Question

solve the problem for linear equation of two variables(word problem)​

Answer 1

Answer:

the answer is 10m

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Step-by-step explanation:

Answer

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Let the length of a rectangle =xm and the breadth of a rectangle =ym Then, 

Area of rectangle =xym2

Condition I :Area is reduced by 8m2, when length =(x−5)m and breadth =(y+3)m

Then, area of rectangle =(x−5)×(y+3)m2 

According to the question, xy−(x−5)×(y+3)=8

⇒xy−(xy+3x−5y−15)=8

⇒xy−xy−3x+5y+15=8

⇒−3x+5y=8−15

⇒3x−5y=7

Condition II:Area is increased by 74m2, when length =(x+3)m and breadth =(y+2)m 

Then, area of rectangle =(x+3)×(y+2)m2 

According to the question, (x+3)×(y+2)−xy=74

⇒(xy+3y+2x+6)−xy=74

⇒xy+2x+3y+6−xy=74

⇒2x+3y=74−

Answer 2

Question :-

The area of a rectangle gets reduced by 8 m², when its length is reduced by 5 m and its breadth is increased by 3 if we increase the length by 3 and breadth by 2, then the area increases by 74 m². Find the dimensions of the rectangle.

Given :–

• Area of a rectangle gets reduced by 8 m² when its length is reduced by 5 m and its breadth is increased by 3.
• If we increase the length by 3 and breadth by 2, then the area increases by 74 m².

To Find :–

• The dimensions of the rectangle (length and breadth)

Solution :–

☯️ Let's consider the length of the rectangle be x and the breadth of rectangle as y.

$\implies \tt \: (x - 5)(y + 3) = 8 \: {m}^{2}. \: . \: . \: ( \bf \: {1}^{st} \: case)$

$\implies \tt \: (x + 3)(y +2) = 74\: {m}^{2} . \: . \: . \: . \:( \bf \: {2}^{nd} \: case)$

Case 1 :-

$\implies \tt \: xy - (x - 5)(y +3) = 8\: {m}^{2} \\ \\ \implies \tt xy - (5y + 3x - 15)= 8{m}^{2} \: \\ \\ \implies \tt \: xy -( 5y - 3x + 15) = 8{m}^{2}  ( \sf \: change \: of \: sign)\\ \\ \implies \tt \: 3x - 5y \: . \: . \: .(eq. \: 1)$

Case 2 :-

$\implies \tt \: xy - (x + 3)(y + 2) = 74\: {m}^{2} \\ \\ \implies \tt xy + (2x + 3y - 6) = 74{m}^{2} \: \\ \\ \implies \tt \: 2x + 3y = 68 {m}^{2} \: . \: . \: .(eq. \: 2)$

A/Q,

By multiplying both the equations,

$\implies \bf \: 15y - 9x = 21 \\ \\ \implies \bf \: 10x + 15y = 340$

Adding both the equations,

$\implies \bf \: 19x = 361 \\ \\ \implies \bf \: x = \cancel\dfrac{361}{19} \\ \\ \implies \bf \underline{ \green{x = 19}} \: \: \: \tt \: (Length)$

Now calculating the breadth of rectangle,

$\implies \bf \: 4y + 6x + 18 = 222 \\ \\ \implies \bf \: 19y + 48 = 230 \\ \\ \implies \bf \:19y = 190 \\ \\ \implies \bf \:y = \cancel\dfrac{190}{19} \\ \\ \implies \bf \underline{ \red{y = 10}} \: \: \: \tt \: (Breadth)$

So,

• $\underline{ \boxed{ \green{ \bf{Length = 19 \: m}}}}$

• $\underline{ \boxed{ \red{ \bf{Breadth = 10 \: m}}}}$

❝Therefore, the length of the rectangle is 19 m and breadth of the rectangle is 10 m respectively.❞