Pair of Linear Equations in Two Variables Test

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Pair of Linear Equations in Two Variables Test - 30

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Question 1
1 / -0

Solve: $$4x+\displaystyle \frac{6}{y}= 15$$ and $$6x-\displaystyle \frac{8}{y}= 14$$. Hence, find $$a$$ if $$y= ax-2$$

A
$$x= 2;y= 26$$ and $$a= 2\displaystyle \frac{5}{3}$$

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

C
$$x= 1;y= 5$$ and $$a= 2\displaystyle \frac{7}{3}$$

D
$$x= 5;y= 1$$ and $$a= 5\displaystyle \frac{1}{8}$$

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:

$$12xy+18=45y.........(3)$$

$$12xy-16=28y.........(4)$$

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

$$8x+6=30 \\ \Rightarrow 8x=30-6 \\ \Rightarrow 8x=24 \\ \Rightarrow x=3$$

Therefore, the solution of the equations is $$x=3,y=2$$.

Now substitute the values of $$x$$ and $$y$$ in $$y=ax-2$$ as shown below:

$$2=3a-2$$
$$3a=2+2$$
$$3a=4$$
$$a=\frac { 4 }{ 3 } =1\frac { 1 }{ 3 }$$

Hence, $$a=1\frac { 1 }{ 3 }$$.
Question 2
1 / -0

Solve the following pair of equations:

$$2x-3y-3= 0$$, $$\displaystyle \frac{2x}{3}+4y+\displaystyle \frac{1}{2}= 0$$

A
$$x= \displaystyle \frac{13}{20},y= -\displaystyle \frac{7}{10}$$

B
$$x= \displaystyle \frac{17}{20},y= -\displaystyle \frac{1}{20}$$

C
$$x= \displaystyle \frac{1}{10},y= -\displaystyle \frac{7}{10}$$

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

Solution

$$2x-3y-3= 0$$ ...... (1)

$$\displaystyle \frac{2x}{3}+4y=-\displaystyle \frac{1}{2}$$ ...... (2)

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

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

Substituting in the 2nd equation
$$ \dfrac{-1}{2}=\dfrac{2}{3} \times \dfrac{3+3y}{2} +4y$$

$$ 5y=-\dfrac{3}{2} \implies y=\dfrac{-3}{10}$$

Put the value of $$y$$ in (3)

$$x=\dfrac{3+3(\dfrac{-3}{10})}{2} \implies x=\dfrac{21}{20}$$

$$\therefore x=\dfrac{21}{20} \quad and \quad y=\dfrac{-3}{10}$$
Question 3
1 / -0

Pooja and Ritu can do a piece of work in $$\displaystyle 17\frac{1}{7}$$ days. If one day work of Pooja is three fourth of one day work of Ritu then find in how many days each alone will do the same work.

A
Pooja in 40 days and Ritu in 30 days

B
Pooja in 60 days and Ritu in 40 days

C
Pooja in 30 days and Ritu in 10 days

D
Pooja in 45 days and Ritu in 20 days

Solution

Let the work done by pooja in one day be $$ 1/x $$ and work done by Ritu in one day be $$1/y.$$
Therefore, according to the question total work done by both in one day,

$$\\ 1/x+1/y=\dfrac{1}{17\dfrac{1}{7}} = 7/120$$.....(1)

Also, one day work of Pooja is three fourth of one day work of Ritu
$$\\ 1/x=3/4y$$

Substituting above equation in equation (1) , we get
$$y=40$$
Substituting $$y=40$$ in equation (1) we get
$$x=30$$
Hence pooja will complete the work in $$40$$ days whereas ritu in $$30$$ days
Question 4
1 / -0

Solve the following equations by substitution method.
$$2x+y=-2;\, \, 3x-y=7$$

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

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

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

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

Solution

$$ 2x+y =-2; \, \, 3x-y =7 $$
$$ y = -2x -2 $$
Put it in second equation.
$$ 3x -(-2x-2) =7 $$
$$ 5x +2 = 7 $$
$$ x =1 $$
$$ y = -2x-2 = -2 (1)-2 = -4 $$
$$ x =1$$ and $$y = -4 $$
Question 5
1 / -0

Solve the following equations by substitution method.
$$2x-3y=14; \, \, 5x+2y=16$$

A
$$x = 7, y= 1$$

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

C
$$x = 6, y= -2$$

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

Solution

$$ 2x-3y=14; \, \, 5x+2y=16 $$

$$ x = \dfrac {14+3y}{2} $$ Put it in second equation.

$$ 5 \times \dfrac {14+3y}{2} +2y =16 $$
$$ 70 +15y +4y = 32 $$
$$ 19y = -38 $$
$$ y = -2 $$

$$ x = \dfrac {14+3y}{2} = \dfrac {14+3(-2)}{2} = 4 $$

$$ x = 4$$ and $$y = -2$$
Question 6
1 / -0

Solve the following equations by substitution method.
$$x-2y+2=0; \, \, x+2y=10$$

A
$$x = 4, y= 3$$

B
$$x = 1, y= 3$$

C
$$x = 9, y= 3$$

D
$$x = 5, y= 3$$

Solution

$$x-2y+2=0 $$             ....(i)
$$ x+2y=10$$        ....(ii)
We pick either of the equations and write one variable in terms of the other.
Let us consider the Equation (i) and write it as
$$x=2y-2$$                 .........(iii)
Substitute the value of $$x=2y-2$$ in Equation (ii). We get
$$ 2y-2+2y=10$$
$$ \Rightarrow 4y=12$$
$$\Rightarrow y=3 $$
Substituting this value of $$y=3$$ in Equation (iii), we get
$$x=2(3)-2=4$$
$$\therefore x=4 , y=3$$
Question 7
1 / -0

Solve the following simultaneous equations by the method of equating coefficients.$$3x-4y=7;\, \, 5x+2y=3$$ .

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

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

C
$$x = 5, y= 2$$

D
$$x = 4, y= -3$$

Solution

$$3x-4y=7$$                              ...(i)
$$5x+2y=3$$                           ...(ii)
On multiplying (i) by 5 and (ii) by 3, we get
$$15x-20y=35$$
$$\underline {\underset {-}15x\underset {-}{+}6y=\underset{-}9}$$
          $$-26y=26$$
$$\therefore y=-1$$
On putting $$y=-1$$ in (i), we get
$$3x-4(-1)=7$$
$$x=1$$
$$\therefore x=1,y=-1$$
Question 8
1 / -0

Find the values of $$(x+y)$$ and $$(x-y) $$without actually solving for $$x$$ and $$y$$.
$$13x+15y=114 \quad and \quad15x+13y=110$$

A
$$x+y= 9, x-y= -2$$

B
$$x+y= 8, x-y= -2$$

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

D
$$x+y= 1, x-y= -2$$

Solution

$$13x+15y=114$$ ...(i)

$$15x+13y=110$$ ...(ii)

On adding (i) and (ii) , we get

$$28x+28y=224$$

$$\Rightarrow x+y=\frac {224}{28} =8$$

On subtracting (i) from (ii) , we get

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

$$\Rightarrow x-y=-\frac 42 =-2$$.
Question 9
1 / -0

Solve the following equations by substitution method:
$$5x-2y=13; \, \, 4x+3y=15$$

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

B
$$x=3,y=1$$

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

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

Solution

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

Let us consider the equation $$5x-2y=13$$ and write it as

$$x=\frac { 13+2y }{ 5 }$$

Substitute the value of $$x$$ in equation $$4x+3y=15$$. We get

$$4\left( \frac { 13+2y }{ 5 } \right) +3y=15\\ \Rightarrow 52+8y+15y=75\\ \Rightarrow 23y=23\\ \Rightarrow y=1$$

Therefore, $$y = 1$$

Substituting this value of $$y$$ in the equation $$5x-2y=13$$, we get

$$5x-2=13$$
$$5x=13+2=15$$
$$x=3$$

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

Solve the following equations by substitution method.
$$3y-2x=9; \, \, 2x+5y=15$$

A
$$x = 0, y= 2$$

B
$$x = 5, y= 2$$

C
$$x = 1, y= 3$$

D
$$x = 0, y= 3$$

Solution

$$ 3y -2x =9; \, \, 2x +5y =15 $$
$$ y = \dfrac {9+2x}{3} $$ Put it in second equation.

$$ 2x + 5 \times \dfrac {9+2x}{3} =15 $$

$$ 6x +45 +10x = 45 $$
$$ 16x =0 $$
$$ \boxed{x =0} $$

$$y = \dfrac {9+2x}{3} = \dfrac {9+2 (0)}{3} = 3 \\ \boxed{y=3}$$

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Pair of Linear Equations in Two Variables Test - 30

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Pair of Linear Equations in Two Variables Test - 30

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Question 1

$$x= 2;y= 26$$ and $$a= 2\displaystyle \frac{5}{3}$$

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

$$x= 1;y= 5$$ and $$a= 2\displaystyle \frac{7}{3}$$

$$x= 5;y= 1$$ and $$a= 5\displaystyle \frac{1}{8}$$

Solution

Question 2

$$x= \displaystyle \frac{13}{20},y= -\displaystyle \frac{7}{10}$$

$$x= \displaystyle \frac{17}{20},y= -\displaystyle \frac{1}{20}$$

$$x= \displaystyle \frac{1}{10},y= -\displaystyle \frac{7}{10}$$

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

Solution

Question 3

Pooja in 40 days and Ritu in 30 days

Pooja in 60 days and Ritu in 40 days

Pooja in 30 days and Ritu in 10 days

Pooja in 45 days and Ritu in 20 days

Solution

Question 4

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

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

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

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

Solution

Question 5

$$x = 7, y= 1$$

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

$$x = 6, y= -2$$

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

Solution

Question 6

$$x = 4, y= 3$$

$$x = 1, y= 3$$

$$x = 9, y= 3$$

$$x = 5, y= 3$$

Solution

Question 7

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

$$x = 2, y= 1$$

$$x = 5, y= 2$$

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

Solution

Question 8

$$x+y= 9, x-y= -2$$

$$x+y= 8, x-y= -2$$

$$x+y= 3, x-y= -2$$

$$x+y= 1, x-y= -2$$

Solution

Question 9

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

$$x=3,y=1$$

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

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

Solution

Question 10

$$x = 0, y= 2$$

$$x = 5, y= 2$$

$$x = 1, y= 3$$

$$x = 0, y= 3$$

Solution

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