Math, asked by tejamp1806, 4 days ago

BOOK 3. List - I (Equations) List - II (Solutions) [ ] a) 2x + y = 8, x + 6y = 15 i) (4,5) 1 b) 5x + 3y = 35, 2x + 4y = 28 ii) itd.) 14 6 - - c) 15x + 4y = 61 and 4x + 15y = 72 A) a - ii, b - iii, c-i B) a - iii, b -ii, c-i iii) (3,2) C) a -i, b - iii, c - i D) a - iii, b - i, c- ii T T​

Answers

Answered by 0alizaykhan0
0

Step-by-step explanation:

12

Step-by-step explanation:

Given that:-

288\div16-[18\div6+[29-(12-4-6) ]]288÷16−[18÷6+[29−(12−4−6)]]

To find:- The value of the expression

In order to solve the expression, we have to use the BODMAS concept.

B - Brackets

O - order

D - division

M - multiplication

A - addition

S - subtraction

As we have,

=> 288\div16-[18\div6+[29-(12-4-6) ]]288÷16−[18÷6+[29−(12−4−6)]]

By further calculation, we get

=> 18-[3+[29-(12-4-6) ]]18−[3+[29−(12−4−6)]]

=> 18-[3+[27]]18−[3+[27]]

=> 18-3018−30

=> -12

Answered by studinesh10212
2

Step-by-step explanation:

search-icon-header

Search for questions & chapters

search-icon-image

Class 10

>>Maths

>>Pair of Linear Equations in Two Variables

>>Algebraic Methods of Solving a Pair of Linear Equations

>>Solve the following pair of equations by

Question

Bookmark

Solve the following pair of equations by cross multiplication rule

5x+3y=35,2x+4y=28

Then

Medium

Solution

verified

Verified by Toppr

Correct option is A)

The equations are

⇒5x+3y−35=0

⇒2x+4y−28=0

By cross multiplication

[3×(−28)−4×(−35)]

x

=

[−35×2−(−28)×5]

y

=

[5×4−2×3]

1

−84+140

x

=

−70+140

y

=

20−6

1

56

x

=

70

y

=

14

1

56

x

=

14

1

⇒14x=56⇒x=4

70

y

=

14

1

⇒14y=70⇒y=5

⇒x=4,y=5

Similar questions