Pair of Linear Equations in Two Variables Test

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Pair of Linear Equations in Two Variables Test - 24

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Weekly Quiz Competition

Question 1
1 / -0

Without actually solving the simultaneous equations given below, decide whether the system has unique solution, no solution or infinitely many solutions.
$$3x+5y=16; 4x-y=6$$

A
No Solution

B
Infinitely many solutions

C
Unique solution

D
Cannot be determined

Solution

As we know the condition for the pair of equations $$a_1x+b_1y+c_1=0$$ and $$a_2x+b_2y+c_2 = 0$$ to have a unique solution is -
$$\displaystyle \frac{a_1}{a_2} \neq \frac{b_1}{b_2} $$

We have $$\ $$ $$\dfrac{3}{4} \neq \dfrac{5}{-1}$$
Hence, Unique Solution
Question 2
1 / -0

Solve the following equation simultaneously using Graphical method:
$$x+2y=5; y=-2x-2$$

Then $$(x,y)$$ is equal to

A
$$(-3, 4)$$

B
$$(4, -2)$$

C
$$(3, 4)$$

D
$$(4, 2)$$

Solution

Given first line $$x+2y=5$$
Take two points $$(5,0)$$ and $$(1,2)$$ and join them by drawing a line.
Another line $$y=-2x-2$$
Take two points $$(0,-2)$$ and $$(-1,0)$$ and join them by drawing a line.
We get the graphs
$$x+2y=5\ \implies $$ brown line
$$2x+y=2\ \implies $$ blue line
the point of intersection is $$(-3,4)$$
Question 3
1 / -0

Without actually solving the simultaneous equations given below, decide whether the system has unique solution, no solution or infinitely many solutions.
$$3y=2-x; 3x=6-9y$$

A
Infinite solutions

B
unique solution

C
no solution

D
Cannot be determined

Solution

As we know the condition for the pair of equations $$a_1x+b_1y+c_1=0$$ and $$a_2x+b_2y+c_2 = 0$$ to have infinite solutions is
$$\displaystyle \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} $$
Here the given system of linear equations is
$$
x + 3y -2=0;\\ 3x+9y -6=0\\ \dfrac { 1 }{ 3 } =\dfrac { 3 }{ 9 } =\dfrac { -2 }{ -6 } ;\\$$
$$\therefore $$ The system has infinite number of solutions.
Question 4
1 / -0

Solve the following simultaneous equations:

$$\displaystyle \frac{1}{x}\, +\, \frac{1}{y}\, =\, 8;\quad \frac{4}{x}\, -\, \frac{2}{y}\, =\, 2$$

A
$$x\, =\,\displaystyle \frac{1}{3}\, ,\, y\, =\, \frac{1}{5}$$

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

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

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

Solution

Given equations are
$$\dfrac { 1 }{ x } +\dfrac { 1 }{ y } = \ 8\\ \dfrac { 4 }{ x } -\dfrac { 2 }{ y }=\ 2 $$

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

So, $$ a+b\ =\ 8$$ ....... $$(1)$$
$$\\ 4a\ -\ 2b\ =\ 2$$ ...... $$(2)$$

Multiply equation $$(1)$$ by $$2$$, we get
$$2a\ +2b\ =\ 16\ $$ ........ $$(3)$$

Adding equations $$(2)$$ and $$(3)$$, we get
$$ 6a=18$$

$$a=\dfrac { 18 }{ 6 }= 3$$

Put $$a=3$$ in $$a+b=8$$
we get $$ b = 5$$

So $$ x = \dfrac { 1 }{ 3 }$$ and $$y= \dfrac { 1 }{ 5 }$$
Question 5
1 / -0

Solve the following simultaneous equations :
$$\displaystyle \frac{16}{x + y}\, +\, \frac{2}{x - y}\, =\, 1;\quad \frac{8}{x + y}\, -\, \frac{12}{x - y}\, =\, 7$$

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

B
$$x = 6 , y = 1$$

C
$$x = 7 , y = 4$$

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

Solution

$$Let \dfrac { 1 }{ x+y } = a; \dfrac { 1 }{ x-y } =b$$
So the equations are
$$16a + 2b = 1$$ ...(1)
$$8a - 12b = 7$$ ...(2)
Eqn (2) $$\times 2 $$
$$\Rightarrow 16a - 24b = 14$$ ....(3)
Subtracting eq(1) and eq(3) we get
$$26b = -13 \Rightarrow b = -\dfrac12$$
Eqn (1) $$\times 6 \Rightarrow 96a +12b = 6$$ ....(4)
Adding eqn(4) and eq(2)
$$104a = 13\\ \therefore a = \dfrac{1}{8}\\ \dfrac { 1 }{ x+y } = \dfrac { 1 }{ 8 } $$
$$x+y = 8$$ ...(5)
$$\dfrac { 1 }{ x-y } = \dfrac { 1 }{ -2 } \\ x-y = -2 $$ ...(6)
Adding eq(5) and eq(6)
$$2x = 6 \Rightarrow x = 3$$
Substitute for $$x$$ in eqn(5)
$$\therefore y = 8-3 = 5$$
Question 6
1 / -0

Solve the following simultaneous equations by substitution method.
$$2x +3y= -4; \, \, x-5y =11$$

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

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

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

D
$$x=1, \, y=2$$

Solution

$$2x +3y= -4$$..........(1)
$$x-5y =11$$
$$\therefore x=11+5y$$ .............(2)
Substituting eq. (2) in eq. (1), we get,
$$2(11+5y)+3y = -4$$
$$22+10y+3y=-4$$
$$22+4 = -13y$$
$$\therefore y = -\dfrac {26}{13}$$

$$y=-2$$
Substituting $$y=-2$$ in (1) we get,
$$x=1$$
Question 7
1 / -0

Find the value of $$p$$ for which the given simultaneous equations have unique solution:
$$3x\, +\, y\, =\, 10;\quad 9x\, +\, py\, =\,23$$

A
$$p = 5$$

B
All values of $$p$$ except $$7$$

C
All values of $$p$$ except $$3$$

D
Cannot be determined

Solution

As we know the condition for the pair of equations $$a_1x+b_1y+c_1=0$$ and $$a_2x+b_2y+c_2 = 0$$ to have a unique solution is -
$$\displaystyle \frac{a_1}{a_2} \neq \frac{b_1}{b_2} $$
Here the given system of linear equation is
$$
3x+y=10;\\ 9x+py=23\\$$
So for unique solution, we have
$$ \dfrac { 3 }{ 9 } \neq \dfrac { 1 }{ p } \\ \Rightarrow p\neq 3\quad
$$
Question 8
1 / -0

Find the value of $$p$$ for which the given simultaneous equations have unique solution:
$$8x\, -\, py\, +\, 7\, =\, 0;\quad 4x\, -\, 2y\, +\, 3\, =\, 0$$

A
All values of $$p$$ except $$4$$

B
$$p = 7$$

C
$$p = 6$$

D
All values of $$p$$ except $$5$$

Solution

As we know the condition for the pair of equations $$a_1x+b_1y+c_1=0$$ and $$a_2x+b_2y+c_2 = 0$$ to have a unique solution is -
$$\displaystyle \frac{a_1}{a_2} \neq \frac{b_1}{b_2} $$
Here the given system linear of equations is
$$
8x-py+7=0;\\ 4x-2y+3=0$$
So for the unique solution
$$\dfrac { 8 }{ 4 } \neq \dfrac { -p }{ -2 } \quad $$
$$\\ \Rightarrow p\neq 4\quad\quad$$
Question 9
1 / -0

Find the values of $$x$$ and $$y$$, if
$$\displaystyle \frac{2}{x}\, +\, \frac{6}{y}\, =\, 13;\quad \frac{3}{x}\, +\, \frac{4}{y}\, =\, 12$$

A
$$x\, =\,\displaystyle \frac{1}{6}\, ,\, y\, =\, \frac{2}{3}$$

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

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

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

Solution

$$\dfrac { 2 }{ x } +\dfrac { 6 }{ y } = 13$$
$$ \dfrac { 3 }{ x } +\dfrac { 4 }{ y } = 12$$
Lets assume, $$\dfrac { 1 }{ x } = a, \dfrac { 1 }{ y } = b$$, then the above equations reduces to
$$ 2a + 6b = 13$$ ...(1)
$$ 3a + 4b = 12 $$ ...(2)
Multiplying (1) by $$3$$ and (2) by $$2$$, we get
$$6a +18b = 39 $$ ...(3)
$$6a + 8b = 24 $$ ...(4)
Subtracting (3) and (4), we get
$$10b = 15\Rightarrow b = \dfrac { 3 }{ 2 } $$
Now, $$ 2a = 13 - 6\times \dfrac { 3 }{ 2 }= 13 - 9 = 4$$
$$\Rightarrow a = 2$$
So, $$x = \dfrac { 1 }{ 2 } ; y = \dfrac { 2 }{ 3 } $$

Hence, option D is the correct answer.
Question 10
1 / -0

Find the values of the $$x+y$$ and $$x-y$$ from the examples given below without solving for x and y

$$4x+3y=24;\, \, 3x+4y=25$$

A
$$7$$ , $$-1$$

B
$$1$$ , $$-7$$

C
$$1$$ , $$7$$

D
$$-7$$ , $$-1$$

Solution

$$\text{The given equations are}$$

$$4x+3y=24$$                           ...(i)
$$3x+4y=25$$                           ...(ii)

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

$$7x+7y=49\Rightarrow x+y=7$$

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

$$x-y=-1$$

$$\therefore x+y=7 \space and \space x-y= -1$$

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Pair of Linear Equations in Two Variables Test - 24

Result

Pair of Linear Equations in Two Variables Test - 24

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SHARING IS CARING

Question 1

No Solution

Infinitely many solutions

Unique solution

Cannot be determined

Solution

Question 2

$$(-3, 4)$$

$$(4, -2)$$

$$(3, 4)$$

$$(4, 2)$$

Solution

Question 3

Infinite solutions

unique solution

no solution

Cannot be determined

Solution

Question 4

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

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

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

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

Solution

Question 5

$$x = 2 , y = 3$$

$$x = 6 , y = 1$$

$$x = 7 , y = 4$$

$$x = 3 , y = 5$$

Solution

Question 6

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

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

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

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

Solution

Question 7

$$p = 5$$

All values of $$p$$ except $$7$$

All values of $$p$$ except $$3$$

Cannot be determined

Solution

Question 8

All values of $$p$$ except $$4$$

$$p = 7$$

$$p = 6$$

All values of $$p$$ except $$5$$

Solution

Question 9

$$x\, =\,\displaystyle \frac{1}{6}\, ,\, y\, =\, \frac{2}{3}$$

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

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

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

Solution

Question 10

$$7$$ , $$-1$$

$$1$$ , $$-7$$

$$1$$ , $$7$$

$$-7$$ , $$-1$$

Solution

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