Pair of Linear Equations in Two Variables Test - 38

Given equations are
$$x + y = 3$$
$$2x + 5y = 12 $$
Since, the system has unique solution at $$x = x_1$$ and $$y=y_1$$
That means the lines intersect at only one point $$(x_{1},y_{1})$$
By putting $$x=x_1$$ and $$y=y_1$$ in equations, we get
$$ x_{1} + y_{1}  = 3 $$     ....$$(1)$$
$$2 x_{1} +5y_{1}  = 12 $$    ....$$(2)$$
From equation $$(1)$$, we  have  
$$ y_1 = 3 - x_{1} $$
Putting the value of $$y_{1}$$ in equation $$(2)$$, 
$$\Rightarrow 2x_{1} + 5\left ( 3 - x_{1} \right ) = 12 $$
$$\Rightarrow 2x_{1} + 15 - 5x_{1} = 12$$
$$\Rightarrow 2x_{1}  - 5x_{1} = 12 - 15 $$
$$\Rightarrow  - 3x_{1} = -3 $$
$$ \Rightarrow x_{1} = \displaystyle \frac{-3}{-3} $$
$$ \Rightarrow x_{1} = 1$$

Let cost of one chair be Rs $$ x$$ and cost of one table be Rs $$ y$$
Now as per the question, 
$$3x+2y = 1850$$      .....(1)
$$5x + 3y = 2850$$    ......(2)
Now mutiplying equation (1) by 5 and equation (2) by 3
$$15x+10y = 9250$$   .....(3)
$$ 15x+9y = 8550$$    ......(4)
Now subtracting equation (4) from equation (3)
$$\Rightarrow y = 700$$
Put this value in eq (1), 
$$\Rightarrow 3x+2\left ( 700 \right ) = 1850$$
$$3x+1400 = 1850$$
$$3x = 450$$
$$ x = 150$$
Now total cost of 1 chair and 1 table = $$x + y $$
$$ =150+700= 850$$
So, total cost of 1 chair and 1 table is Rs 850

The equations are
$$ 2x-3y-7=0$$
$$  \left (a +b  \right )x-\left (a+b-3  \right )y-(4a+b)=0$$
Here, $$a_{1}=2,b_{1}=-3,c_{1}=-7$$
$$ a_{2}=a+b ,b_{2}=-\left (a+b-3  \right ),c_{2}=-(4a+b)$$
The system of linear equations has infinite solutions
$$\therefore \displaystyle \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$$
$$\Rightarrow \displaystyle \frac{2}{a+b}=\frac{-3}{-\left (a+b-3  \right )}=\frac{-7}{-\left (4a+b  \right )}$$
$$\Rightarrow \displaystyle \frac{2}{a+b}=\frac{3}{\left (a+b-3  \right )}$$
$$\Rightarrow  2a+2b-6=3a+3b$$
$$\Rightarrow b=-a-6 $$
Again, we have
$$\Rightarrow \displaystyle \frac{3}{\left (a+b-3  \right )}=\frac{7}{\left (4a+b  \right )}$$
$$\Rightarrow 12a+3b=7a+7b-21$$
$$\Rightarrow 5a-4b=-21$$
Putting $$b=-a-6 $$
$$\Rightarrow 5a-4\left ( -a-6 \right )=-21\Rightarrow 9a=-45$$
$$\Rightarrow a=-5$$
Putting  $$ a=-5   \  in \    b=-a-6$$
$$\Rightarrow  b=-\left (-5  \right )-6\Rightarrow b=-1$$
$$\Rightarrow  a=-5,b=-1$$

The equations are

$$\ x-3y-4=0$$
$$\Rightarrow a_{1}=1,b_{1}=-3,c_{1}=-4$$
$$\ 3x+2y-1=0$$
$$\Rightarrow a_{2}=3,b_{2}=2,c_{2}=-1$$

 comparing  $$\frac{a_{1}}{a_{2}} and \frac{b_{1}}{b_{2}}$$

We have $$\ \frac{1}{3}\neq\frac{-3}{2}$$

$$\Rightarrow  \frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}$$

$$\Rightarrow   System  \  has \   unique \   solution$$

If the system has unique solution then it is said to be consistent system 

 

The equations are
$$2x+4y-15=0 $$
$$ x+2y-4=0 $$
Here, $$ a_{1}=2,b_{1}=4,c_{1}=-15$$
and $$ a_{2}=1,b_{2}=2,c_{2}=-4$$
$$\displaystyle \frac{a_{1}}{a_{2}}=\frac{2}{1}=2$$
$$\displaystyle \frac{b_{1}}{b_{2}}=\frac{4}{2}=2$$
$$\displaystyle \frac{c_{1}}{c_{2}}=\frac{-15}{-4}$$

$$\Rightarrow \displaystyle \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}\neq\frac{c_{1}}{c_{2}}$$
$$\Rightarrow$$ The system of equations has no solution.
$$\therefore$$ Graphical representation of  the given pair of equations is parallel lines
 

The equations are
$$3x-y+8=0$$
$$ 6x-ky+16=0$$
Here, $$ a_{1}=3,b_{1}=-1,c_{1}=8$$
and $$ a_{2}=6,b_{2}=-k,c_{2}=16$$

The equation will represent coincident lines only when they have infinite number of solutions.

$$ \displaystyle \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$$

$$\Rightarrow \displaystyle \frac{3}{6}=\frac{-1}{-k}=\frac{8}{16} $$

Taking,
$$\displaystyle \frac{3}{6}=\frac{-1}{-k}$$

$$\Rightarrow \displaystyle \frac{1}{2}=\displaystyle \frac{1}{k}$$

$$\Rightarrow k=2 $$

Let  the  cost  of  one  pencil  be  Rs. $$x$$ and the cost  of  one  pen  be  Rs. $$y$$
According   to   the   first   condition
$$5x+7y=50....eq\left (  1\right )$$
$$\Rightarrow x=\displaystyle \frac{50-7y}{5}$$
According   to   the   second   condition
$$7x+5y=46....eq\left ( 2 \right )$$
Put the value of $$x$$ in $$eq\left ( 2 \right )$$
$$\Rightarrow \displaystyle \frac{7\left ( 50-7y \right )}{5}+5y=46$$
$$\Rightarrow 350-49y+25y=230\Rightarrow y=5$$

The linear equations are
$$ 6x-3y+10=0$$
$$ 2x-y+9=0$$
Here, $$a_{1}=6,b_{1}=-3,c_{1}=10$$
$$a_{2}=2,b_{2}=-1,c_{2}=9$$
$$\therefore \displaystyle \frac{a_{1}}{a_{2}}=\frac{6}{2}=3$$
$$\displaystyle \frac{b_{1}}{b_{2}}=\frac{-3}{-1}=3$$
$$\displaystyle \frac{c_{1}}{c_{2}}=\frac{10}{9} $$
$$\Rightarrow \displaystyle \frac{a_{1}}{a_{2}}=\frac{b_{2}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$$
$$\therefore$$ The lines represented by the given equations are parallel.
 

The equation are
$$\  x-2y-3=0$$
$$\Rightarrow a_{1}=1,b_{1}=-2,c_{1}=-3$$
$$\ 3x-6y-1=0$$
$$\Rightarrow a_{2}=3,b_{2}=-6,c_{2}=-1$$

comparing  $$\rightarrow      \dfrac{a_{1}}{a_{2}},\dfrac{b_{1}}{b_{2}},\dfrac{c_{1}}{c_{2}}$$

We have $$\ \dfrac{1}{3}=\dfrac{2}{6}\neq\dfrac{-3}{-1}$$
$$\Rightarrow \dfrac{a_{1}}{a_{2}}=\dfrac{b_{1}}{b_{2}}\neq \dfrac{c_{1}}{c_{2}}$$

from the  above relation we can say that this $$\ System  \  has \   no \   solution$$

If the system has no solution, then it is said to be the inconsistent system.
 

The equation are
$$ x-y=0.9$$
$$\Rightarrow x=0.9+y$$      .....(1)
$$\displaystyle \frac{11}{x+y}=2$$
$$\Rightarrow 2x+2y=11$$     .....(2)
Put the value of $$x$$ in eq(2)
$$\Rightarrow 2\left (.9+y  \right )+2y=11$$
$$\Rightarrow 1.8+2y+2y=11$$
$$\Rightarrow 4y=9.2$$
$$\Rightarrow y=2.3$$
$$\Rightarrow x=.9+y\Rightarrow x=.9+2.3=3.2$$
$$\therefore   x=3.2 ,y=2.3$$