Pair of Linear Equations in Two Variables Test

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Pair of Linear Equations in Two Variables Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

Solve the following pair of equations :
$$y = 2x - 6$$
$$y = 0$$

A
$$x =3$$

B
$$x =4$$

C
$$x =-6$$

D
$$x =-4$$

Solution

Given equations are $$y=2x-6$$ ....(1)
and $$y=0$$ ....(2)
Put the value of equation (2) in equation (1), we get
$$0=2x-6$$
$$\Rightarrow 2x=6$$
$$\Rightarrow x=3$$
Hence, option A is the correct answer.
Question 2
1 / -0

Solve the following pair of equations:
$$3\, -\, (x\, -\, 5)\, =\, y\, +\, 2$$
$$2\, (x\, +\, y)\, =\, 4\, -\, 3y$$

A
$$x\, =\, \displaystyle \frac{16}{7}\, ;\, y\, =\, \displaystyle -\frac{4}{5}$$

B
$$x\, =-\, \displaystyle \frac{4}{5}\, ;\, y\, =\, \displaystyle -\frac{17}{8}$$

C
$$x\, =\, \displaystyle \frac{18}{5}\, ;\, y\, =\, \displaystyle -\frac{4}{5}$$

D
$$x\, =\, \displaystyle \frac{26}{3}\, ;\, y\, =\, \displaystyle -\frac{8}{3}$$

Solution

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

Let us consider the equation $$3-(x-5)=y+2$$ and write it as

$$3-x+5=y+2$$
or $$-x-y=-6$$
or $$x+y=6$$
or $$y=6-x...........(1)$$

Now the equation $$2(x+y)=4-3y$$ can be rewritten as:

$$2x+2y=4-3y$$
or $$2x+5y=4$$

Substitute the value of $$y$$ in the equation $$2x+5y=4$$. We get

$$2x+5(6-x)=4$$

i.e. $$2x+30-5x=4$$

i.e. $$-3x=-26$$

i.e. $$x=\dfrac { 26 }{ 3 }$$

Therefore, $$x=\dfrac { 26 }{ 3 }$$

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

$$y=6-\dfrac { 26 }{ 3 } =\dfrac { 18-26 }{ 3 } =-\dfrac { 8 }{ 3 }$$

Hence, the solution is $$x=\dfrac { 26 }{ 3 } ,y=-\dfrac { 8 }{ 3 }$$.
Question 3
1 / -0

Solve : $$\displaystyle \frac{3}{x+y}+\displaystyle \frac{2}{x-y}= 2$$ and $$\displaystyle \frac{9}{x+y}-\displaystyle \frac{4}{x-y}= 1$$

A
$$x= \displaystyle \frac{2}{5}; y= \displaystyle \frac{7}{2}$$

B
$$x= \displaystyle \frac{5}{2}; y= \displaystyle \frac{1}{2}$$

C
$$x= \displaystyle \frac{6}{7}; y= \displaystyle \frac{4}{3}$$

D
$$x= \displaystyle \frac{7}{3}; y= \displaystyle \frac{1}{2}$$

Solution

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

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

$$\therefore x=\dfrac52, y=\dfrac12$$
Question 4
1 / -0

Solve for $$x$$ and $$y$$:
$$mx \, -\, ny\, =\, m^{2}\, +\, n^{2}$$, $$x\, -\, y\, =\, 2n$$

A
$$x\, =\, m\, +\, n\,;\, y\, =\, m\, -\, n$$

B
$$x\, =\, m\, -\, n\, ;\, y\, =\, mn\, -\, n$$

C
$$x\, =\, m\, +\, mn\, ;\, y\, =\, m\, +\, n$$

D
$$x\, =\, mn\, -\, n\, ;\, y\, =\, m\, -\, n$$

Solution

The equations given are:

$$mx-ny=m^2+n^2...........(1)$$

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

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

Let us consider the equation 2 and write it as

$$x=2n+y$$

Substitute the value of $$x$$ in equation 1. We get

$$m\left( 2n+y \right) -ny=m^{ 2 }+n^{ 2 }\\ \Rightarrow 2mn+my-ny=m^{ 2 }+n^{ 2 }\\ \Rightarrow y\left( m-n \right) =m^{ 2 }+n^{ 2 }-2mn\\ \Rightarrow y\left( m-n \right) =\left( m-n \right) ^{ 2 }\\ \Rightarrow y=m-n$$

Therefore, $$y = m-n$$

Substituting this value of $$y$$ in the equation $$x=2n+y$$, we get

$$x=2n+y=2n+m-n=m+n$$

Hence, the solution is $$x = m+n, y = m-n$$.
Question 5
1 / -0

Solve the following pair of equations:
$$2x\, -\, 3y\, -\, 3\, =\, 0$$
$$\displaystyle \frac{2x}{3}\, +\, 4y\, +\, \displaystyle \frac{1}{2}\, =\, 0$$

A
$$x\, =\, \displaystyle \frac{1}{2}\, ;\, y\, =\, \displaystyle -\frac{7}{6}$$

B
$$x\, =\, \displaystyle \frac{21}{20}\, ;\, y\, =\, \displaystyle -\frac{3}{10}$$

C
$$x\, =\, \displaystyle \frac{14}{5}\, ;\, y\, =\, \displaystyle -\frac{6}{11}$$

D
$$x\, =\, \displaystyle \frac{18}{19}\, ;\, y\, =\, \displaystyle -\frac{16}{10}$$

Solution

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

Let us consider the equation $$2x-3y-3=0$$ and write it as

$$y=\cfrac { 2x-3 }{ 3 }.......(1)$$

Substitute the value of $$y$$ in the equation $$\cfrac { 2x }{ 3 } +4y+\cfrac { 1 }{ 2 } =0$$. We get

$$\cfrac { 2x }{ 3 } +4y+\cfrac { 1 }{ 2 } =0\\ \Rightarrow \cfrac { 2x }{ 3 } +4\left( \cfrac { 2x-3 }{ 3 } \right) +\dfrac { 1 }{ 2 } =0\\ \Rightarrow \cfrac { 2x }{ 3 } +\cfrac { 8x-12 }{ 3 } +\cfrac { 1 }{ 2 } =0\\ \Rightarrow \cfrac { 2x+8x-12 }{ 3 } +\dfrac { 1 }{ 2 } =0\\ \Rightarrow \cfrac { 10x-12 }{ 3 } +\dfrac { 1 }{ 2 } =0\\ \Rightarrow \cfrac { 2(10x-12)+3 }{ 6 } =0\\ \Rightarrow \cfrac { 20x-24+3 }{ 6 } =0\\ \Rightarrow 20x=21\\ \Rightarrow x=\cfrac { 21 }{ 20 }$$

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

$$y=\dfrac { 2\left( \dfrac { 21 }{ 20 } \right) -3 }{ 3 } =\cfrac { 42-60 }{ 60 } =-\cfrac { 18 }{ 60 } =-\cfrac { 3 }{ 10 }$$

Hence, the solution is $$x=\cfrac { 21 }{ 20 } ,y=-\cfrac { 3 }{ 10 }$$.
Question 6
1 / -0

Solve, graphically, the following pairs of equations:
$$2x + y = 23$$
$$4x - y = 19$$

A
$$x = 7, y = 7$$

B
$$x = 7, y = 3$$

C
$$x = 7, y = 6$$

D
$$x = 7, y = 9$$

Solution
Question 7
1 / -0

Solve the following pair of equations:
$$\displaystyle \frac{a}{x}\, -\, \displaystyle \frac{b}{y}\, =\, 0$$
$$\displaystyle \frac{ab^{2}}{x}\, +\, \displaystyle \frac{a^{2}b}{y}\, =\, a^{2} \, +\, b^{2}$$

A
$$x\, =\, ab$$ and $$y\, =\, b$$

B
$$x\, =\, a$$ and $$y\, =\, b$$

C
$$x\, =\, b$$ and $$y\, =\, a$$

D
$$x\, =\, b-a$$ and $$y\, =\, ab$$

Solution

Given equations are
$$\dfrac {a}{x}-\dfrac {b}{y}=0$$ ....(1)
and $$\dfrac {ab^2}{x}+\dfrac {a^2b}{y}=a^2+b^2$$ ....(2)
Multiply equation (1) by $$a^2$$, we get
$$\dfrac {a^3}{x}-\dfrac {a^2b}{y}=0$$ ....(3)
Add equations (2) and (3),
$$\dfrac {1}{x}\left (ab^2+a^3\right)=a^2+b^2$$
$$\Rightarrow \dfrac {1}{x}[a(a^2+b^2)]=a^2+b^2$$
$$\Rightarrow x=a$$
Substitute this value in equation (1), we get
$$\dfrac {a}{a}-\dfrac {b}{y}=0$$
$$\Rightarrow 1-\dfrac {b}{y}=0$$
$$\Rightarrow y=b$$
Therefore, solution is $$x=a, y=b$$.
Question 8
1 / -0

Solve :
$$x\, +\, y\, =\, 2xy$$
$$x\, -\, y\, =\, 6xy$$

A
$$x\, =\, -\displaystyle \frac{1}{2}$$ and $$y\, =\, \displaystyle \frac{1}{7}$$

B
$$x\, =\, -\displaystyle \frac{1}{2}$$ and $$y\, =\, \displaystyle \frac{1}{4}$$

C
$$x\, =\, -\displaystyle \frac{1}{2}$$ and $$y\, =\, \displaystyle \frac{1}{2}$$

D
$$x\, =\, -\displaystyle \frac{1}{2}$$ and $$y\, =\, \displaystyle \frac{1}{5}$$

Solution

Let $$ x + y = 2xy $$ --- $$(1)$$
$$ x - y = 6xy $$ ----- $$(2)$$
Adding equations $$(1)$$ and $$(2)$$, we get
$$ 2x = 8xy$$
$$ \Rightarrow y = \dfrac {1}{4} $$
Substituting $$ y = \dfrac {1}{4} $$ in the equation $$ (1) $$, we get
$$ x + \dfrac {1}{4} = 2(x)( \dfrac {1}{4}) $$
$$\therefore x = -\dfrac {1}{2} $$
Question 9
1 / -0

Solve graphically, the following pairs of equations:
$$\displaystyle \frac{x + 1}{4} = \frac{2}{3}(1 - 2y)$$

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

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

B
$$x = 7, y = -1$$

C
$$x = 5, y = -1$$

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

Solution

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

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

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

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

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

$$\implies x=\dfrac{5-16y}{3}$$
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 $$(7,-1)$$
Question 10
1 / -0

Solve graphically, the following pairs of equations:
$$3x + 7y = 27$$
$$8 - y = \displaystyle \frac{5}{2}x$$

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

B
$$x = 7, y = 2$$

C
$$x = 4, y = 1$$

D
$$x = 4, y = 6$$

Solution

$$3x+7y=27$$

$$\implies 3x=27-7y$$

$$\implies x=\dfrac{27-7y}{3}$$

Let, $$y=3$$, then $$x=\dfrac{27-7(3)}{3}=2$$

Similarly, when $$y=6, x=-5$$

$$\therefore$$ the coordinates are $$(2,3),(-5,6)$$

$$8-y=\dfrac{5}{2}x$$

$$\implies 8-\dfrac{5}{2}x=y$$

Let, $$x=2$$, then $$y=8-\dfrac{5}{2}\times 2=3$$

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

$$\therefore$$ the coordinates are $$(2,3),(4,-2)$$

$$\therefore x=2,y=3$$

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Pair of Linear Equations in Two Variables Test - 34

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Pair of Linear Equations in Two Variables Test - 34

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SHARING IS CARING

Question 1

$$x =3$$

$$x =4$$

$$x =-6$$

$$x =-4$$

Solution

Question 2

$$x\, =\, \displaystyle \frac{16}{7}\, ;\, y\, =\, \displaystyle -\frac{4}{5}$$

$$x\, =-\, \displaystyle \frac{4}{5}\, ;\, y\, =\, \displaystyle -\frac{17}{8}$$

$$x\, =\, \displaystyle \frac{18}{5}\, ;\, y\, =\, \displaystyle -\frac{4}{5}$$

$$x\, =\, \displaystyle \frac{26}{3}\, ;\, y\, =\, \displaystyle -\frac{8}{3}$$

Solution

Question 3

$$x= \displaystyle \frac{2}{5}; y= \displaystyle \frac{7}{2}$$

$$x= \displaystyle \frac{5}{2}; y= \displaystyle \frac{1}{2}$$

$$x= \displaystyle \frac{6}{7}; y= \displaystyle \frac{4}{3}$$

$$x= \displaystyle \frac{7}{3}; y= \displaystyle \frac{1}{2}$$

Solution

Question 4

$$x\, =\, m\, +\, n\,;\, y\, =\, m\, -\, n$$

$$x\, =\, m\, -\, n\, ;\, y\, =\, mn\, -\, n$$

$$x\, =\, m\, +\, mn\, ;\, y\, =\, m\, +\, n$$

$$x\, =\, mn\, -\, n\, ;\, y\, =\, m\, -\, n$$

Solution

Question 5

$$x\, =\, \displaystyle \frac{1}{2}\, ;\, y\, =\, \displaystyle -\frac{7}{6}$$

$$x\, =\, \displaystyle \frac{21}{20}\, ;\, y\, =\, \displaystyle -\frac{3}{10}$$

$$x\, =\, \displaystyle \frac{14}{5}\, ;\, y\, =\, \displaystyle -\frac{6}{11}$$

$$x\, =\, \displaystyle \frac{18}{19}\, ;\, y\, =\, \displaystyle -\frac{16}{10}$$

Solution

Question 6

$$x = 7, y = 7$$

$$x = 7, y = 3$$

$$x = 7, y = 6$$

$$x = 7, y = 9$$

Solution

Question 7

$$x\, =\, ab$$ and $$y\, =\, b$$

$$x\, =\, a$$ and $$y\, =\, b$$

$$x\, =\, b$$ and $$y\, =\, a$$

$$x\, =\, b-a$$ and $$y\, =\, ab$$

Solution

Question 8

$$x\, =\, -\displaystyle \frac{1}{2}$$ and $$y\, =\, \displaystyle \frac{1}{7}$$

$$x\, =\, -\displaystyle \frac{1}{2}$$ and $$y\, =\, \displaystyle \frac{1}{4}$$

$$x\, =\, -\displaystyle \frac{1}{2}$$ and $$y\, =\, \displaystyle \frac{1}{2}$$

$$x\, =\, -\displaystyle \frac{1}{2}$$ and $$y\, =\, \displaystyle \frac{1}{5}$$

Solution

Question 9

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

$$x = 7, y = -1$$

$$x = 5, y = -1$$

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

Solution

Question 10

$$x = 2, y = 3$$

$$x = 7, y = 2$$

$$x = 4, y = 1$$

$$x = 4, y = 6$$

Solution

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