Pair of Linear Equations in Two Variables Test

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Pair of Linear Equations in Two Variables Test - 29

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Question 1
1 / -0

Solve the following pair of equations:

$$\displaystyle \frac{9}{x}-\displaystyle \frac{4}{y}= 8$$, $$\displaystyle \frac{13}{x}+\displaystyle \frac{7}{y}=101$$

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

B
$$x= \displaystyle \frac{1}{4};y= \displaystyle \frac{1}{7}$$

C
$$x= \displaystyle \frac{5}{4};y= \displaystyle \frac{2}{5}$$

D
$$x= \displaystyle \frac{3}{2};y= \displaystyle \frac{6}{5}$$

Solution

The equation $$\dfrac { 9 }{ x } -\dfrac { 4 }{ y } =8$$ can be solved as:

$$\dfrac { 9 }{ x } -\dfrac { 4 }{ y } =8$$

$$ \Rightarrow \dfrac { 9y-4x }{ xy } =8$$

$$ \Rightarrow 9y-4x=8xy\quad .........(1)$$

The equation $$\dfrac { 13 }{ x } +\dfrac { 7 }{ y } =101$$ can be solved as:

$$\dfrac { 13 }{ x } +\dfrac { 7 }{ y } =101$$

$$ \Rightarrow \dfrac { 13y+7x }{ xy } =101$$

$$ \Rightarrow 13y+7x=101xy\quad .........(2)$$

Multiply the equation 1 by $$7$$ and equation 2 by $$4$$. Then we get the equations:

$$63y-28x=56xy.........(3)$$

$$52y+28x=404xy.........(4)$$

Add equations 3 and 4 to eliminate $$x$$, because the coefficients of $$x$$ are the opposite. So, we get

$$(63y+52y)+(-28x+28x)=56xy+404xy$$

i.e. $$115y=460xy$$

i.e. $$115=460x$$

i.e. $$x=\dfrac { 115 }{ 460 } =\dfrac { 1 }{ 4 }$$

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

$$9y-4\left( \dfrac { 1 }{ 4 } \right) =8\left( \dfrac { 1 }{ 4 } \right) y\\ \Rightarrow 9y-1=2y\\ \Rightarrow 9y-2y=1\\ \Rightarrow 7y=1\\ \Rightarrow y=\dfrac { 1 }{ 7 }$$

Hence, the solution of the equations is $$x=\dfrac { 1 }{ 4 } ,y=\dfrac { 1 }{ 7 }$$.
Question 2
1 / -0

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

A
$$(-4, 3)$$

B
$$(4, -3)$$

C
$$(-4, -3)$$

D
$$(4, 3)$$

Solution

The solution set of equation $$\frac { x }{ 2 } -\frac { y }{ 3 } =3$$ or $$3x-2y=18$$ is as follows:

If $$x=0$$ then

$$3\times 0-2y=18\\ \Rightarrow -2y=18\\ \Rightarrow y=-9$$

If $$x=4$$ then

$$3\times 4-2y=18\\ \Rightarrow 12-2y=18\\ \Rightarrow y=-3$$

The required points are $$(0,-9),(4,-3)$$.

Also, the solution set of equation $$x+y=1$$ is as follows:

If $$x=0$$ then

$$0+y=1\\ \Rightarrow y=1$$

If $$x=4$$ then

$$4+y=1\\ \Rightarrow y=1-4=-3$$

The required points are $$(0,1),(4,-3)$$.

Plot all the above points in the graph. Since the point of intersection in the graph is $$(4,-3)$$.

Hence the solution is $$x=4,y=-3$$.
Question 3
1 / -0

Solve the following pairs of linear equations, graphically :
$$\displaystyle\,2x \,-\, 3y\, =\, -6\,$$ and $$x\, -\,\dfrac{y}{2}\,=\,1$$

A
$$(3, \,2)$$

B
$$(3, \,3)$$

C
$$(4, \,4)$$

D
$$(3, \,4)$$

Solution

Given,
$$ 2x - 3y = - 6 $$
$$ \Rightarrow y = \dfrac {2x}{3} + 2 $$
Substituting different values of $$ x $$ to get corresponding values of $$ y $$, we get
$$ ( 0, 2), (3,4), (6,6), (-3, 0) $$
Plotting these points on the graph, we get the graph of the equation, $$ 2x - 3y = - 6 $$
Similarly, given second equation,
$$ x - \dfrac {y}{2} = 1 $$
$$ \Rightarrow y = 2(x-1) $$
Substituting different values of $$ x $$ to get corresponding values of $$ y $$, we get
$$ ( 1, 0), (0,-2), (2,2), (-1,-4), (3,4) $$
Plotting these points on the graph, we get the graph of the equation, $$ y = 2(x-1) $$.
The point of intersection of the graphs of the lines is the solution of the equations which is $$ (3,4) $$.
Question 4
1 / -0

Solve: $$\displaystyle \frac{3}{x}-\displaystyle \frac{2}{y}= 0$$ and $$\displaystyle \frac{2}{x}+\displaystyle \frac{5}{y}= 19$$. Hence, find $$a$$ if $$y= ax+3$$.

A
$$x=\displaystyle \frac{2}{3};y= \displaystyle \frac{5}{3}$$ and $$a= 6\displaystyle \frac{6}{5}$$

B
$$x=\displaystyle \frac{3}{4};y= \displaystyle \frac{1}{3}$$ and $$a= -4\displaystyle \frac{7}{3}$$

C
$$x=\displaystyle \frac{7}{2};y= \displaystyle \frac{8}{3}$$ and $$a= -7\displaystyle \frac{2}{9}$$

D
$$x=\displaystyle \frac{1}{2};y= \displaystyle \frac{1}{3}$$ and $$a= -5\displaystyle \frac{1}{3}$$

Solution

Let $$\dfrac{1}{x} = u$$ and $$\dfrac{1}{y} = v$$
So, the equations will reduce to
$$3u - 2v = 0$$ ...$$(1)$$
$$2u + 5v = 19$$ ...$$(2)$$

Multiplying $$(1)$$ by $$5$$ and $$(2)$$ by $$2$$
$$15u - 10v = 0$$ ...$$(3)$$
$$4u + 10v = 38$$ ...$$(4)$$

Adding $$(3)$$ and $$(4)$$
$$19u = 28$$
$$u = 2$$

Substituting $$u = 2$$ in $$(1)$$, we get
$$6 - 2v = 0$$
$$v = 3$$

$$u = \dfrac{1}{x}$$ So, $$x = \dfrac{1}{u} = \dfrac{1}{2}$$
$$v = \dfrac{1}{y}$$ So, $$y = \dfrac{1}{v} = \dfrac{1}{3}$$

$$x = \dfrac{1}{2}, y = \dfrac{1}{3}$$

$$y = ax + 3$$
$$\dfrac { 1 }{ 3 } =\dfrac { a }{ 2 } +3\\ \dfrac { a }{ 2 } = -\dfrac { 8 }{ 3 } \\ a = \dfrac { -16 }{ 3 } $$

Option $$D$$
Question 5
1 / -0

Solve the following pair of equations:

$$\displaystyle \frac{6}{x}+\displaystyle \frac{4}{y}= 20, \displaystyle \frac{9}{x}-\displaystyle \frac{7}{y}= 10.5$$

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

B
$$x= \displaystyle \frac{2}{5};y= \displaystyle \frac{2}{3}$$

C
$$x= \displaystyle \frac{5}{7};y= \displaystyle \frac{1}{3}$$

D
$$x= \displaystyle \frac{2}{5};y= \displaystyle \frac{1}{7}$$

Solution

Given equations are

$$\dfrac { 6 }{ x } +\dfrac { 4 }{ y } =20\quad ...(1)\\ \dfrac { 9 }{ x } -\dfrac { 7 }{ y } =10.5\quad ...(2)$$

Let $$\dfrac{1}{x} = a$$ and $$\dfrac{1}{y} = b$$

So, the original equations will reduce to
$$6a + 4b = 20$$ ...$$(1)$$
$$9a - 7b = 10.5$$ ...$$(2)$$

Multiplying $$(1)$$ by $$7$$ and $$(2)$$ by $$4$$, we get
$$42a + 28b = 140$$ ...$$(3)$$
$$36a - 28b = 42$$ ...$$(4)$$

Adding $$(3)$$ and $$(4)$$,
$$78a = 182$$

$$a = \dfrac{182}{78} = \dfrac{14}{6} = \dfrac{7}{3}$$

Substituting $$a = \dfrac{7}{3}$$ in $$(1)$$, we get $$b$$ as
$$14 + 4b = 20$$

$$4b = 6$$

$$b = \dfrac{6}{4} = \dfrac{3}{2}$$

Now, $$a = \dfrac{1}{x}$$, so $$x = \dfrac{1}{a} = \dfrac{3}{7}$$

$$b = \dfrac{1}{y}$$, so $$y = \dfrac{1}{b} = \dfrac{2}{3}$$

$$x = \dfrac{3}{7}, y = \dfrac{2}{3}$$

Option $$A$$
Question 6
1 / -0

Solve: $$\displaystyle \frac{34}{3x+4y}+\displaystyle \frac{15}{3x-2y}= 5$$ and $$\displaystyle \frac{25}{3x-2y}-\displaystyle \frac{8.50}{3x+4y}= 4.5$$

A
$$x= 1;y= 4$$

B
$$x= 3;y= 2$$

C
$$x= 5;y= 8$$

D
$$x= 7;y= 6$$

Solution

On putting $$\dfrac1{3x+4y}=X$$ and $$\dfrac1{3x-2y}=Y$$ in given equations, we get

$$34X+15Y=5$$                      ...(i)

$$25Y-\dfrac{17}2X=4.5\Rightarrow -17X+50Y=9$$            ...(ii)

On multiplying (i) by 1 and (ii) by 2 and adding, we get

$$34X+15Y=5$$
$$\underline {-34X+100Y=18}$$
             $$115Y=23$$

$$\therefore Y=\dfrac{23}{115}=\dfrac15$$

On putting $$Y=\dfrac15$$ in (i), we get
$$34X+15\times\dfrac15=5\Rightarrow 34X=2\Rightarrow X=\dfrac1{17}$$

$$\therefore 3x+4y=17$$                                 ...(iii)
     $$3x-2y=5$$                                  ...(iv)

On subtracting (iv) from (iii), we get
$$6y=12\Rightarrow y=2$$

On putting $$y=2$$ in (iv), we get
$$3x-4=5\Rightarrow x=\dfrac93=3$$

Therefore, $$x=3,y=2$$
Question 7
1 / -0

Solve the following pair of equations:
$$4x+\displaystyle \frac{6}{y}= 15$$ and $$6x-\displaystyle \frac{8}{y}=14$$

A
$$x= 5;y= 6$$

B
$$x= 3;y= 2$$

C
$$x= 7;y= 4$$

D
$$x= 0;y= 6$$

Solution

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

$$8x+6=30$$

i.e. $$8x=30-6=24$$

i.e. $$x=3$$

Hence, the solution of the equations is $$x=3 ,y=2$$.
Question 8
1 / -0

Solve the following pair of equations :
$$x-y= 0.9$$ and $$\displaystyle \frac{11}{2\left ( x+y \right )}= 1$$

A
$$x= 1.5;y= 4$$

B
$$x= 3;y= 2.5$$

C
$$x= 5.2;y= 0.3$$

D
$$x= 3.2;y= 2.3$$

Solution

$$x=y+0.9$$
Substituting in the 2nd equation:
$$11=2(y+0.9+y)$$
$$\dfrac{11}{2}=2y+0.9\\5.5-0.9=2y\\4.6=2y\\ \therefore y=2.3$$

and
$$x=2.3+0.9=3.2$$
Therefore, $$x=3.2$$ and $$y=2.3$$
Question 9
1 / -0

Solve: $$\displaystyle \frac{20}{x+y}+\displaystyle \frac{3}{x-y}= 7$$ and $$\displaystyle \frac{8}{x-y}-\displaystyle \frac{15}{x+y}= 5$$

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

B
$$x= 4;y= -4$$

C
$$x= 3;y= 2$$

D
$$x= 1;y= 6$$

Solution

Let $$ \dfrac {20}{x+y} + \dfrac {3}{x-y} = 7 $$ --- (1)

and $$ \dfrac {8}{x-y} - \dfrac {15}{x+y} = 5 $$ --- (2)

Multiplying equation $$ (1) $$ with $$ 3 $$ we get, $$ \dfrac {60}{x+y} + \dfrac {9}{x-y} = 21 $$ ----- equation $$ (3) $$
Multiplying equation $$ (2) $$ with $$ 4 $$ we get, $$ \cfrac {32}{x-y} - \dfrac {60}{x+y} = 20 $$ ----- equation $$ (4) $$
Adding equations $$ 3 $$ and $$ 4 $$, we get $$ \dfrac {41}{x-y} = 41

=> x - y = 1 $$ ---- (5)
Substituting $$ x-y = 1 $$ in the equation $$ (1) $$, we get $$
\dfrac {20}{x+y} + \dfrac {3}{1} = 7 => \dfrac {20}{x+y} = 4$$
=>$$ x +y= 5 $$ --- (6)
Adding equations $$ 5 $$ and $$ 6 $$, we get $$ 2x = 6 => x = 3 $$ ---- (5)
Substituting $$ x = 3 $$ in the equation $$ (5) $$, we get $$ 3-y = 1 => y = 2 $$
Question 10
1 / -0

The sides of an equilateral triangle are given by $$x+3y$$; $$3x+2y-2$$ and $$4x+\displaystyle \frac{1}{2}y+1$$ respectively. Find the lengths of the sides of the triangle.

A
$$15$$ units each

B
$$17$$ units each

C
$$12$$ units each

D
$$11$$ units each

Solution

Since it is an equilateral triangle, the lengths of the sides are equal
So, $$x+3y = 3x+2y-2 = 4x+\dfrac { 1 }{ 2 } y+1$$
$$ \therefore x + 3y = 3x+2y -2\\ \therefore y = 2x - 2 $$
$$ 2x - y = 2$$ ...(1)
Also, $$ 4x+\dfrac { 1 }{ 2 } y+1 = x+3y$$
$$\therefore 8x + y + 2 = 2x +6y\\ 6x + 2 = 5y$$
$$ 6x - 5y = -2$$ ...(2)
Multiplying eq(1) by 5 we get,
$$10x - 5y = 10$$ ...(3)
Subtracting eq(2) from eq(3) we get,
$$4x = 12 \Rightarrow x = 3$$
$$\therefore y = 2x -2 = 4$$
The length of one side of equilateral triangle = $$ x +3y $$ =$$ 3 + 3(4) = 15 $$units

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Re-Attempt Weekly Quiz Competition

Pair of Linear Equations in Two Variables Test - 29

Result

Pair of Linear Equations in Two Variables Test - 29

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SHARING IS CARING

Question 1

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

$$x= \displaystyle \frac{1}{4};y= \displaystyle \frac{1}{7}$$

$$x= \displaystyle \frac{5}{4};y= \displaystyle \frac{2}{5}$$

$$x= \displaystyle \frac{3}{2};y= \displaystyle \frac{6}{5}$$

Solution

Question 2

$$(-4, 3)$$

$$(4, -3)$$

$$(-4, -3)$$

$$(4, 3)$$

Solution

Question 3

$$(3, \,2)$$

$$(3, \,3)$$

$$(4, \,4)$$

$$(3, \,4)$$

Solution

Question 4

$$x=\displaystyle \frac{2}{3};y= \displaystyle \frac{5}{3}$$ and $$a= 6\displaystyle \frac{6}{5}$$

$$x=\displaystyle \frac{3}{4};y= \displaystyle \frac{1}{3}$$ and $$a= -4\displaystyle \frac{7}{3}$$

$$x=\displaystyle \frac{7}{2};y= \displaystyle \frac{8}{3}$$ and $$a= -7\displaystyle \frac{2}{9}$$

$$x=\displaystyle \frac{1}{2};y= \displaystyle \frac{1}{3}$$ and $$a= -5\displaystyle \frac{1}{3}$$

Solution

Question 5

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

$$x= \displaystyle \frac{2}{5};y= \displaystyle \frac{2}{3}$$

$$x= \displaystyle \frac{5}{7};y= \displaystyle \frac{1}{3}$$

$$x= \displaystyle \frac{2}{5};y= \displaystyle \frac{1}{7}$$

Solution

Question 6

$$x= 1;y= 4$$

$$x= 3;y= 2$$

$$x= 5;y= 8$$

$$x= 7;y= 6$$

Solution

Question 7

$$x= 5;y= 6$$

$$x= 3;y= 2$$

$$x= 7;y= 4$$

$$x= 0;y= 6$$

Solution

Question 8

$$x= 1.5;y= 4$$

$$x= 3;y= 2.5$$

$$x= 5.2;y= 0.3$$

$$x= 3.2;y= 2.3$$

Solution

Question 9

$$x= 4;y= -7$$

$$x= 4;y= -4$$

$$x= 3;y= 2$$

$$x= 1;y= 6$$

Solution

Question 10

$$15$$ units each

$$17$$ units each

$$12$$ units each

$$11$$ units each

Solution

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