Question

Write an equation of a line passing through the point representing solution of the pair of linear equation x+y=2 and 2x-y=1.How many such lines can be drawn?(plz explain how to write an equation)

Answer 1

I hope it helps u the graph is rough but u must draw it on fair

Answer 2

Answer:

A linear equation exists an equation in which the highest power of the variable is always 1. It exists even known as a one-degree equation.

Step-by-step explanation:

provided pair of linear equations exist

$x+y-2=0$

and$2 x-y-1=0$

On comparing with$a x+b y+c=0$

Here, $a_{1}=1,$

$b_{1}=1$

$c_{1}=-2$

And $\mathrm{a}_{2}=2$

$\mathrm{~b}_{2}=-1$

$\mathrm{c}_{2}=-1$

Then,

$a_{1} / a_{2}=1 / 2$

$b_{1} / b_{2}=-1$

$c_{1} / c_{2}=2$

since $a_{1} / a_{2} \neq b_{1} / b_{2}$.

So, both lines intersect at a point. Thus, the pair of equations contains a unique explanation. Therefore, these equations exist consistent.

Now, $x+y=2$ or $y=2-x$

If $x=0$ then $y=2$ and if $x=2$ then $y=0$

X              0      2

Y               2      0

Points       A      B

and

If $x=0$ then $y=-1$if-then$x=1 / 2$  $y=0$ and

if $x=1$ then $y=1$

x                o         1/2            1

Y                -1          0             1

Points         C          D            E

Plotting the points $A(2,0)$  $B(0,2)$, we get the straight line $A B$. Plotting the points $C(0,-1)$  $D(1 / 2,0)$, we get the straight line $C D$.

The lines$A B$ ,$C D$ intersect at$E(1,1)$.

Therefore,

Infinite lines can pass via the Intersection point of linear equations$x+y=2$ and $2 x-y-1=0$ i.e.

$E(1,1)$ like as $y=x, x+2 y=3$, $x+y=2$and so on.

The graphical representation is shown below,

#SPJ2