Math, asked by vk6175400, 7 months ago

Equ:
bu
2. From the choices given below, choose the equation whose graphs are given in Fig. 4.7
Fig. 4.8.
For Fig. 4.7
For Fig. 4.8
(i) y = 2x
(1) y = 2x + 2
(ü) x + y = 0
(ü) y = r-2
ü) y = 2x + 4
(ii) y = -x + 2
(iv) 2 + 3y = 7x
(iv) x - 2y = 6
have
hint in
Y
4
(-1, 3) 3
2 (0,2)
2
1
(2,0)
(-1,1) 1 (0,0)
X4
X'
-2 -10
1 2 3
1,-1)
2-19
1 2 3
-21
-2
-3
Y'
Fig. 4.7
Fig. 4.8
• 78
Linear Equatio​

Answers

Answered by janvisapra27
0

Answer:

y = 2x

(1) y = 2x + 2

(ü) x + y = 0

(ü) y = r-2

ü) y = 2x + 4

(ii) y = -x + 2

(iv) 2 + 3y = 7x

(iv) x - 2y = 6

have

hint in

Y

4

(-1, 3) 3

2 (0,2)

2

1

(2,0)

(-1,1) 1 (0,0)

X4

X'

-2 -10

1 2 3

1,-1)

2-19

1 2 3

-21

-2

-3

Y'

Fig. 4.7

Fig. 4.8

• 78

Linear Equation

Answered by anjupurohit6
4

Step-by-step explanation:

ANSWER

In the given fig. (i), the solutions of the equation are (−1,1),(0,0) and (1,−1).

∴ the equation which satisfies these solutions is the correct equation.

Equation (ii) x+y=0, satisfies these solutions.

Proof:

If we put the value of x=−1 and y=1 in the equation x+y=0

=x+y=−1+1=0

∴L.H.S=R.H.S

if we put the the value of x = 0 and y = 0

=x+y=0+0=0

∴L.H.S=R.H.S

If we put the value of x=1 and y=−1

=x+y=1+(−1)=1–1=0

L.H.S=R.H.S

Hence, option (ii) x+y=0 is correct.

In the given Fig.(ii) the solutions of the equation are (−1,3),(0,2) and (2,0).

∴The equation which satisfies these solutions is the correct equation.

Equation (iii) y=−x+2 , satisfies these solutions.

Proof:

If we put the value of x=−1 and y=3 in the equation y=−x+2

y=−x+2

3=−(−1)+2

3=3

∴L.H.S=R.H.S

if we put the the value of x=0 and y=2

y=−x+2

2=−0+2

2=2

∴L.H.S=R.H.S

If we put the value of x=2 and y=0

y=−x+2

0=−2+2

0=0

L.H.S=R.H.S

Hence, option (iii) y=−x+2 is correct.

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