

C
$$x= 4;y= 6$$

D
$$x= 1;y= 7$$

Solution

Let $$\dfrac{1}{x} = a$$ and $$\dfrac{1}{y} = b$$
We have, $$8a -9b = 1$$ ...$$(1)$$
$$10a +6b = 7$$ ...$$(2)$$
Multiplying $$(1)$$ by $$10$$ and $$(2)$$ by $$8$$
We get,
$$80a - 90b = 10$$
$$80a + 48b = 56$$
Subtracting, we get $$-138b = -46$$
$$\therefore b = \dfrac{1}{3}$$.
Substituting in $$(1)$$ or $$(2)$$, we get $$a = \dfrac{1}{2}$$

So, $$x =2$$ and $$y = 3$$.
Option $$A$$
Question 6
1 / -0

Solve the following pair of simultaneous equations:
$$\displaystyle \frac{1}{x}\, +\, \frac{1}{y}\, =\, 5\,;\, \frac{1}{x}\, -\, \frac{1}{y}\, =\, 1$$

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

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

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

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

Solution

Let $$\dfrac{1}{x} = u$$ and $$\dfrac{1}{y} = v$$
So, it becomes $$u + v = 5$$ and $$u - v = 1$$
Adding the both equations, $$2u = 6$$ and $$u = 3$$.
Substituting $$u$$ in either of the equations, we get $$v = 2$$.
So, $$\dfrac{1}{x} = 3$$ and $$\dfrac{1}{y} = 2$$

$$x = \dfrac{1}{3}$$ and $$y = \dfrac{1}{2}$$.

Option $$C$$
Question 7
1 / -0

Solve the following pair of simultaneous equations:
$$4x\, -\, 3y\, =\, 8$$
$$3x\, -\, 4y\, =\, - 1$$

A
$$x=3,y=-1$$

B
$$x=5,y=4$$

C
$$x=-4,y=2$$

D
$$x=7,y=-6$$

Solution

Multiply the equation $$4x-3y=8$$ by $$3$$ and equation $$3x-4y=-1$$ by $$4$$ to make the coefficients of $$x$$ equal. Then we get the equations:

$$12x-9y=24.........(1)$$

$$12x-16y=-4.........(2)$$

Subtract Equation (2) from Equation (1) to eliminate $$x$$, because the coefficients of $$x$$ are the same. So, we get

$$(12x-12x)+(-9y+16y)=24+4$$

i.e. $$7y =28$$

i.e. $$y =4$$

Substituting this value of $$y$$ in the equation $$4x-3y=8$$, we get

$$4x-12=8$$

i.e. $$4x=20$$

i.e. $$x=5$$

Hence, the solution of the equations is $$x = 5, y =4$$.
Question 8
1 / -0

Solve the following pair of simultaneous equations:
$$\displaystyle\,3x\, +\, \frac{1}{y}\, =\, 13\, ;\, \frac{2}{y}\, -\, x\, =\, 5$$

A
$$\displaystyle \frac{2}{5}$$ and $$4$$

B
$$3$$ and $$\displaystyle \frac{1}{4}$$

C
$$7$$ and $$\displaystyle \frac{1}{7}$$

D
$$3$$ and $$\displaystyle \frac{3}{4}$$

Solution

The equation $$3x +\frac { 1 }{ y } =13$$ can be solved as:

$$3x+\frac { 1 }{ y } =13\\ \Rightarrow \frac { 3xy+1 }{ y } =13\\ \Rightarrow 3xy+1=13y\quad .........(1)$$

The equation $$\frac { 2 }{ y }-x =5$$ can be solved as:

$$\frac { 2 }{ y }-x =5\\ \Rightarrow \frac { 2-xy }{ y } =5\\ \Rightarrow 2-xy=5y\quad .........(2)$$

Multiply the equation 2 by $$3$$. Then we get:

$$-3xy+6=15y.........(3)$$

Add equations 1 and 3 to eliminate $$xy$$, because the coefficients of $$xy$$ are opposite. So, we get

$$(3xy-3xy)+(1+6)=13y+15y$$

i.e. $$28y=7$$

i.e. $$y=\frac { 1 }{ 4 }$$

Substituting this value of $$y$$ in the equation 1, we get

$$3x\left( \frac { 1 }{ 4 } \right) +1=13\left( \frac { 1 }{ 4 } \right) \\ \Rightarrow \frac { 3 }{ 4 } x+1=\frac { 13 }{ 4 } \\ \Rightarrow \frac { 3 }{ 4 } x=\frac { 13 }{ 4 } -1\\ \Rightarrow \frac { 3 }{ 4 } x=\frac { 9 }{ 4 } \\ \Rightarrow 3x=9\\ \Rightarrow x=3$$

Hence, the solution of the equations is $$x=3,y=\frac { 1 }{ 4 }$$.
Question 9
1 / -0

Solve the following pair of simultaneous equations:
$$\displaystyle \frac{6}{x}\, -\, \frac{2}{y}\, =\, 1\,;\, \frac{9}{x}\, -\, \frac{6}{y}\,=\, 0$$

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

B
$$x = 6, y = -4$$

C
$$x = =1, y = 2$$

D
$$x = 3, y = 2$$

Solution

The equation $$\cfrac { 6 }{ x } -\cfrac { 2 }{ y } =1$$ can be solved as:

$$\cfrac { 6 }{ x } -\cfrac { 2 }{ y } =1\\ \Rightarrow \cfrac { 6y-2x }{ xy } =1\\ \Rightarrow 6y-2x=xy\quad .........(1)$$

The equation $$\cfrac { 9 }{ x } -\cfrac { 6 }{ y } =0$$ can be solved as:

$$\cfrac { 9 }{ x } -\cfrac { 6 }{ y } =0\\ \Rightarrow \cfrac { 9y-6x }{ xy } =0\\ \Rightarrow 9y-6x=0\quad .........(2)$$

Multiply the equation 1 by $$3$$.Then we get:

$$18y-6x=3xy.........(3)$$

Subtract equation 2 from equation 3 to eliminate $$x$$, because the coefficients of $$x$$ are same. So, we get

$$(18y-9y)+(-6x+6x)=3xy-0$$

i.e. $$9y=3xy$$

i.e. $$9=3x$$

i.e. $$x=3$$

Substituting this value of $$x$$ in the equation 1, we get

$$6y-6=3y$$

i.e. $$3y=6$$

i.e. $$y=2$$

Hence, the solution of the equations is $$x=3 ,y=2$$.
Question 10
1 / -0

Solve the following pair of simultaneous equations:
$$\displaystyle \frac{x}{3}\, =\, \frac{y}{2}\,;\, \frac{2x}{3}\, -\, \frac{y}{2}\, =\, 2$$

A
$$(-3 , 4)$$

B
$$(1 , 9)$$

C
$$(6 , 4)$$

D
$$(3 , 8)$$

Solution

The equation $$\frac { x }{ 3 } =\frac { y }{ 2 }$$ can be rewritten as:

$$\frac { x }{ 3 } =\frac { y }{ 2 } \\ \Rightarrow 2x=3y\\ \Rightarrow 2x-3y=0\quad .........(1)$$

The equation $$\frac { 2x }{ 3 } -\frac { y }{ 2 }=2$$ can be solved as:

$$\frac { 2x }{ 3 } -\frac { y }{ 2 } =2\\ \Rightarrow \frac { 4x-3y }{ 6 } =2\\ \Rightarrow 4x-3y=12\quad .........(2)$$

Subtract Equation 2 from equation 1 to eliminate $$y$$, because the coefficients of $$y$$ are the same. So, we get

$$(4x-2x)+(-3y+3y)=12-0$$

i.e. $$2x=12$$

i.e. $$x=6$$

Substituting this value of $$x$$ in the equation 1, we get

$$12-3y=0$$

i.e. $$3y=12$$

i.e. $$y=4$$

Hence, the solution of the equations is $$x =6, y =4$$.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Pair of Linear Equations in Two Variables Test - 26

Result

Pair of Linear Equations in Two Variables Test - 26

Score

-

Rank

-

SHARING IS CARING

Question 1

$$a = 3, b = 2$$

$$a = -15, b = 12$$

$$a = 8, b = 15$$

$$a = -7, b = 3$$

Solution

Question 2

$$(2 , 7)$$

$$(-3 , 3)$$

$$(4 , 9)$$

$$(-6 , 2)$$

Solution

Question 3

$$ 2\dfrac{1 }{2 } $$ women

$$ 5\dfrac{1 }{3 } $$ women

$$ 5\dfrac{2 }{3 } $$ women

$$ \dfrac{3}{2 } $$ women

Question 4

$$x=2$$ , $$y=1$$

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

$$x=3$$ , $$y=2$$

$$x=7$$ , $$y=-8$$

Solution

Question 5

$$x= 2;y= 3$$

$$x= 7;y= 5$$

$$x= 4;y= 6$$

$$x= 1;y= 7$$

Solution

Question 6

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

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

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

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

Solution

Question 7

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

$$x=5,y=4$$

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

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

Solution

Question 8

$$\displaystyle \frac{2}{5}$$ and $$4$$

$$3$$ and $$\displaystyle \frac{1}{4}$$

$$7$$ and $$\displaystyle \frac{1}{7}$$

$$3$$ and $$\displaystyle \frac{3}{4}$$

Solution

Question 9

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

$$x = 6, y = -4$$

$$x = =1, y = 2$$

$$x = 3, y = 2$$

Solution

Question 10

$$(-3 , 4)$$

$$(1 , 9)$$

$$(6 , 4)$$

$$(3 , 8)$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.