Pair of Linear Equations in Two Variables Test

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Pair of Linear Equations in Two Variables Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

Solve: $$3\left ( 2x+y \right )= 7xy$$ and $$3\left ( x+3y \right )= 11xy$$;
where, $$x\neq 0, y\neq 0$$

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

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

C
$$x= 4; y= 2$$

D
$$x= 3; y= 4$$

Solution

Dividing both the equations by xy, we get:
$$ \dfrac3x+\dfrac6y=7;\\ \dfrac3y+\dfrac9x=11 $$
Multiplying the 2nd equation by -2 and Adding
$$ \dfrac{ -15 }{x}= -15;\\ y= \dfrac32 \quad and \quad x=1 $$
Question 2
1 / -0

Solve: $$4x+\displaystyle \frac{6}{y}= 15$$ and $$6x-\displaystyle \frac{8}{y}= 14$$. Hence find the value of $$k$$, if $$y= kx-2$$.

A
$$x= 4, y= 1$$ and $$k= \displaystyle \frac{2}{3}$$

B
$$x= 5, y= 3$$ and $$k= \displaystyle \frac{3}{5}$$

C
$$x= 3, y= 2$$ and $$k= \displaystyle \frac{4}{3}$$

D
$$x= 1, y= 2$$ and $$k= \displaystyle \frac{9}{2}$$

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

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

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

Now Substitute the values of $$x$$ and $$y$$ in the equation $$y=kx-2$$ as follows:

$$2=3k-2$$
$$3k=4$$
$$\Rightarrow k=\frac { 4 }{ 3 }$$

Hence, $$k=\frac { 4 }{ 3 }$$.
Question 3
1 / -0

Solve the following simultaneous equations:
$$12x+15y+18= 0$$, $$18x-7y+86= 0$$

A
$$x=-4;\, y=2$$

B
$$x=1;\, y=0$$

C
$$x=-3;\, y=5$$

D
$$x=4;\, y=1$$

Solution

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

Let us consider the equation $$12x+15y+18=0$$ and write it as

$$x=\frac { -18-15y }{ 12 }..........(1)$$

Substitute the value of $$x$$ in Equation $$18x-7y=-86$$. We get

$$18\left( \frac { -18-15y }{ 12 } \right) -7y=-86\\ \Rightarrow -54-45y-14y=-172\\ \Rightarrow -59y=-118\\ \Rightarrow y=2$$

Therefore, $$y = 2$$

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

$$x=\frac { -18-30 }{ 12 } =-\frac { 48 }{ 12 } =-4$$

Hence, the solution is $$x = -4, y = 2$$.
Question 4
1 / -0

Solve the following pair of equations:
$$3x\, -\, y\, =\, 23$$
$$\displaystyle \frac{x}{3}\, +\, \displaystyle \frac{y}{4}\, =\, 4$$

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

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

C
$$x\, =\, 9\, ;\, y\, =\, 4$$

D
$$x\, =\, 7\, ;\, y\, =\, 13$$

Solution

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

Let us consider the equation $$3x-y=23$$ and write it as

$$y=3x-23...........(1)$$

Substitute the value of $$y$$ in Equation $$\frac { x }{ 3 } +\frac { y }{ 4 } =4$$ or $$4x+3y=48$$. We get

$$4x+3(3x-23)=48$$

i.e. $$4x+9x-69=48$$

i.e. $$13x=117$$

i.e. $$x=9$$

Therefore, $$x=-9$$

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

$$y=27-23$$

i.e. $$y=4$$

Hence, the solution is $$x = 9, y =4$$.
Question 5
1 / -0

Solve the following pair of equations:
$$13 + 2y = 9x$$, $$3y = 7x$$

A
$$x=6$$; $$y =-7$$

B
$$x=5$$ ; $$ =13$$

C
$$x=-7$$; $$y =1$$

D
$$x=3$$; $$y =7$$

Solution

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

Let us consider the equation $$3y=7x$$ and write it as

$$x=\frac { 3 }{ 7 }y..........(1)$$

Substitute the value of $$x$$ in the equation $$13+2y=9x$$ or $$9x-2y=13$$. We get:

$$9\left( \frac { 3 }{ 7 } y \right) -2y=13\\ \Rightarrow 27y-14y=91\\ \Rightarrow 13y=91\\ \Rightarrow y=7$$

Therefore, $$y = 7$$

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

$$x=\frac { 3 }{ 7 } \times 7=3$$

Hence, the solution is $$x = 3, y = 7$$.
Question 6
1 / -0

Solve the following pairs of linear (simultaneous) equation by the method of elimination:
$$x + y = 7$$, $$5x + 12y = 7$$

A
$$x =13$$ and $$y=5$$

B
$$x=11$$ and $$y=-4$$

C
$$x=7 $$ and $$y =-2$$

D
$$x=5$$ and $$y =7$$

Solution

Multiply the equation $$x+y=7$$ by $$5$$ to make the coefficients of $$x$$ equal. Then we get the equations:

$$5x+5y=35.........(1)$$

$$5x+12y=7.........(2)$$

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

$$(5x-5x)+(5y-12y)=35-7$$

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

i.e. $$y = -4$$

Substituting this value of $$y$$ in the equation $$x+y=7$$, we get

$$x-4=7$$

i.e. $$x=11$$

Hence, the solution of the equations is $$x = 11, y = -4$$.
Question 7
1 / -0

If $$1$$ is added to each of the two certain numbers, their ratio is $$1:2$$; and if $$5$$ is subtracted from each of the two numbers, their ratio becomes $$5:11$$. Find the numbers.

A
$$35$$ and $$70$$

B
$$35$$ and $$71$$

C
$$35$$ and $$72$$

D
$$35$$ and $$73$$

Solution

Let the numbers be $$ x $$ and $$ y $$

Given, if $$1 $$ is added to the numbers, then ratio $$ = \dfrac {x + 1}{y + 1} = \dfrac {1}{2} $$
$$ => 2x + 2 = y + 1 $$
$$ => 2x - y = -1 $$ ...(1)

Also, if $$ 5 $$ is subtracted from the numbers, then ratio $$ = \dfrac {x - 5}{y - 5} = \dfrac {5}{11} $$
$$ => 11x - 55 = 5y - 25 $$
$$ => 11x - 5y = 30 $$ ...(2)
Multiplying equation $$(1) $$ with $$ 5 $$ we get, $$ 10x - 5y = -5 $$ ----- equation $$(3) $$
Subtracting equation $$ (3) $$from $$ (1) $$, we get $$ x = 35 $$
Substituting $$ x = 35 $$ in the equation $$ (1) $$, we get $$ 70 - y = -1 => y = 71$$
Hence, the numbers are $$ 35 $$ and $$ 71 $$
Question 8
1 / -0

Solve the set of equations: $$3\left ( 2u+v \right )= 7uv$$ and $$3\left ( u+3v \right )= 11uv$$

A
$$u= 1; v= \displaystyle \frac{3}{2}$$

B
$$u= 2; v= \displaystyle \frac{1}{2}$$

C
$$u= 2; v= \displaystyle \frac{5}{4}$$

D
$$u= 5; v= \displaystyle \frac{7}{4}$$

Solution

The equation $$3(2u+v)=7uv$$ can be rewritten as $$6u+3v=7uv$$ and the equation $$3(u+3v)=11uv$$ can be rewritten as $$3u+9v=11uv$$. Then we get the equations:

$$6u+3v=7uv.........(1)$$

$$3u+9v=11uv.........(2)$$

Multiply equation 2 by $$2$$:

$$6u+18v=22uv.........(3)$$

Subtract Equation 1 from equation 3 to eliminate $$u$$, because the coefficients of $$u$$ are the same. So, we get

$$(6u-6u)+(18v-3v)=22uv-7uv$$

i.e. $$-15v=-15uv$$

i.e. $$u=1$$

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

$$6+3v=7v$$

i.e. $$4v=6$$

i.e. $$v=\frac { 3 }{ 2 }$$

Hence, the solution of the equations is $$u=1,v=\frac { 3 }{ 2 }$$.
Question 9
1 / -0

Solve the following pair of equations :
$$7x + 6y = 71$$
$$5x - 8y = -23$$

A
$$x = 2; y = 7$$

B
$$x = 5; y = 6$$

C
$$x = 4; y = 9$$

D
$$x = 1; y = 8$$

Solution

Given equations are $$ 7x + 6y = 71 $$ --- (1)

and $$ 5x - 8y = -23 $$ --- (2)

Multiplying equation $$

(1) $$ with $$ 4 $$ we get, $$ 28x + 24y = 284 $$ ----- equation $$

(3) $$
Multiplying equation $$

(2) $$ with $$ 3 $$ we get, $$ 15x - 24y = - 69 $$ ----- equation $$ (4)

$$
Adding equations $$ (4) $$ and $$ (3) $$, we get $$ 43x = 215 => x = 5 $$
Substituting $$x = 5 $$ in the equation $$ (2) $$, we get
$$ 5(5) - 8y = -23 => y = 6 $$
Question 10
1 / -0

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

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

B
$$x\, =\, \displaystyle \frac{3}{4}\; \, y\, =\, \displaystyle \frac{1}{6}$$

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

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

Solution

Let $$\dfrac{1}{x}=a$$ and $$\dfrac{1}{y}=b$$
$$\displaystyle \frac{9}{x}\, -\, \displaystyle \frac{4}{y}\, =\, 8$$ ......(i)
$$\Rightarrow 9a-4b=8$$ .....(ii)
$$\displaystyle \frac{13}{x}\, +\, \displaystyle \frac{7}{y}\, =\, 101$$ ......(iii)
$$\Rightarrow 13a+7b=101$$ .....(iv)
Multipliying equation (ii) by $$7$$ and (iv) by $$4$$, we get
$$63a-28b=56$$
$$52a+28b=404$$

$$115a =460$$
$$\Rightarrow a =\dfrac{460}{115}$$
$$\Rightarrow a =4$$
Putting the value in (ii), we get
$$9a-4b=8$$
$$\Rightarrow 9\times 4-4b=8$$
$$\Rightarrow -4b=8-36$$
$$\Rightarrow -4b=-28$$
$$\Rightarrow b=7$$
Now, resubstituting the value, we get
$$a=\dfrac{1}{x}=4$$ $$\Rightarrow x=\dfrac{1}{4}$$
Again, $$b=\dfrac{1}{y}=7$$ $$\Rightarrow y=\dfrac{1}{7}$$

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Re-Attempt Weekly Quiz Competition

Pair of Linear Equations in Two Variables Test - 33

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Pair of Linear Equations in Two Variables Test - 33

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SHARING IS CARING

Question 1

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

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

$$x= 4; y= 2$$

$$x= 3; y= 4$$

Solution

Question 2

$$x= 4, y= 1$$ and $$k= \displaystyle \frac{2}{3}$$

$$x= 5, y= 3$$ and $$k= \displaystyle \frac{3}{5}$$

$$x= 3, y= 2$$ and $$k= \displaystyle \frac{4}{3}$$

$$x= 1, y= 2$$ and $$k= \displaystyle \frac{9}{2}$$

Solution

Question 3

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

$$x=1;\, y=0$$

$$x=-3;\, y=5$$

$$x=4;\, y=1$$

Solution

Question 4

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

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

$$x\, =\, 9\, ;\, y\, =\, 4$$

$$x\, =\, 7\, ;\, y\, =\, 13$$

Solution

Question 5

$$x=6$$; $$y =-7$$

$$x=5$$ ; $$ =13$$

$$x=-7$$; $$y =1$$

$$x=3$$; $$y =7$$

Solution

Question 6

$$x =13$$ and $$y=5$$

$$x=11$$ and $$y=-4$$

$$x=7 $$ and $$y =-2$$

$$x=5$$ and $$y =7$$

Solution

Question 7

$$35$$ and $$70$$

$$35$$ and $$71$$

$$35$$ and $$72$$

$$35$$ and $$73$$

Solution

Question 8

$$u= 1; v= \displaystyle \frac{3}{2}$$

$$u= 2; v= \displaystyle \frac{1}{2}$$

$$u= 2; v= \displaystyle \frac{5}{4}$$

$$u= 5; v= \displaystyle \frac{7}{4}$$

Solution

Question 9

$$x = 2; y = 7$$

$$x = 5; y = 6$$

$$x = 4; y = 9$$

$$x = 1; y = 8$$

Solution

Question 10

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

$$x\, =\, \displaystyle \frac{3}{4}\; \, y\, =\, \displaystyle \frac{1}{6}$$

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

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

Solution

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