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
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
Step-by-step explanation:
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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
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Solve the following pair of equations by cross multiplication rule
5x+3y=35,2x+4y=28
Then
Medium
Solution
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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