Free Pair of Linear Equations in Two Variables 03 Practice Test - 10th Grade 

The given equation is

$3x=2y-2.$

After rearranging the equation, we get

$\Rightarrow 3x-2y+2=0$

On comparing the above equation with $ax+by+c=0$, we get

a = 3, b = -2 and c = 2

$\therefore$ The values of a, b and c are 3, -2 and 2 respectively.

But the equation can also be written as

$(-1)\times (3x-2y+2)=(-1)\times 0$
$\Rightarrow -3x+2y-2=0$

On comparing the above equation with $ax+by+c=0$, we get

a = -3, b = 2 and c = -2

$\therefore$ The values of a, b and c are -3, +2 and -2 respectively.

Hence, we have two sets of values for a, b and c.

The line intersects the $yaxis$ when the value of x coordinate is 0.

Thus, on $yaxis,x=0.$

Substituting this value in the given equation:

$\Rightarrow 5y=10$

$\Rightarrow y=2$

Thus, the line intersects the $yaxis$ at point (0,2).

Let the 2 parts of 54 be x and y

 x+y = 54 ....(i)

and 10x + 22y = 780 ...(ii)

Multiply (i) by 10 , we get 

10 x + 10 y = 540 ...(iii)

 Subtracting  (ii) from (iii)

 - 12y = - 240

    y  =  20

Subsituting y = 20 in x + y = 54 ,
$⟹$ x + 20 = 54
$⟹$x = 34 

Hence, x = 34 and y = 20

A pair of linear equations in two variables which has a solution, is said to be consistent.

Here the lines representing the given linear equations are meeting only at point (-2, 3.5) giving a unique solution. Therefore, it is a consistent pair of linear equations.

Now, a pair of linear equations in two variables is said to be dependent when the lines representing them are coinciding i.e. will have many solutions.

Here, the lines have only one solution (-2, 3.5). Therefore, they aren't dependent.

Let, p = $\frac{1}{2+x}$ and  q = $\frac{1}{y-4}$

Thus, given equations reduce to,

$5p+q=2$  ... (i)

$6p-3q=1$   ... (ii)

On multiplying (i) by 3 and then adding it to (ii), we get

$15p+3q=6$

$6p-3q=1$

_______________

   $21p=7$ 

$\Rightarrow p=\frac{1}{3}$ 

and  on substituting this value in (ii) we get

$6\frac{1}{3}-3q=1$

$\Rightarrow q=\frac{1}{3}$

Now, $p=\frac{1}{2+x}$  

$\Rightarrow \frac{1}{3}=\frac{1}{2+x}$ 

$\Rightarrow x=1$

We know that $q=\frac{1}{y-4}$

$\Rightarrow \frac{1}{3}=\frac{1}{y-4}$ 

$\Rightarrow y=7$

Two linear equations in same two variables are called pair of linear equation in two variables.

$12x+4y+3=0and4x+4y+4=0$
are linear equations in same two variables.

We have $y-3x+1=0$

For $x=3$
$\Rightarrow y-3\left(3\right)+1=0$
$\therefore y=8$

For $x=-4$
$\Rightarrow y-3(-4)+1=0$
$\therefore y=-13$

For $x=2$
$\Rightarrow y-3\left(2\right)+1=0$
$\therefore y=5$

Thus, the values of $y$ if $x=[3,-4,2]$  are 8, -13 and 5 respectively.

Let the line intersects at point (x, y).

We know that if a line intersects the x-axis then the  y coordinate of the point at which it intersects will be 0.

$\Rightarrow$ y = 0

On substituting y = 0 in the given equation, we get

$6x+0=9$

$\Rightarrow x=\frac{3}{2}$

Thus, the point at which the given line intersects the x-axis is $(\frac{3}{2},0).$

The standard form of a linear equation is $ax+by+c=0.$

The given equation is $y=7x-4.$

It can be rewritten as $-7x+y+4=0.$

$\Rightarrow a=-7,b=1andc=4.$

But the equation can also be written as,

$7x-y-4=0.$

$\Rightarrow a=+7,b=-1andc=-4.$