Question

Draw the graph of the equation x+2y-3=0 From the graph, find the value of y when (i) x=2 (ii) x=−2​

Answer 1

EXPLANATION.

Graph of the equation.

⇒ x + 2y - 3 = 0.

As we know that,

Put the value of x = 0 in the equation, we get.

⇒ (0) + 2y - 3 = 0.

⇒ 2y = 3.

⇒ y = 3/2.

⇒ y = 1.5.

Their Co-ordinates = (0,1.5).

Put the value of y = 0 in the equation, we get.

⇒ x + 2(0) - 3 = 0.

⇒ x - 3 = 0.

⇒ x = 3.

Their Co-ordinates = (3,0).

To find value of y :

⇒ (1) = When x = 2.

Put the value of x = 2 in the equation, we get.

⇒ (2) + 2y - 3 = 0.

⇒ 2y - 1 = 0.

⇒ 2y = 1.

⇒ y = 1/2.

(2) = When x = - 2.

Put the value of x = - 2 in the equation, we get.

⇒ (-2) + 2y - 3 = 0.

⇒ - 2 + 2y - 3 = 0.

⇒ 2y - 5 = 0.

⇒ 2y = 5.

⇒ y = 5/2.

Answer 2

★ Given,

• A linear equation x + 2y - 3 = 0 ------(1)

It can be written as :

$\bf \: y = \frac{3 - x}{2} \: \: \: - - - (2)$

★ We have to draw the graph and find the value of y when (i) x = 2 (ii) x = −2

Solution :

$\underline{ \bf \: drawing \: graph}$

→ To draw a linear graph we need at least two co-ordinates which lies on the equation.

→ Finding 1st co-ordinates :

Let, x = 1 and put this value in eqn.(1)

$\bf \: 1 + 2y - 3 = 0 \\ \rm \implies \: 2y - 2 = 0 \\ \rm \implies \: 2y = 2 \\ \rm \implies \: y = \frac{2}{2} \\ \rm \implies \: y = 1$

So, co-ordinates is (1,1)

→ Finding 2nd co-ordinates :

Let, x = - 1 and put this value in eqn. (1)

$\rm \: - 1 + 2y - 3 = 0 \\ \rm \implies \: 2y - 4 = 0 \\ \rm \implies \: 2y = 4 \\ \rm \implies \: y = \frac{4}{2} \\ \rm \implies \: y = 2$

So, co-ordinates is (-1,2)

Now if we add this two points we will get the graph of the equation.

$\underline{ \bf \: finding \: value \: of \: \: y}$

To find value of y we have to put the value of x in eqn.(2)

${ \boxed{\boxed{\begin{array}{c | c} \bf \: value \: of \: \: x& \bf \: value \: of \: y \\ \\ \underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } & \: \underline{\bf \: y = \frac{3 - x}{2 }} \\ \\ \pink{ \bf \:(i)\:\:\: x = 2}& \pink{\bf \: y = \frac{3 - 2}{2} = \frac{1}{2}} \\ \underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }&\underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ \\ \blue{\bf \:(ii)\:\: x = - 2}& \blue{\bf \: y = \frac{ 3 + 2}{2} = \frac{5}{2} } \end{array}}}}$