Question

राज्य शैक्षणिक संशोधन व प्रशिक्षण परिषद ,महाराष्ट्र,पुणे -30
इ10 वी गणित भाग 1
प्रश्नपेढी
प्रकरण : दोन चलातील रेषीय समीकरणे
Q.1A) 1 गुणाचे MCQ
1. 4x + 5y = 19 या समीकरणाचा आलेख काढण्यासाठी x = 1 असताना y ची किंमत 3
असेल.
A)4
B3
C)2
D)-3
2. दिलेल्या दोन समीकरणांसाठी D = 26, D, = -39 आणि D= 13 असल्यास x =?
A)2
B)-3
C)-2
D)3​

Answer 1

Given :  4x + 5y = 19 या समीकरणाचा आलेख काढण्यासाठी x = 1 असताना

Equation 4x + 5y = 19 and x = 1

To Find : y   किंमत -

Value of y

A)4

B3

C)2

D)-3

Solution:

4x + 5y = 19

x = 1

=> 4(1) + 5y = 19

=> 4 + 5y = 19

Subtract 4 from both sides

=> 4  - 4 + 5y = 19 - 4

=> 5y = 15

Divide by 5 on both sides

=> 5y/5 = 15/5

=> y = 3

Value of y is 3 if x = 1

x = 1 असताना y ची किंमत  3 असेल.

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Answer 2

SOLUTION

TO CHOOSE THE CORRECT OPTION

1. For the line 4x + 5y = 19 the value of y when x = 1

A) 4

B) 3

C) 2

D) - 3

For the equations with variables x and y

2. $\sf{D_x = 26 \: , \: D_y = - 39 \: , \: D = 13}$

A) 2

B) - 3

C) - 2

D) 3

EVALUATION

1. For the line 4x + 5y = 19 when x = 1

( 4 × 1 ) + 5y = 19

$\implies \sf{4 + 5y = 19}$

$\implies \sf{4 + 5y = 19}$

$\implies \sf{5y = 15}$

$\implies \sf{y = 3}$

Hence the correct option is D) 3

2. Here it is given that for the equations with variables x and y

$\sf{D_x = 26 \: , \: D_y = - 39 \: , \: D = 13}$

Hence

$\displaystyle \sf{x = \frac{D_x}{D} = \frac{26}{13} = 2 }$

$\displaystyle \sf{y = \frac{D_y}{D} = \frac{ - 39}{13} = - 3 }$

Hence the required value of x = 2

Hence the correct option is A) 2

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